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Operator Generic Fundamentals Foundation - Physics © Copyright 2016 – Rev 2 Operator Generic Fundamentals 2 Terminal Learning Objectives At the completion of this training session, the trainee will demonstrate mastery of this topic by passing a written exam with a grade of ≥ 80 percent on the following Terminal Learning Objectives (TLOs): 1. Convert between units of measure associated with the English and System Internationale (SI) measuring systems. 2. Describe vector quantities, and how they are represented. 3. Solve resultant vector problems. 4. Describe the measurement of force and its relationship to free body diagrams. 5. Apply Newton's laws of motion to a body at rest. 6. Calculate the change in velocity when two objects collide, with respect to the conservation of momentum. 7. Describe energy, work, and power in mechanical systems. © Copyright 2016 – Rev 2 TLOs Operator Generic Fundamentals 3 Units of Measure TLO 1 – Convert between units of measure associated with the English and System Internationale (SI) measuring systems. 1.1 Describe the three fundamental dimensions: length, mass, and time. 1.2 List standard units of the fundamental dimensions for each of the following systems: a. International System of Units (SI) b. English System 1.3 Differentiate between fundamental and derived measurements. 1.4 Convert between English and SI units of mass and length. 1.5 Convert time measurements between the following: Years, weeks, days, hours, minutes, seconds. © Copyright 2016 – Rev 2 TLO 1 Operator Generic Fundamentals 4 Three Fundamental Dimensions ELO 1.1 – Describe the three fundamental dimensions: length, mass, and time. • Mass: amount of material present in an object – Description of "how much" material makes up an object – Not the same thing as weight, even though units are similar • Weight (derived unit): measurement that describes the force of gravity on "mass" of an object © Copyright 2016 – Rev 2 ELO 1.1 Operator Generic Fundamentals 5 Three Fundamental Dimensions • Length: distance between two points – Needed to locate position of a point in space ⇒ describe size of a physical object or system – When measuring a length of pipe, ends of pipe are two points ⇒ distance between two points is length • Time: duration between two instants – Typically measured in: seconds, minutes, or hours © Copyright 2016 – Rev 2 ELO 1.1 Operator Generic Fundamentals 6 Three Fundamental Dimensions Knowledge Check Which of the following is NOT a fundamental dimension? A. Weight B. Length C. Mass D. Time Correct answer is A. © Copyright 2016 – Rev 2 ELO 1.1 Operator Generic Fundamentals 7 Units of Measure ELO 1.2 – List standard units of the fundamental dimensions for each of the following systems: International System of Units (SI) and English System. • Units define magnitude of a measurement • For example: length – Could be centimeter or meter, each of which describes a different magnitude of length – Could also be inches, miles, fathoms, or other units • Units must be specified when same physical quantity may be measured using a variety of different units © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 8 Unit Systems • Two unit systems in use – English units – International System of Units (SI) English System • Used in U.S. • Various units for the fundamental dimensions or measurements © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 9 Unit Systems English Units of Measure Length Mass Time Inch Ounce Second* Foot* Pound* Minute Yard Ton Hour Mile Day Month Year * denotes standard unit of measure © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 10 Unit Systems System Internationale (SI) • SI system is made up of two related systems – Meter-kilogram-second (MKS) system – Centimeter-gram-second (CGS) system • Simpler to use than English system because they use a decimalbased system in which prefixes are used to denote powers of ten © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 11 SI versus English System • SI based on powers of ten: – One kilometer = 1,000 meters – One centimeter = one one-hundredth of a meter • English system has odd units of conversion: – Mile = 5,280 feet – Inch = one twelfth of a foot © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 12 MKS Units of Measure • MKS system used primarily for calculations in Physics MKS Units of Measure Length Mass Time Millimeter Milligram Second* Meter* Gram Minute Kilometer Kilogram* Hour Day Month Year * denotes standard unit of measure © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 13 CGS Units of Measure • Both MKS and CGS systems are used in Chemistry CGS Units of Measure Length Mass Time Millimeter Milligram Second* Meter* Gram Minute Kilometer Kilogram* Hour Day Month Year * denotes standard unit of measure © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 14 Dimensions Of Familiar Objects and Events Approximate Lengths of Familiar Objects Object Length (meters) Diameter of earth's orbit around the sun 2 x 1011 Football field 0.91 x 102 (100 yards) Diameter of a dime 2 x 10-2 Thickness of a window pane 1 x 10-3 Thickness of paper 1 x 10-4 © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 15 Dimensions Of Familiar Objects and Events Approximate Masses of Familiar Objects Object Mass (kilograms) Earth 6 x 1024 House 2 x 105 Car 2 x 103 Dime 3 x 10-3 Postage stamp 5 x 10-8 © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 16 Dimensions Of Familiar Objects and Events Approximate Time Durations of Familiar Events Event Time (seconds) Age of earth 2 x 1017 Human life span 2 x 109 Earth's rotation around the sun 3 x 107 Earth's rotation on its axis 8.64 x 104 Time between heart beats 1 © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 17 Units of Measure Knowledge Check The measurement system most used in the United States when dealing with physics is: A. foot – pound – second B. mks C. cgs D. foot – pound – minute Correct answer is A. © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 18 Derived Measurements ELO 1.3 – Differentiate between fundamental and derived measurements. • Most physical quantities have units that are combinations of the three fundamental dimensions (length, mass, time) • When these dimensions or measurements are combined, they produce derived units – Derived from one or more fundamental measurements – Can be combination of same or different units © Copyright 2016 – Rev 2 ELO 1.3 Operator Generic Fundamentals 19 Area and Volume • Area – Product of two lengths (e.g., width x length for a rectangle) – Units of length squared ⇒ in2 or m2 1 m x 1 m = 1 m2 4 in x 2 in = 8 in2 • Volume – Product of three lengths (e.g., length x width x depth for a rectangular solid) – Units of length cubed, ⇒ in3 or m3 – MKS and CGS unit systems have a specific unit for volume ⇒ liter – 1 liter = 1,000 cm3 – 2 in X 3 in X 5 in = 30 in3 © Copyright 2016 – Rev 2 ELO 1.3 Operator Generic Fundamentals 20 Density, Velocity and Acceleration • Density – Measure of the mass of an object per unit volume – Units of mass divided by length cubed ⇒ kg/m3 or lbs/ft3 • Velocity – Change in length per unit time – Units ⇒ km/h or ft/s • Acceleration – Measure of change in velocity or velocity per unit time – Units ⇒cm/s2 or ft/s2 © Copyright 2016 – Rev 2 ELO 1.3 Operator Generic Fundamentals 21 Fundamental and Derived Measurements Knowledge Check Match the following terms with the correct dimensions of measurement. A. A measure of the change in velocity per unit time 1. is centimeters per second per second (cm/s2) or feet per second per second (ft/s2) Density B. The product of three lengths (e.g., length x width x depth for a rectangular solid) is (in3) or cubic meters (m3) 2. Acceleration C. Mass of an object per unit volume is kilograms per cubic meter (kg/m3) or pounds per cubic foot (lbs/ft3) 3. Velocity D. Length per unit time is kilometers per hour (km/hr) or feet per second (ft/s) 4. Volume Correct answers are: 1-A, 2-B, 3-C, 4-D. © Copyright 2016 – Rev 2 ELO 1.3 Operator Generic Fundamentals 22 Need for Unit Conversions ELO 1.4 – Convert between English and SI Systems of mass and length. • Personnel may encounter both English and SI systems of units in their work • To apply measurements from the SI system to the English system and vice versa – It is necessary to develop relationships of known equivalents (Conversion Factors) • Equivalents can then be used to convert from given units to desired units © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 23 Conversion Factors • Conversion Factors – Based on relationships of equivalents from different measurement systems – Applied to given measurement to convert it to required units – Equivalent relationships between different units of measurement are defined in Conversion Tables • Example Conversion Factors: 1 yard = 0.9144 meters 1 kilogram = 2.205 pounds mass (lbm) 1 hour = 3,600 seconds © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 24 Typical Conversion Factors Converting Table Measurement Unit of Measurement Conversion Measurement Time 60 seconds 60 minutes = 1 minute = 1 hour Length 1 yard 12 inches 5,280 feet 1 meter 1 inch = 0.9144 meter = 1 foot = 1 mile = 3.281 feet = 0.0254 meter Mass 1 lbm 2.205 lbm 1 kg = 0.4535 kg = 1 kg = 1,000 g Area 1 ft2 10.764 ft2 1 yd2 1 mile2 = 144 in2 = 1 m2 = 9 ft2 = 3.098 x 106 yd2 Volume 7.48 gal 1 gal 1l = 1 ft3 = 3.785 l = 1,000 cm3 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 25 Performing Unit Conversions Convert 5 feet to inches • First select appropriate equivalent relationship from the conversion table on slide 24: 1 𝑓𝑜𝑜𝑡 = 12 𝑖𝑛𝑐ℎ𝑒𝑠 • Conversion is basically a multiplication by 1 • Divide both sides of equation 1 ft = 12 inches by 1 foot to obtain the following: 1 𝑓𝑡 1 𝑓𝑡 = 12 𝑖𝑛𝑐ℎ𝑒𝑠 1 𝑓𝑜𝑜𝑡 5 𝑓𝑒𝑒𝑡 = 5 𝑓𝑒𝑒𝑡 × © Copyright 2016 – Rev 2 or 1= 12 𝑖𝑛𝑐ℎ𝑒𝑠 1 𝑓𝑜𝑜𝑡 12 𝑖𝑛𝑐ℎ𝑒𝑠 = 5 × 12 𝑖𝑛𝑐ℎ𝑒𝑠 = 60 𝑖𝑛𝑐ℎ𝑒𝑠 1 𝑓𝑜𝑜𝑡 ELO 1.4 Operator Generic Fundamentals 26 Unit Conversion - Examples Example 1: Convert 795 m to ft. • Solution 1: – Step 1: Select the equivalent relationship from the conversion table (See slide 24) 1 𝑚𝑒𝑡𝑒𝑟 (𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑢𝑛𝑖𝑡𝑠) = 3.281 𝑓𝑡 (𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑢𝑛𝑖𝑡𝑠) – Step 2: Divide to obtain the factor 1 as a ratio 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑢𝑛𝑖𝑡𝑠 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑢𝑛𝑖𝑡𝑠 1= 3.281 𝑓𝑡 1𝑚 – Step 3: Multiply the quantity by the ratio: 795 𝑚 × 3.281 𝑓𝑡 795 𝑚 3.281 𝑓𝑡 = × = 795 × 3.281 𝑓𝑡 1𝑚 1 1𝑚 = 2,608.395 𝑓𝑡 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 27 Unit Conversion - Multiple Steps • If an equivalent relationship between given units and desired units cannot be found in conversion tables – multiple conversion factors must be used • Conversion is performed in several steps until measurement is in desired units • Given measurement must be multiplied by each conversion factor (ratio) • After common units have been canceled out, answer will be in desired units © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 28 Unit Conversion - Examples Example 2: Convert 2.91 square miles to square meters • Solution 2: – Step 1: Select equivalent relationship from conversion table on slide 24. o There is no direct conversion shown for square miles to square meters ⇒ multiple conversions will be necessary o Following conversions will be used: 1 𝑠𝑞 𝑚𝑖𝑙𝑒 = 3.098 × 106 𝑠𝑞 𝑦𝑑 1 𝑠𝑞 𝑦𝑑 = 9 𝑠𝑞 𝑓𝑡 10.764 𝑠𝑞 𝑓𝑡 = 1 𝑠𝑞 𝑚 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 29 Unit Conversion - Examples • Solution 2 (continued): – Step 2: Express relationship as a ratio (desired unit/present unit): 3.098 × 106 𝑠𝑞 𝑦𝑑 1= 1 𝑠𝑞 𝑚𝑖𝑙𝑒 – Step 3: Multiply quantity by ratio: 3.098 × 106 𝑠𝑞 𝑦𝑑 2.91 𝑠𝑞 𝑚𝑖𝑙𝑒𝑠 × = 9.015 × 106 𝑠𝑞 𝑦𝑑 1 𝑠𝑞 𝑚𝑖𝑙𝑒 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 30 Unit Conversion - Examples • Solution 2 (continued): – Step 4: Repeat steps until value is in desired units 9 𝑓𝑡 2 1= 1 𝑦𝑑 2 2 9 𝑓𝑡 7 2 9.015 × 106 𝑦𝑑 2 × = 8.114 × 10 𝑓𝑡 1 𝑦𝑑 2 1 𝑚2 1= 10.764 𝑓𝑡 2 2 7 2 2 1 𝑚 (8.114 × 10 𝑚 )(1 𝑓𝑡 ) 7 2 8.114 × 10 𝑓𝑡 × = 10.764 𝑓𝑡 2 10.764 𝑓𝑡2 8.114 × 107 𝑚2 = 10.764 = 7.538 × 106 𝑚2 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 31 Unit Conversion - Examples • It is possible to perform all conversions in a single equation as long as all appropriate conversion factors are included: 6 2 2 2 3.098 × 10 𝑦𝑑 9 𝑓𝑡 1 𝑚 2.91 𝑚𝑖 2 × × × 1 𝑚𝑖 2 1 𝑦𝑑 2 10.764 𝑓𝑡 2 2.91 × (3.098 × 106 )(9)(1 𝑚2 ) = 10.764 8.114 × 107 𝑚2 = 10.764 = 7.538 × 106 𝑚2 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 32 Unit Conversion - Examples Example 3: A Swedish firm is producing a valve that is to be used by an American supplier. The Swedish firm uses the MKS system for all machining. To conform to the MKS system, how will the following measurements be listed? Valve stem = 57.20 in. Valve inlet and outlet I.D. = 22.00 in. O.D. = 27.50 in. • Solution 3: – Valve stem: 57.20 𝑖𝑛.× 0.0254 © Copyright 2016 – Rev 2 𝑚 = 1.453 𝑚 𝑖𝑛. ELO 1.4 Operator Generic Fundamentals 33 Unit Conversion - Examples • Solution 3 (continued): – Valve inlet and outlet: 𝑚 𝐼. 𝐷. 22.00 𝑖𝑛. × 0.0254 = 0.559 𝑚 𝑖𝑛. 𝑚 𝑂. 𝐷. 27.50 𝑖𝑛. × 0.0254 = 0.699 𝑚 𝑖𝑛. © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 34 Common Conversion Factors Common Conversion Factors for Units of Mass g kg t lbm 1 gram 1 0.001 10-5 2.2046 x 10-3 1 kg 1,000 1 0.001 2.2046 1 metric ton (t) 106 1,000 1 2,204.6 1 pound-mass (lbm) 453.59 0.45359 4.5359 x 10-4 1 1 slug 14,594 14.594 0.014594 32.174 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 35 Common Conversion Factors Conversion Factors for Common Units of Length cm m km in. ft mi 1 0.01 10-5 0.3937 0.032808 6.2137 x 10-6 1 meter 100 1 0.001 39.37 3.2808 6.2137 x 10-4 1 kilometer 105 1,000 1 39,370 3,208.8 0.62137 1 inch 2.5400 0.025400 2.54 x 10-5 1 0.083333 1.5783 x 10-5 1 foot 30.480 0.30480 3.0480 x 10-4 12.000 1 1.8939 x 10-4 1 mile 1.6093 x 105 1,609.3 1.6093 63,360 5,280 1 1 centimeter © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 36 Conversion between English and SI Systems Knowledge Check Match the following dimensions to their equivalents. A. 3,280.8 feet 1. 1 mile per hour B. 453,590 grams 2. 1 lbm C. 0.017 hours 3. 1 minute D. 1.47 feet per second 4. 1 kilometer Correct answers are: 1-D, 2-B, 3-C, 4-A. © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 37 Common Conversion Factors ELO 1.5 – Convert time measurements between the following: years, weeks, days, hours, minutes, and seconds. Conversion Factors for Common Units of Time sec min hr Day Year 1 second 1 0.017 2.7 x 10-4 1.19 x 10-5 3.1 x 10-8 1 minute 60 1 0.017 6.9 x 10-4 1.9 x 10-6 1 hour 3,600 60 1 4.16 x 10-2 1.14 x 10-4 1 day 86,400 1,440 24 1 2.74 x 10-3 1 year 3.15 x 107 5.26 x 105 8,760 365 1 © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 38 Converting Time Units Converting Time Units Demonstration Step Convert three years into seconds. Result The conversion factor goes directly from years to seconds. 7 1 𝑦𝑒𝑎𝑟 = 3.15 × 10 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 1. Select the appropriate relationship from the conversion table. 2. Express the relationship as 3.15 × 107 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 a ratio (desired units/present 1 𝑦𝑒𝑎𝑟 units). 7 3. Multiply the quantity by the ratio. 3.15 × 10 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 3 𝑦𝑒𝑎𝑟𝑠 1 𝑦𝑒𝑎𝑟 7 = 9.45 × 10 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 4. Repeat the steps until the value is in the desired units. Multiple steps are not necessary in this case. © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 39 Converting Time Units Knowledge Check Calculate the number of minutes in 3.2 years. A. 164,000 B. 168,000 C. 1,640,000 D. 1,680,000 Correct answer is D. © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 40 Vector Quantities TLO 2 – Describe vector quantities and how they are represented. 2.1 Define the following as they relate to vectors: a. Scalar quantity b. Vector quantity c. Vector component d. Resultant 2.2 Describe methods for identifying vectors in written material. © Copyright 2016 – Rev 2 TLO 2 Operator Generic Fundamentals 41 Vector Terminology ELO 2.1 – Define the following as they relate to vectors: scalar quantity, vector quantity, vector component, and resultant. • Scalars - quantities that have magnitude only ⇒ independent of direction • Vectors - have both magnitude and direction – Indicated graphically using an arrow o Length of a vector (arrow) represents magnitude o Arrow shows direction of vector • Resultant - single vector which represents combined effect of two or more other vectors (called components) • Component – two or more vectors that contribute to a net result, or “resultant” vector – Can be determined either graphically or by using trigonometry © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 42 Scalar Quantities • Scalar Quantity – Quantity that has magnitude only – No directional component • Include – Time – Speed – Temperature – Volume – Density – Mass – Energy © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 43 Vector Quantities • Vector quantity has both – Magnitude – Direction • Magnitude – Size of a vector – Referred to as a vector's displacement – Can be thought of as the scalar portion of vector – Represented by length of vector © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 44 Vector Quantities • Direction of vector indicates how vector is oriented relative to some reference axis • Giving direction to scalar A makes it a vector – Vector A is oriented in the NE quadrant with a direction of 45° north of the E-W axis • Length of A is representative of its magnitude or displacement Figure: Vector Reference Axis © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 45 Scalar vs. Vector Example • Car moving at 80 kilometers per hour – 80 km/hour only refers to car's speed; no indication of direction car is moving – Scalar quantity • Car traveling at 80 km/hour due east – Indicates velocity of car, magnitude (80 km/hour) and direction (due east) – Vector quantity © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 46 Diagramming a Vector Quantity • Straight line drawn to show unit of length – Length represents magnitude of vector • Arrow drawn on one end of line – Arrow represents direction of vector • Description of Vector – Simple straight lines used to illustrate direction and magnitude of quantities – Have a starting point at one end (tail) and an arrow at opposite end (head) Figure: Simple Vector © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 47 Vector Quantities • All have magnitude and direction, includes: – Displacement – Velocity – Acceleration – Force – Momentum – Magnetic field strength © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 48 Vector Quantities Knowledge Check Select all that are true. A. Vectors have direction and magnitude. B. Scalars have direction and magnitude. C. Temperature is a vector quantity. D. Force is a vector quantity. Correct answers are A and D. © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 49 Vector Identification ELO 2.2 – Describe methods for identifying vectors in written material. • Vector quantities are often represented by boldfaced letter – For example: A, B, C, R – Particular quantities may be predefined: F - force, V - velocity, and A - acceleration • Sometimes represented as: 𝑨 𝑩 𝑪 𝑹 • Written vector quantities must include specific magnitude and direction – 50 km/hour due north – 50 lbf at 90° © Copyright 2016 – Rev 2 ELO 2.2 Operator Generic Fundamentals 50 Identifying Vectors Knowledge Check Which of the following is not a way to represent a vector? A. As a quantity with both magnitude and direction (50 km/hr, due north) B. As a quantity (30 mi/hr) C. As a bold, capital letter with an arrow over it D. As a bold capital letter for predefined items, such as F for force Correct answer is B. © Copyright 2016 – Rev 2 ELO 2.2 Operator Generic Fundamentals 51 Solving Vector Problems TLO 3 – Solve resultant vector problems. 3.1 Given a magnitude and direction, graph a vector using the rectangular coordinate system. 3.2 Determine components of a vector from a resultant vector. 3.3 Add vectors using the following methods: a. Graphical b. Component addition c. Analytical © Copyright 2016 – Rev 2 TLO 3 Operator Generic Fundamentals 52 Graphing Vectors ELO 3.1 – Given a magnitude and direction, graph a vector using the rectangular coordinate system. • Vector quantities graphically represented using rectangular coordinate system • Two-dimensional system that uses an x-axis and a y-axis – x-axis is horizontal straight line – y-axis is a vertical straight line, perpendicular to x-axis Figure: Rectangular Coordinate System © Copyright 2016 – Rev 2 ELO 3.1 Operator Generic Fundamentals 53 Rectangular Coordinate System • Point of origin - intersection of axes • Each axis marked off in equal divisions in all four directions from point of origin • On horizontal axis (x), – Values to right of origin are positive (+) – Values to left of origin are negative (-) • Very important to use same units (divisions) on both axes © Copyright 2016 – Rev 2 Figure: Rectangular Coordinate System ELO 3.1 Operator Generic Fundamentals 54 Rectangular Coordinate System • Rectangular coordinate system creates four infinite quadrants – Quadrant I - located above and to right of origin – Quadrant II - located above and to left of origin – Quadrant III - located to left and below origin – Quadrant IV - located below and to right of origin Figure: Rectangular Coordinate System © Copyright 2016 – Rev 2 ELO 3.1 Operator Generic Fundamentals 55 Displaying Vectors Graphically • Using a ruler and graph paper, a rectangular coordinate system should be laid out • Label x- and y-axes • Equal divisions should be marked off in all four directions – Divisions to right and above point of origin labeled positive (+) – Divisions to left and below point of origin labeled negative (-) © Copyright 2016 – Rev 2 ELO 3.1 Operator Generic Fundamentals 56 Graphing Vectors • Beginning at point of origin a line segment of proper length is drawn along x-axis, in positive direction – Line segment represents vector magnitude, or displacement • Arrow is placed at head of vector to indicate direction • Tail of vector located at point of origin © Copyright 2016 – Rev 2 Figure: Displaying Vectors Graphically - Magnitude ELO 3.1 Operator Generic Fundamentals 57 Displaying Vectors That Do Not Fall on X or Y-axes • Tail located at point of origin • Head location: – If coordinates (x, y) are given, values can be plotted to locate vector head – If vector described in degrees, line segment can be rotated counterclockwise from x-axis to proper orientation Figure: Displaying Vectors Graphically - Direction © Copyright 2016 – Rev 2 ELO 3.1 Operator Generic Fundamentals 58 Using Directional Coordinates • Because x- and y-axes define direction, conventional directional coordinates may be used to identify x- and y-axes • Degrees may also be used to identify x- and y-axes Figure: Directional Coordinates © Copyright 2016 – Rev 2 Figure: Degree Coordinates ELO 3.1 Operator Generic Fundamentals 59 Graphing Vectors Knowledge Check The graph of a vector is: A. a straight line with an arrow is drawn on one end. The length of the line represents the magnitude of the vector, and the arrow represents the direction of the vector. B. intended to convey direction only. Notes are needed to convey magnitude. C. indicative of a quantity with magnitude but no direction. D. a means of showing the direction of the force, but is not drawn to scale. Correct answer is A. © Copyright 2016 – Rev 2 ELO 3.1 Operator Generic Fundamentals 60 Vector Components ELO 3.2 – Determine components of a vector from a resultant vector. • Components are vectors, when added, yield the resultant vector • Shown in previous example, traveling 3 km north and then 4 km east yields a resultant displacement of 5 km 37° north of east • Shows that component vectors of any two non-parallel directions can be obtained for any resultant vector in the same plane © Copyright 2016 – Rev 2 ELO 3.2 Figure: Vector Addition – Not in Straight Line Operator Generic Fundamentals 61 Plotting Component Vectors • Component vectors determined by plotting on a rectangular coordinate system • Example - Resultant vector of 5 units at 53° can be broken down into its respective x and y magnitudes – x value of 3 and y value of 4 can be determined using trigonometry or graphically – Magnitudes and position can be expressed by one of several conventions including: o (3,4) o (x = 3, y = 4) o (3 at 0° , 4 at 90°) and (5 at 53°) © Copyright 2016 – Rev 2 ELO 3.2 Operator Generic Fundamentals 62 Plotting Component Vectors • If resultant vector given (instead of component coordinates), components can be determined graphically – Resultant vector plotted first on rectangular coordinates – Vector coordinates projected to axis (dashed lines) – Length along x-axis is Fx – Length along y-axis is Fy © Copyright 2016 – Rev 2 Figure: Vector Components ELO 3.2 Operator Generic Fundamentals 63 Component Vector Example • For resultant vector shown, determine component vectors given: FR = 50 N at 53° Figure: Component Vectors © Copyright 2016 – Rev 2 ELO 3.2 Operator Generic Fundamentals 64 Component Vector Example: Solution • First, project a perpendicular line from the head of FR to xaxis and a similar line to y-axis • Where projected lines meet axes determines magnitude of component vectors: 30 N at 0° (Fx) 40 N at 90° (Fy) Figure: Component Vectors © Copyright 2016 – Rev 2 ELO 3.2 Operator Generic Fundamentals 65 Determining Component Vectors Using Trigonometry • Trigonometry may also be used to determine vector components • Relationship between an acute angle of a right triangle and its sides is given by three ratios: – Sine – Cosine – Tangent © Copyright 2016 – Rev 2 ELO 3.2 Operator Generic Fundamentals 66 Right Triangle Relationships sin 𝜃 = 𝑎 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 = 𝑐 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 cos 𝜃 = 𝑏 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = 𝑐 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑎 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 tan 𝜃 = = 𝑐 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 Figure: Right Triangle © Copyright 2016 – Rev 2 ELO 3.2 Operator Generic Fundamentals 67 Calculating Component Vectors Example 1 • Determine component vectors, Fx and Fy, for FR = 50 N at 53°, using trigonometric functions Figure: Component Vectors © Copyright 2016 – Rev 2 ELO 3.2 Operator Generic Fundamentals 68 Calculating Component Vectors Example 1 • Fx is calculated as follows: 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 cos 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝐹𝑥 cos 𝜃 = 𝐹𝑅 or 𝐹𝑥 𝐹𝑥 𝐹𝑥 𝐹𝑥 = 𝐹𝑅 cos 𝜃 = (50 𝑁)(cos 53°) = (50 𝑁)(0.6018) = 30 𝑁 𝑜𝑛 𝑥– 𝑎𝑥𝑖𝑠 © Copyright 2016 – Rev 2 Figure: Component Vectors ELO 3.2 Operator Generic Fundamentals 69 Calculating Component Vectors Example 1 • Fy is calculated as follows: 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 sin 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝐹𝑦 sin 𝜃 = 𝐹𝑅 or 𝐹𝑦 = 𝐹𝑅 sin 𝜃 𝐹𝑦 = (𝐹𝑅 )(sin 𝜃) 𝐹𝑦 = (50 𝑁)(sin 53°) 𝐹𝑦 = (50 𝑁)(0.7986) 𝐹𝑦 = 40 𝑁 𝑜𝑛 𝑦– 𝑎𝑥𝑖𝑠 © Copyright 2016 – Rev 2 Figure: Component Vectors ELO 3.2 Operator Generic Fundamentals 70 Calculating Component Vectors Example 1 • Components for FR are – Fx = 30 N at 0° – Fy = 40 N at 90° • This result is identical to result obtained using graphic method Figure: Component Vectors © Copyright 2016 – Rev 2 ELO 3.2 Operator Generic Fundamentals 71 Calculating Component Vectors Example 2 • What are the component vectors, given FR = 80 N at 220°? Figure: FR = 80 N at 220° © Copyright 2016 – Rev 2 ELO 3.2 Operator Generic Fundamentals 72 Calculating Component Vectors Example 2 • Fx is calculated as follows: 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑐𝑜𝑠 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝐹𝑥 cos 𝜃 = 𝐹𝑅 or 𝐹𝑥 = 𝐹𝑅 cos 𝜃 𝐹𝑥 = (80 𝑁)(cos 220° ) 𝐹𝑥 = (80 𝑁)(−0.766) 𝐹𝑥 = −61 𝑁 𝑎𝑡 0° or 61 𝑁 𝑎𝑡 180° © Copyright 2016 – Rev 2 ELO 3.2 Figure: FR = 80 N at 220° Operator Generic Fundamentals 73 Calculating Component Vectors Example 2 • Fy is calculated as follows: 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 sin 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 sin 𝜃 = 𝐹𝑦 𝐹𝑅 or 𝐹𝑦 𝐹𝑦 𝐹𝑦 𝐹𝑦 = = = = 𝐹𝑅 sin 𝜃 (80 𝑁)(sin 220°) 80 𝑁 −0.6428 −51 𝑁 𝑎𝑡 90° or 51 𝑁 𝑎𝑡 270° © Copyright 2016 – Rev 2 ELO 3.2 Figure: FR = 80 N at 220° Operator Generic Fundamentals 74 Calculating Component Vectors Example 2 • Components for FR: 𝐹𝑥 = 61 𝑁 𝑎𝑡 180° 𝐹𝑦 = 51 𝑁 𝑎𝑡 270° Figure: FR = 80 N at 220° © Copyright 2016 – Rev 2 ELO 3.2 Operator Generic Fundamentals 75 Determining Components of a Vector Knowledge Check A vector begins at (2, 3) and ends at (-1, 7). Select all of the statements about this vector that are true. A. The vector has a magnitude of 5. B. The vector has a direction of 120°. C. The vector has a direction of 127°. D. The vector has a horizontal component of magnitude 3, direction 180°, and a vertical component of magnitude 4, direction 90°. Correct answers are A, B, and C. © Copyright 2016 – Rev 2 ELO 3.2 Operator Generic Fundamentals 76 Methods Used To Add Vectors ELO 3.3 – Add vectors using the following methods: graphical, component addition, and analytical. • Several methods have been developed to add vectors • In this presentation, the graphical, component addition and analytical methods will be explained – Either the graphical method or the component addition method will provide a fairly accurate result – If a higher degree of accuracy is required, an analytical method using geometric and trigonometric functions is required © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 77 Graphic Method Of Vector Addition • Equipment needed to use Graphic Method of Vector Addition: – Standard linear (non-log) graph paper – Ruler – Protractor – Writing instrument (pencil) • Method utilizes a five-step process © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 78 Graphic Method Of Vector Addition • Step 1 - Plot first vector (F1) on rectangular (x-y) axes – Ensure same scale used on both axes – Place tail (beginning) of first vector at origin of axes Figure: Plotting Vector on Rectangular Axes © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 79 Graphic Method Of Vector Addition • Step 2 - Draw second vector (F2) connected to end of first vector – Start tail of second vector at head of first vector – Ensure second vector drawn to scale – Ensure proper angular orientation of second vector with respect to axes of graph © Copyright 2016 – Rev 2 Figure: Plotting Vector F2 ELO 3.3 Operator Generic Fundamentals 80 Graphic Method Of Vector Addition • Step 3 - Add other vectors sequentially – Add one vector at a time – Always start tail of new vector at head of previous vector – Draw all vectors to scale and with proper angular orientation © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 81 Graphic Method Of Vector Addition • Step 4 - When all given vectors have been drawn, draw and label a resultant vector, FR , from point of origin of axes to head of final vector – Tail of resultant vector is tail of first vector drawn – Head of resultant vector is at head of last vector drawn Figure: Plotting Resultant Vector © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 82 Graphic Method Of Vector Addition • Step 5 - Determine magnitude and direction of resultant vector – Measure displacement and angle directly from graph using ruler and protractor – Determine components of resultant by projection onto x- and y-axes Figure: Plotting Resultant Vector © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 83 Component Addition Method • Component Addition Method - addition of vector coordinates on a rectangular (x, y) coordinate system • Vector components added along each axis to determine magnitude and direction of resultant © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 84 Component Addition Method • Coordinates locate specific point in system – Specific point is head of vector – Head can be located by counting units along x-axis and units along y-axis – Example: Point has coordinates (4,3) o x component = +4 o y component = +3 © Copyright 2016 – Rev 2 Figure: Vector Component Addition Method ELO 3.3 Operator Generic Fundamentals 85 Component Addition Method • Four-step method: – Step 1 - Determine x- and y-axes components of all original vectors – Step 2 - Mathematically combine all x-axis components (+𝑥 𝑎𝑡 180° = −𝑥 𝑎𝑡 0° ) – Step 3 - Mathematically combine all y-axis components (+𝑦 𝑎𝑡 270° = −𝑦 𝑎𝑡 90°) – Step 4 - Resulting (x, y) components are the (x, y) components of the resulting vector © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 86 Component Addition Method Example 1 • Find the resultant vector: – 𝐹1 = (4,10) – 𝐹2 = (−6,4) – 𝐹3 = (2, −4) – 𝐹4 = (10, −2) • Step 1: Determine x- and y-axis components of all four original vectors 𝑥– 𝑎𝑥𝑖𝑠 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 = 4, −6, 2, 10 𝑦– 𝑎𝑥𝑖𝑠 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 = 10, 4, −4, −2 © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 87 Component Addition Method Example 1 (continued) • Step 2: Mathematically combine all x-axis components 𝐹𝑥 = 4 + (−6) + 2 + 10 𝐹𝑥 = 4 − 6 + 2 + 10 𝐹𝑥 = 10 • Step 3: Mathematically combine all y-axis components 𝐹𝑦 = 10 + 4 + (−4) + (−2) 𝐹𝑦 = 10 + 4 − 4 − 2 𝐹𝑦 = 8 © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 88 Component Addition Method Example 1 (continued) • Step 4: Express resultant vector – Resultant components from previous additions are coordinates of resultant 𝐹𝑥 = 10, 𝐹𝑦 = 8 𝐹𝑅 = (10, 8) © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 89 Component Addition Method Example 2 • Determine the resultant, FR, given: – 𝐹1 = 30 𝑁 𝑎𝑡 0°, 10 𝑁 𝑎𝑡 90° – 𝐹2 = 50 𝑁 𝑎𝑡 0°, 50 𝑁 𝑎𝑡 90° – 𝐹3 = 45 𝑁 𝑎𝑡 180°, 30 𝑁 𝑎𝑡 90° – 𝐹4 = 15 𝑁 𝑎𝑡 0°, 50 𝑁 𝑎𝑡 270° • Follow the sequence used in first example, remembering that x at 180° is -x at 0°, and y at 270° is -y at 90° 𝐹𝑥 = 30 + 50 + (−45) + 15 = 50 𝑁 𝐹𝑦 = 10 + 50 + 30 + (−50) = 40 𝑁 𝐹𝑅 = 50 𝑁 𝑎𝑡 0°, 40 𝑁 𝑎𝑡 90° © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 90 Analytical Method Of Vector Addition • Trigonometric functions are most accurate method for determining vector addition – Graphic and component addition methods of obtaining resultant described previously can be hard to use and time consuming – Accuracy is a function of the scale used in making the diagram and how carefully the vectors are drawn • Analytical method can be simpler and far more accurate © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 91 Review of Mathematical Functions • Pythagorean Theorem is used to relate lengths sides of right triangles – In any right triangle, square of length of hypotenuse equals sum of squares of lengths of other two sides Figure: Right Triangle – Pythagorean Theorem 𝑐2 = 𝑎2 + 𝑏2 𝑜𝑟 𝑐 = 𝑎2 + 𝑏 2 © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 92 Review of Mathematical Functions 𝑎 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 sin 𝜃 = = 𝑐 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑏 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 cos 𝜃 = = 𝑐 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑎 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 tan 𝜃 = = 𝑐 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 • Trigonometric functions: – Cos used to solve for Fx – Sin used to solve for Fy – Tan normally used to solve for θ o Sin and cos may also be used © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 93 Review of Mathematical Functions • On a rectangular coordinate system, sine values of θ are positive (+) in quadrants I and II and negative (-) in quadrants III and IV • Cosine values of θ are positive (+) in quadrants I and IV and negative (-) in quadrants II and III • Tangent values are positive (+) in quadrants I and III and negative (-) in quadrants II and IV © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 94 Review of Mathematical Functions • When mathematically solving for tan θ, calculators will specify angles in quadrants I and IV only • Actual angles may be in quadrants II and III • Each problem should be analyzed graphically in order to ensure a realistic solution • Quadrant II and III angles may be obtained by adding or subtracting 180° from value calculated © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 95 Using the Analytical Method • A man walks 3 km in one direction, then turns 90° and continues to walk for an additional 4 km • In what direction and how far is he from his starting point? – First, draw a simple sketch Figure: Hypotenuse and Angle © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 96 Using the Analytical Method • His net displacement is found using: 𝑅= 𝑎2 + 𝑏 2 𝑅= 32 + 42 𝑅 = 25 𝑅 = 5 𝑘𝑚 • His direction (angle of displacement) is found using tan: 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 tan 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑎 tan 𝜃 = 𝑏 4 tan 𝜃 = 3 tan 𝜃 = 1.33 𝜃 = tan−1 1.33 𝜃 = 53° © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 97 Using the Analytical Method • His new location is 5 km at 53° from his starting point Figure: Hypotenuse and Angle © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 98 Resultant of Several Vectors Example 1 • Three forces, (F1, F2, and F3) act on an object • Find resultant force (FR) Figure: Force Acting on an Atomic Nucleus © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 99 Resultant of Several Vectors Example 1 (continued) • Step 1: Draw x and y coordinates and three forces from point of origin or center of object – Component vectors and angles have been added to drawing to aid in discussion Figure: Individual Force Vectors – Nucleus Example © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 100 Resultant of Several Vectors Example 1 (continued) • Step 2: Resolve each vector into its rectangular components: Vector Angle x component y component 𝐹1𝑦 = 𝐹1 sin 𝜃1 F1 θ1 𝐹1𝑥 = 𝐹1 cos 𝜃1 F2 θ2 𝐹2𝑥 = 𝐹2 cos 𝜃2 𝐹2𝑦 = 𝐹2 sin 𝜃2 F3 θ3 𝐹3𝑥 = 𝐹3 cos 𝜃3 𝐹3𝑦 = 𝐹3 sin 𝜃3 Figure: Individual Vectors – Nucleus Example © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 101 Resultant of Several Vectors Example 1 (continued) • Step 3: Sum the x and y components. 𝐹𝑅𝑥 = Σ𝐹𝑥 = 𝐹1𝑥 + 𝐹2𝑥 + 𝐹3𝑥 𝐹𝑅𝑦 = Σ𝐹𝑦 = 𝐹1𝑦 + 𝐹2𝑦 + 𝐹3𝑦 (“Σ” means summation) • Step 4: Calculate magnitude of FR 𝐹𝑅 = 2 2 𝐹𝑅𝑥 + 𝐹𝑅𝑦 • Step 5: Calculate angle of displacement 𝐹𝑅𝑦 tan 𝜃 = 𝐹𝑅𝑥 𝐹 −1 𝑅𝑦 𝜃 = tan 𝐹𝑅𝑥 © Copyright 2016 – Rev 2 Figure: Individual Vectors – Nucleus Example ELO 3.3 Operator Generic Fundamentals 102 Resultant of Several Vectors Example 2 • Given three forces acting on an object, determine magnitude and direction of resultant force FR – 𝐹1 = 90 𝑁 𝑎𝑡 39° – 𝐹2 = 50 𝑁 𝑎𝑡 120° – 𝐹3 = 125 𝑁 𝑎𝑡 250° © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 103 Resultant of Several Vectors Example 2 (continued) • Step 1: Draw vectors – Draw x and y coordinate axes on a sheet of paper – Draw F1, F2, and F3 from point of origin o Not necessary to be totally accurate in placing vectors in drawing o Approximate location in correct quadrant is all that is necessary – Label drawing © Copyright 2016 – Rev 2 ELO 3.3 Figure: Individual Force Vectors – Nucleus Example Operator Generic Fundamentals 104 Resultant of Several Vectors Example 2 (continued) • Step 2: Resolve each force into its rectangular coordinates Force Magnitude Angle x component 𝐹1𝑦 90 N 𝐹1𝑥 = 90 cos 39° 𝐹1𝑥 = (90) (0.777) 𝐹1𝑥 = 69.9 𝑁 𝐹1𝑦 = 56.6 𝑁 𝐹2𝑥 = 50 cos 120° 𝐹2𝑦 = (50) sin 120° 𝐹2𝑥 = (50)(−0.5) 𝐹2𝑦 = (50)(0.866) 𝐹2𝑥 = −25 𝑁 𝐹2𝑦 = 43.3 𝑁 𝐹3𝑥 = (125) cos 250° 𝐹3𝑦 = (125) sin 250° 𝐹3𝑥 = 125)(−0.342) 𝐹3𝑦 = (125)(−0.94) 𝐹3𝑥 = −42.8 𝑁 𝐹3𝑦 = −117.5 𝑁 F1 F2 F3 © Copyright 2016 – Rev 2 50 N 125 N 39° 120° 250° ELO 3.3 y component = 90 sin 39° 𝐹1𝑦 = (90)(0.629) Operator Generic Fundamentals 105 Resultant of Several Vectors Example 2 (continued) • Step 3: Sum the x and y components 𝐹𝑅𝑥 = 𝐹1𝑥 + 𝐹2𝑥 + 𝐹3𝑥 𝐹𝑅𝑥 = 69.9 𝑁 + (−25 𝑁) + (−42.8 𝑁) 𝐹𝑅𝑥 = 2.1 𝑁 𝐹𝑅𝑦 = 𝐹1𝑦 + 𝐹2𝑦 + 𝐹3𝑦 𝐹𝑅𝑦 = 56.6 𝑁 + 43.3 𝑁 + (−117.5𝑁) 𝐹𝑅𝑦 = −17.6 𝑁 © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 106 Resultant of Several Vectors Example 2 (continued) • Step 4: Calculate magnitude of FR. 𝐹𝑅 = 𝐹𝑅 = 𝐹𝑥2 + 𝐹𝑦2 2.1 2 + −17.6 2 𝐹𝑅 = 314.2 𝐹𝑅 = 17.7 𝑁 © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 107 Resultant of Several Vectors Example 2 (continued) • Step 5: Calculate angle of displacement 𝐹𝑅𝑦 tan 𝜃 = 𝐹𝑅𝑥 −17.6 tan 𝜃 = 2.1 tan 𝜃 = −8.381 𝜃 = tan−1 −8.381 𝜃 = −83.2° • 𝐹𝑅 = 17.7 𝑁 at −83.2° 𝑜𝑟 276.8° Note: A negative angle means a clockwise rotation from the zero axis. © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 108 Adding Vectors Knowledge Check Determine the resultant of the following vectors: F1, magnitude 6, angle 150° F2, magnitude 11, angle 240° F3, magnitude 5, 0° A. Magnitude 8.7, 228° B. Magnitude 7.9, 210° C. Magnitude 11.1, 196° D. Magnitude 7.9, 196° Correct answer is D. © Copyright 2016 – Rev 2 ELO 3.3 Operator Generic Fundamentals 109 Measurement of Force TLO 4 – Describe the measurement of force and its relationship to free body diagrams. 4.1 Describe the following types of forces: tensile, compressive, frictional, centripetal, and centrifugal. 4.2 Describe the factors that affect the magnitude of the static and kinetic friction force. 4.3 Describe force as it applies to an object and its velocity. 4.4 Describe weight as it applies to an object and its velocity. 4.5 Given the mass of an object and a value for gravity, calculate the weight of the object. 4.6 Describe the purpose of a free-body diagram, and given all necessary information, construct a free-body diagram. 4.7 Describe the conditions necessary for a body to be in force equilibrium. © Copyright 2016 – Rev 2 TLO 4 Operator Generic Fundamentals 110 Tensile and Compressive Forces ELO 4.1 – Describe the following types of forces: tensile, compressive, frictional, centripetal, and centrifugal. • Applied force on an object tends to pull the object apart, defines tension • Applied force tends to compress the object, it is in compression • Ropes, cables, etc, attached to bodies only support tensile loads, and therefore exhibit tension when placed on free-body diagrams • Fluid forces are almost always compressive forces © Copyright 2016 – Rev 2 ELO 4.1 Operator Generic Fundamentals 111 Centripetal Force • An object moving at constant speed in a circle is not in equilibrium • Although magnitude of linear velocity is not changing, direction of velocity is continually changing • A change in direction requires acceleration, an object moving in a circular path has a constant acceleration towards center of circular path © Copyright 2016 – Rev 2 ELO 4.1 Operator Generic Fundamentals 112 Centripetal Force • Force is required to cause acceleration (Newton's second law of motion) 𝐹 = 𝑚𝑎 • To have constant acceleration towards center of circular path, there must be a net force acting towards center • Force is known as Centripetal force Figure: Centripetal Force • Without centripetal force object will move in a straight line © Copyright 2016 – Rev 2 ELO 4.1 Operator Generic Fundamentals 113 Centrifugal Force • Appears to oppose direction of motion, acts on an object that follows a curved path – Appears to be a force directed away from center of circular path – Actually a fictitious force – An apparent force that is used to describe forces present due to an object's rotation © Copyright 2016 – Rev 2 Figure: Apparent Centrifugal Force ELO 4.1 Operator Generic Fundamentals 114 Centrifugal Force • Not an actual force • Can be proven not an actual force by cutting string – Plane will fly off in a straight line that is tangent to circle at velocity it had the moment string was cut – If an actual centrifugal force was present, plane would not fly away in a line tangent to circle, but would fly directly away from the circle © Copyright 2016 – Rev 2 ELO 4.1 Figure: Loss of Centripetal Force Operator Generic Fundamentals 115 Friction ELO 4.2 – Describe the factors that affect the magnitude of the static and kinetic friction force. • Results from two surfaces in contact, where one of the surfaces is attempting to move parallel to, or over other surface – Dry friction (sometimes called Coulomb friction) – Fluid friction © Copyright 2016 – Rev 2 ELO 4.2 Operator Generic Fundamentals 116 Fluid Friction • Develops between layers of fluid moving at different velocities – Used in considering problems involving flow of fluids through pipes – Important to thermodynamics and fluid flow © Copyright 2016 – Rev 2 ELO 4.2 Operator Generic Fundamentals 117 Dry Friction • Laws of dry friction are best understood by the following experiment – A block of weight W is placed on a horizontal plane surface – Forces acting on block are its weight (W) and the normal force (N) of surface Figure: Frictional Forces © Copyright 2016 – Rev 2 ELO 4.2 Operator Generic Fundamentals 118 Dry Friction • Weight has no horizontal component • Normal force of surface also has no horizontal component – Reaction is therefore normal to surface and is represented by N in part (a) of previous figure Figure: Frictional Forces © Copyright 2016 – Rev 2 ELO 4.2 Operator Generic Fundamentals 119 Dry Friction • Horizontal force P is applied to block as shown in part (b) of previous figure – If P is small, block will not move – Some other horizontal force must therefore exist that balances P Figure: Frictional Forces © Copyright 2016 – Rev 2 ELO 4.2 Operator Generic Fundamentals 120 Dry Friction • This balancing force is the static-friction force (F) – Resultant of a great number of forces acting over entire surface of contact between block and plane • Nature of these forces is not known exactly – Assumed that forces are due to irregularities of surfaces in contact and to molecular action Figure: Frictional Forces © Copyright 2016 – Rev 2 ELO 4.2 Operator Generic Fundamentals 121 Dry Friction • If force P is increased, friction force F also increases, continuing to oppose P, until its magnitude reaches a certain maximum value FM • If P is further increased, friction force cannot balance it anymore, and block starts sliding • As soon as block has been set in motion, magnitude of F drops FM to a lower value FK – less interpenetration between irregularities of surfaces in contact • Block continues sliding with increasing velocity (i.e., it accelerates) – Friction force, FK – kinetic-friction force, remains approximately constant Figure: Frictional Forces © Copyright 2016 – Rev 2 ELO 4.2 Operator Generic Fundamentals 122 Static-Friction and Kinetic-Friction Force Static-Friction Force • Maximum value FM of static-friction force is proportional to normal component N of reaction of surface 𝐹𝑀 = µ𝑆 𝑁 • µS - coefficient of static friction (a constant) Kinetic-Friction Force • Magnitude FK of kinetic-friction force may be expressed as: 𝐹𝐾 = µ𝐾𝑁 • µK - coefficient of kinetic friction (a constant) © Copyright 2016 – Rev 2 ELO 4.2 Operator Generic Fundamentals 123 Magnitudes of Coefficients of Friction • µS and µK do not depend upon area of surfaces in contact – Both coefficients depend strongly on nature of surfaces in contact – Also depend upon the exact condition of surfaces o Value is seldom known with accuracy greater than 5 percent Note: Frictional forces are always opposite in direction to motion (or impending motion) of object © Copyright 2016 – Rev 2 ELO 4.2 Operator Generic Fundamentals 124 Friction Force Knowledge Check Select all statements about the friction force that are true. A. The direction of the friction force is opposite the direction of motion. B. The coefficient of kinetic friction between two surfaces is greater than the coefficient of static friction for the same two surfaces. C. Friction force is proportional to the normal force. D. When an object is not moving, the friction force is zero. Correct answers are A and C. © Copyright 2016 – Rev 2 ELO 4.2 Operator Generic Fundamentals 125 Types of Forces Knowledge Check Match the following terms to the appropriate definitions. A. Tensile forces only 1. Ropes and cables B. Cause of circular motion 2. Fluid forces C. Contact between objects 3. Friction D. Compressive forces only 4. Centripetal Correct answers are: 1-A, 2-D, 3-C, 4-B. © Copyright 2016 – Rev 2 ELO 4.1 Operator Generic Fundamentals 126 Force and Velocity ELO 4.3 – Describe weight as it applies to an object and its velocity. • Force - vector quantity that tends to produce an acceleration of a body in the direction of its application – Changing body's velocity causes body to accelerate – Mathematically defined by Newton's second law of motion: 𝐹 = 𝑚𝑎 – Where F = force on object (Newton or lbf) m = mass of object (kg or lbm) a = acceleration of object (m/sec2 or ft/sec2) © Copyright 2016 – Rev 2 ELO 4.3 Operator Generic Fundamentals 127 Force • Characterized by its point of application, its magnitude, and its direction • Two or more forces may act upon an object without affecting its state of motion • For example, a book resting upon a table has a downward force acting on it caused by gravity and an upward force exerted on it from table top – These two forces cancel and net force on book is zero – Can be verified by observing that no change in state of motion has occurred © Copyright 2016 – Rev 2 ELO 4.3 Operator Generic Fundamentals 128 Force and Velocity Knowledge Check Select all the true statements. A. Application of force always causes acceleration. B. Force is a vector quantity. C. When an object remains at rest, there is no force applied to the object. D. When an object remains at rest, the net force on the object is zero. Correct answers are B and D. © Copyright 2016 – Rev 2 ELO 4.3 Operator Generic Fundamentals 129 Weight ELO 4.4 – Describe weight as it applies to an object and its velocity. • Force can be thought of simply as a push or pull, but is more clearly defined as any action on a body that tends to change the velocity of the body • Weight is a force exerted on an object due to the object's position in a gravitational field • Newton's laws of motion are some of the fundamental concepts used in the study of force and weight © Copyright 2016 – Rev 2 ELO 4.4 Operator Generic Fundamentals 130 Weight • Special application of concept of force – Defined as force exerted on an object by gravitational field of earth, or more specifically, pull of the earth on body 𝑊 = 𝑚𝑔 or 𝑊 = 𝑚𝑔 𝑔𝑐 (English system) • W = weight (N or lbf) • m = mass (kg or lbm) of object • g = local acceleration of gravity – on earth = 9.8 m/sec2 or 32.17 ft/sec2 • gc = a conversion constant employed to facilitate the use of Newton's second law of motion with English system of units, equal to 32.17 ftlbm/lbf-sec2 – gc has same numerical value as acceleration of gravity at sea level © Copyright 2016 – Rev 2 ELO 4.4 Operator Generic Fundamentals 131 Weight • Mass of a body is same regardless of whether on the moon, earth, etc. • Weight of a body depends upon local acceleration of gravity – Weight of an object is less on the moon than on earth because local acceleration of gravity is less on the moon than on earth © Copyright 2016 – Rev 2 ELO 4.4 Operator Generic Fundamentals 132 Weight – Examples • Calculate the weight of a person with a mass of 84 kg. 𝑊 = 𝑚𝑔 • Calculate the weight of a person with a mass of 84 kg on the moon. Gravity on the moon is 1.63 m/sec2. 𝑚 = 84 𝑘𝑔 9.81 𝑠𝑒𝑐 2 𝑊 = 𝑚𝑔 𝑚 𝑊 = 84 𝑘𝑔 1.63 𝑠𝑒𝑐 2 2 = 8.24 × 10 𝑁 𝑊 = 1.37 × 103 𝑁 © Copyright 2016 – Rev 2 ELO 4.4 Operator Generic Fundamentals 133 Gravitational Force Review • Any object dropped will accelerate as it falls, not in physical contact with any other body • Gravitational force – Concept that one body, exerts a force on another body, even though they are far apart – Gravitational attraction of two objects depends upon mass of each and distance between them (Newton's Law of Gravitation) © Copyright 2016 – Rev 2 ELO 4.4 Operator Generic Fundamentals 134 Weight Knowledge Check Select all the true statements. A. The mass of a body is the same, wherever the body is located. The weight of a body, however, depends upon the local acceleration due to gravity. B. Weight is a function of mass only, and does not vary based on location. C. Weight is a vector quantity. D. Weight is a scalar quantity. Correct answers are A and C. © Copyright 2016 – Rev 2 ELO 4.4 Operator Generic Fundamentals 135 Calculating Weight ELO 4.5 – Given the mass of an object and a value for gravity, calculate the weight of the object. • Calculate the weight of an object by the following four steps: – Determine which system (English or SI) is used, and select the appropriate formula – Determine mass of the object – Determine local acceleration due to gravity – Complete formula and calculate the weight © Copyright 2016 – Rev 2 ELO 4.5 Operator Generic Fundamentals 136 Calculating Weight - Example Calculate the weight of an object with mass of 262 kg. • Step 1: Determine which system is used (English or metric) and choose the correct formula – Mass is given in kg, metric system is being used, and the formula is 𝑊 = 𝑚𝑔 • Step 2: Determine the mass of the object – Mass is 262 kg • Step 3: Determine the local acceleration due to gravity – Local acceleration due to gravity is 9.8 m/sec2 © Copyright 2016 – Rev 2 ELO 4.5 Operator Generic Fundamentals 137 Calculating Weight - Example • Step 4: Complete the formula and calculate the weight of the object. 𝑊 = 𝑚𝑔 = 262 𝑘𝑔 𝑚 9.8 𝑠𝑒𝑐 2 𝑘𝑔– 𝑚 = 2.6 × 10 𝑠𝑒𝑐 2 3 3 = 2.6 × 10 𝑁 © Copyright 2016 – Rev 2 ELO 4.5 Operator Generic Fundamentals 138 Calculating Weight Knowledge Check An object weighs 144 pounds-mass on earth. What would it weigh on a planet with acceleration due to gravity of 14.9 ft/sec2? A. 66.6 pounds-mass B. 66.6 kilograms C. 2.19 x 102 pounds-mass D. 2.19 x 102 kilograms Correct answer is A. © Copyright 2016 – Rev 2 ELO 4.5 Operator Generic Fundamentals 139 Free-Body Diagrams Introduction ELO 4.6 – Describe the purpose of a free-body diagram, and given all necessary information, construct a free-body diagram. • To study the effect of forces on a body, isolate the body and determine all forces acting upon it • The diagram with the representation of all external forces acting on it is called a Free-Body Diagram • The isolation of a body is the tool that clearly separates cause and effect and focuses attention to the application of a selected principle © Copyright 2016 – Rev 2 ELO 4.6 Operator Generic Fundamentals 140 Simple Free-Body Diagram • Book resting on table • Two forces are acting on book to keep it stationary – Weight (W) of book exerts a force downward on table – Normal force (N) is exerted upward on book by table ⇒equal to weight of book – A normal force is defined as any perpendicular force with which any two surfaces are pressed against each other © Copyright 2016 – Rev 2 Figure: Book on a Table ELO 4.6 Operator Generic Fundamentals 141 Constructing a Free-Body Diagram • Step 1. Determine which body/bodies to be isolated – Body chosen will usually involve one or more of desired unknown quantities • Step 2. Isolate body or combination of bodies chosen with a diagram that represents its complete external boundaries • Step 3. Represent all forces that act on isolated body as applied in their proper positions in diagram of isolated body – Do not show forces that object exerts on anything else, since these forces do not affect object itself • Step 4. Indicate the choice of coordinate axes directly on diagram. Pertinent dimensions may also be represented for convenience © Copyright 2016 – Rev 2 ELO 4.6 Operator Generic Fundamentals 142 Constructing a Free-Body Diagram • Free-body diagram serves purpose of focusing accurate attention on action of external forces; therefore, diagram should not be cluttered with excessive information • Force arrows should be clearly distinguished from other arrows to avoid confusion © Copyright 2016 – Rev 2 ELO 4.6 Operator Generic Fundamentals 143 Free-Body Diagram – Example 1 • Car is being towed by a force of some magnitude • Construct a free-body diagram showing all forces acting on car Figure: Car © Copyright 2016 – Rev 2 ELO 4.6 Operator Generic Fundamentals 144 Free-Body Diagram – Solution 1 • Car is chosen and isolated • All forces acting on car are represented with proper coordinate axes – Fapp – Force applied to tow car – Fk – Frictional force that opposes applied force due to weight of car and nature of surfaces (car's tires and road surface) – W – Weight of car – N – Normal force acting on car © Copyright 2016 – Rev 2 Figure: Free-Body Diagram: Car Under Tow ELO 4.6 Operator Generic Fundamentals 145 Calculating Weight Knowledge Check Select all the true statements for the diagram: A. If the hanging object is at rest, the net forces applied to it must be zero. B. Cable T2 has no force applied to it. C. Cable T2 has a horizontal force pulling the hanging object to the left but no vertical force applied. D. Cable T1 has a vertical force to match the weight of the hanging object, and a horizontal force pulling the hanging object to the right. Correct answers are A, C, and D. Figure: Hanging Object © Copyright 2016 – Rev 2 ELO 4.6 Operator Generic Fundamentals 146 Force Equilibrium ELO 4.7 – Describe the conditions necessary for a body to be in force equilibrium. • Knowledge of the forces required to maintain an object in equilibrium is essential in understanding the nature of bodies at rest and in motion © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 147 Net Force • When forces act on an object, result may be a change in object's state of motion • If certain conditions exist, forces may combine to maintain a state of equilibrium or balance • To determine if a body is in equilibrium, overall effect of all forces acting on it must be assessed • All forces that act on an object result in essentially one force that influences object's motion • Force which results from all forces acting on body defined as Net Force • Important to remember that forces are vector quantities • When analyzing various forces, must account for both magnitude (displacement) as well as direction in which force is applied – Best done using a free-body diagram © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 148 Book Resting on Table • Book remains stationary resting on table because table exerts normal force upward equal to weight of book ⇒ net force on book is zero • If force applied to book and effect of friction is neglected, net force will be equal to applied force – Book will move in direction of applied force Figure: Net Force Acting on Book on Table © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 149 Book Resting on Table • Free-body diagram (a) below shows that weight (W) of book is canceled by normal force (N) of table since they are equal in magnitude but opposite in direction • Resultant (net) force is therefore equal to applied force (FAPP) Figure: Net Force Acting on Book on Table © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 150 Equilibrium and Inertia • An object in equilibrium is considered to be in a state of balance ⇒ net force on object is equal to zero – If vector sum of all forces acting on an object is equal to zero, then object is in equilibrium • Newton's first law of motion describes equilibrium and the effect of force on a body that is in equilibrium – "An object remains at rest (if originally at rest) or moves in a straight line with a constant velocity if the net force on it is zero.“ • Newton's first law of motion is also called the “Law of Inertia” • Inertia - tendency of a body to resist a change in its state of motion © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 151 First Condition of Equilibrium • Consequence of Newton's first law • May be written in vector form: – "A body will be in translational equilibrium if and only if the vector sum of forces exerted on a body by the environment equals zero.“ • Example – if three forces act on a body, it is necessary for the following to be true for body to be in equilibrium: 𝐹1 + 𝐹2 + 𝐹3 = 0 • Equation may also be written: Σ𝐹 = 0 – Sum includes all forces exerted on body by its environment © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 152 First Condition of Equilibrium • The vanishing of this vector sum is a necessary condition, called the First Condition of Equilibrium – Must be satisfied in order to ensure translational equilibrium • In three dimensions (x, y, z), component equations of First Condition of Equilibrium are: Σ𝐹𝑥 = 0 Σ𝐹𝑦 = 0 Σ𝐹𝑧 = 0 • Condition applies to objects in motion with constant velocity and to bodies at rest or in static equilibrium (referred to as Statics) © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 153 Equilibrium Example • Object has a weight of 125 kg • Object is suspended by cables as shown • Calculate tension (T1 ) in cable at 30° with the horizontal • Tension in a cable is force transmitted by cable • Tension at any point in cable can be measured by removing a suitable length of cable and inserting a spring scale © Copyright 2016 – Rev 2 ELO 4.7 Figure: Hanging Object Operator Generic Fundamentals 154 Equilibrium Example • Object and its supporting cables are motionless (in equilibrium) ⇒ net force acting on intersection of cables is zero – Sum of x-components of T1, T2 and T3 is zero Σ𝐹𝑥 = 𝑇1𝑥 + 𝑇2𝑥 + 𝑇3𝑥 = 0 – Sum of y-components also zero Σ𝐹𝑦 = 𝑇1𝑦 + 𝑇2𝑦 + 𝑇3𝑦 = 0 • Tension T3 equal to weight of the object - 125 N Figure: Free-Body Diagram: Hanging Object © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 155 Equilibrium Example • x and y components of tensions can be found using trigonometry (sine function) • Substituting known values into equation: 𝛴𝑭𝑦 = 𝑻1 sin 30° + 𝑻2 sin 180° + 𝑻3 sin 270° = 0 𝛴 𝑻1 0.5 + 𝑻2 0 + 125 𝑁 −1 = 0 0.5 𝑻1 − 125 𝑁 = 0 0.5 𝑻1 = 125 𝑁 𝑻1 = 250 𝑁 Figure: Free-Body Diagram: Hanging Object © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 156 Equilibrium Example • A simpler method to solve this problem involves assigning a sign convention to free-body diagram and examining direction of forces • By choosing (+) as upward direction and (-) as downward direction: – Upward component of T1 is + T1 sin 30° – Tension T3 is -125 N – T2 has no y-component Figure: Free-Body Diagram: Hanging Object © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 157 Equilibrium Example • Using same equation as before: Σ𝐹𝑦 = (𝑇1 sin 30°) − 125 𝑁 = 0 0.5 𝑇1 = 125 𝑁 𝑇1 = 250 𝑁 © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 158 Equilibrant • If the sum of all forces acting upon a body is equal to zero, that body is said to be in force equilibrium • If the sum of all the forces is not equal to zero, any force or system of forces capable of balancing the system is defined as an equilibrant © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 159 Equilibrant Example • A 900 kg car is accelerating (on a frictionless surface) at a rate of 2 m/sec2 • What force must be applied to the car to act as an equilibrant for this system? • First, a free-body diagram is drawn Figure: Free-Body Diagram: Accelerating Car • A force, F2 , MUST be applied in opposite direction to F1 such that sum of all forces acting on car is zero Σ 𝐹𝑜𝑟𝑐𝑒𝑠 = 𝐹1 + 𝐹2 + 𝑁 + 𝑊 = 0 © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 160 Equilibrant Example • Since car remains on surface, forces N and W are in equal and opposite directions • Force F2 must be applied in an equal and opposite direction to F1 in order for forces to be in equilibrium 𝐹2 = 𝐹1 = 𝑚𝑎 = (900 𝑘𝑔 × 2 𝑚/sec2) = 1,800 𝑘𝑔– 𝑚/sec2 = 1,800 𝑁𝑒𝑤𝑡𝑜𝑛𝑠 © Copyright 2016 – Rev 2 Figure: Free-Body Diagram: Accelerating Car ELO 4.7 Operator Generic Fundamentals 161 Force Equilibrium Knowledge Check Assuming the crate in the diagram is at rest, which of the following statements about forces on the crate is false? A. Any horizontal force applied must be less than the force caused by static friction, or the crate would move. B. The crate is in force equilibrium. C. No horizontal force can be applied, or the crate would not be at rest. D. The normal force applied is equal to the force due to gravity. © Copyright 2016 – Rev 2 ELO 4.7 Operator Generic Fundamentals 162 Newton’s Laws of Motion TLO 5 – Apply Newton's laws of motion to a body at rest. 5.1 Describe Newton’s 1st, 2nd, and 3rd laws of motion. 5.2 Describe Newton’s law of universal gravitation. © Copyright 2016 – Rev 2 TLO 5 Operator Generic Fundamentals 163 Newton’s Laws of Motion ELO 5.1 – Describe Newton's 1st, 2nd, and 3rd laws of motion. • Newton's First Law of Motion - "an object remains at rest (if originally at rest) or moves in a straight line with constant velocity if the net force acting on it is zero.“ • Newton’s Second Law - "the acceleration of a body is proportional to the net (i.e., sum or resultant) force acting on it and in the direction of that net force.“ – Establishes the relationship between force, mass, and acceleration – Mathematically: 𝐹 = 𝑚𝑎 F = force (Newton = 1 kg-m/sec2, or lbf) m = mass (kg or lbm) a = acceleration (m/sec2 or ft/sec2) © Copyright 2016 – Rev 2 ELO 5.1 Operator Generic Fundamentals 164 Newton’s Second Law • 𝐹 = 𝑚𝑎 can be used to calculate an object’s weight at the earth’s surface – F is force, or weight, caused by gravitational acceleration of earth acting on mass, m, of object – Acceleration designated, g, which equals 9.8 m/sec2 or 32.17 ft/sec2 (g - gravitational acceleration constant) • For weight: 𝐹 = 𝑚𝑔 © Copyright 2016 – Rev 2 ELO 5.1 Operator Generic Fundamentals 165 Newton’s Third Law • Newton's Third Law of Motion - “if a body exerts a force on a second body, the second body exerts an equal and opposite force on the first” – “For every action there is an equal and opposite reaction" • Forces always occur in pairs of equal and opposite forces – Example: Downward force exerted on desk by pencil is accompanied by upward force of equal magnitude exerted on pencil by desk • Principle holds for all forces, variable or constant, regardless of their source © Copyright 2016 – Rev 2 ELO 5.1 Operator Generic Fundamentals 166 Newton’s Laws of Motion Knowledge Check State Newton’s Second Law of Motion. A. A particle with a force acting on it has an acceleration proportional to the magnitude of the force and in the direction of that force. B. For every action there is an equal but opposite reaction. C. An object remains at rest (if originally at rest) or moves in a straight line with constant velocity if the net force on it is zero. D. Motion of an object is determined by the size and shape of the object, not the mass of the object. Correct answer is A. © Copyright 2016 – Rev 2 ELO 5.1 Operator Generic Fundamentals 167 Newton’s Universal Law of Gravitation ELO 5.2 – Describe Newton's law of universal gravitation. • Newton’s Universal Law Of Gravitation - "Each and every mass in the universe exerts a mutual, attractive gravitational force on every other mass in the universe. For any two masses, the force is directly proportional to the product of the two masses and is inversely proportional to the square of the distance between them." © Copyright 2016 – Rev 2 ELO 5.2 Operator Generic Fundamentals 168 Universal Law of Gravitation • Newton expressed the universal law of gravitation using equation: 𝑚1 𝑚2 𝐹 = 𝐺 𝑟2 – 𝐹 = force of attraction (Newton = 1kg-m/sec2 or lbf) – 𝐺 = universal constant of gravitation (6.67 x 10-11 m3/kg-sec2 or 3.44 x 10-8 ft3/slug-sec2) – 𝑚1 = mass of first object (kg or lbm) – 𝑚2 = mass of second object (kg or lbm) – r = distance between centers of the two objects (m or ft) © Copyright 2016 – Rev 2 ELO 5.2 Operator Generic Fundamentals 169 Universal Law of Gravitation • Value of g (gravitational acceleration constant), at surface of earth can be determined using universal law of gravitation – Value is known to be 9.8 m/sec2 (or 32.17 ft/sec2) – Can be calculated using: 𝐹 = 𝐺 © Copyright 2016 – Rev 2 ELO 5.2 𝑚1 𝑚2 𝑟2 Operator Generic Fundamentals 170 Calculating the Gravitational Acceleration Constant • First, assume that earth is much larger than an object and that the object resides on surface of earth – Value of r will be equal to radius of earth • Second, understand that force of attraction (F) for the object is equal to object's weight (F) • Setting Equations equal to each other yields: 𝐹 = 𝐺 𝑀𝑒 𝑚1 𝑟2 = 𝑚1 𝑎 – Me = mass of the earth (5.95 x 1024 kg) – m1 = mass of the object – r = radius of the earth (6.367 x 106 m) © Copyright 2016 – Rev 2 ELO 5.2 Operator Generic Fundamentals 171 Calculating the Gravitational Acceleration Constant • Mass (m1) of object cancels, and value of (g) can be determined as follows since a = g by substituting (g) for (a) in previous equation 𝑀𝑒 𝑔=𝐺 2 𝑟 3 𝑚 𝑔 = 6.67 × 10−11 𝑘𝑔– 𝑠𝑒𝑐 2 5.97 × 1024 𝑘𝑔 6.371 × 106 𝑚 2 𝑚 𝑔 = 9.81 𝑠𝑒𝑐 2 © Copyright 2016 – Rev 2 ELO 5.2 Operator Generic Fundamentals 172 Calculating the Gravitational Acceleration Constant • If object is significant distance from earth, can demonstrate that (g) is not a constant value but varies with distance (altitude) from earth – Object at an altitude of 30 km (18.63 mi). Value of (g) is: 𝑟 = 30,000 𝑚 + 6.371 × 106 𝑚 = 6.401 × 106 𝑚 3 𝑚 𝑔 = 6.67 × 10−11 𝑘𝑔– 𝑠𝑒𝑐 2 𝑚 𝑔 = 9.72 𝑠𝑒𝑐 2 © Copyright 2016 – Rev 2 5.97 × 1024 𝑘𝑔 6.401 × 106 𝑚 2 ELO 5.2 Operator Generic Fundamentals 173 Calculating the Gravitational Acceleration Constant • Height of 30 km only changes (g) from 9.8 m/sec2 to 9.7 m/sec2 – Even smaller change for objects closer to earth • (g) is normally considered constant value since most calculations involve objects close to surface of earth © Copyright 2016 – Rev 2 ELO 5.2 Operator Generic Fundamentals 174 Universal Gravitation Knowledge Check Select all statements about the Law of Universal Gravitation that are true. A. The law applies to every mass in the universe. B. The force due to gravity is mutual. C. The force due to gravity is directly proportional to the distance between the bodies. D. The force due to gravity is attractive. Correct answers are A, B, and D. © Copyright 2016 – Rev 2 ELO 5.2 Operator Generic Fundamentals 175 Momentum TLO 6 – Calculate the change in velocity when two objects collide, with respect to the conservation of momentum. 6.1 Describe momentum and conservation of momentum. 6.2 Using the conservation of momentum, calculate the velocity of an object (or objects) following a collision of two objects. © Copyright 2016 – Rev 2 TLO 6 Operator Generic Fundamentals 176 Momentum ELO 6.1 – Describe momentum and conservation of momentum. • A measure of the motion of a moving body • An understanding of momentum and the conservation of momentum are essential tools in solving physics problems © Copyright 2016 – Rev 2 ELO 6.1 Operator Generic Fundamentals 177 Momentum • Momentum is a basic and widely applicable concept of physics – Measure of motion of a moving body – Result of product of body's mass and velocity: 𝑃 = 𝑚𝑣 P = momentum of the object (kg-m/sec or ft-lbm/sec) 𝑚 = mass of the object (kg or lbm) 𝑣 = velocity of the object (m/sec or ft/sec) • Momentum is a vector quantity, results from velocity of object • If different momentum quantities are to be added as vectors, direction of each momentum must be taken into account • To simplify understanding of momentum, only straight line motions considered in this presentation © Copyright 2016 – Rev 2 ELO 6.1 Operator Generic Fundamentals 178 Momentum Example • Calculate momentum of 7 kg bowling ball rolling down lane at 7 m/sec 𝑃 = 𝑚𝑣 𝑝 = 7.0 𝑘𝑔 𝑚 7.0 𝑠 𝑘𝑔– 𝑚 𝑝 = 49 𝑠 © Copyright 2016 – Rev 2 ELO 6.1 Operator Generic Fundamentals 179 Force and Momentum • Direct relationship between force and momentum: – Rate at which momentum changes with time is equal to net force applied to an object – Directly from Newton's second law of motion: 𝐹 = 𝑚𝑎 – Special case of Newton's second law for a constant force which gives rise to a constant acceleration • Acceleration is rate at which velocity changes with time ⇒ We know that, 𝐹 = 𝑚𝑎 and since, 𝑎 = then, 𝐹 = 𝑚 © Copyright 2016 – Rev 2 ∆𝑣 ∆𝑡 ∆𝒗 ∆𝑡 ELO 6.1 Operator Generic Fundamentals 180 Force and Momentum • Acceleration is rate at which velocity changes with time 𝐹 = 𝑚𝑎 ∆𝑣 𝑎= ∆𝑡 ∆𝑣 𝐹=𝑚 ∆𝑡 𝑚𝑣 − 𝑚𝑣𝑜 𝐹= 𝑡 − 𝑡𝑜 © Copyright 2016 – Rev 2 • Substituting P for mv and P0 for mv0 𝑃 − 𝑃𝑜 𝐹= 𝑡 − 𝑡𝑜 Or ∆𝑝 𝐹= ∆𝑡 ELO 6.1 Operator Generic Fundamentals 181 Force and Momentum Example The velocity of a rocket must be increased by 35 m/sec to achieve proper orbit around the earth. If the rocket has a mass of 5,000 kg and it takes 9 seconds to reach orbit, calculate the required thrust (force) to achieve this orbit. • Initial velocity (vo) and final velocity (v) are unknown • Do know change in velocity (v-vo) - 35 m/sec • Equation used can be used to find the solution 𝐹=𝑚 ∆𝑣 ∆𝑡 𝑚 35 3 𝑠 𝐹 = (5.00 × 10 𝑘𝑔) 9𝑠 4 𝑘𝑔– 𝑚 𝐹 = 1.94 × 10 2 𝑠𝑒𝑐 4 𝐹 = 1.94 × 10 𝑁 © Copyright 2016 – Rev 2 ELO 6.1 Operator Generic Fundamentals 182 Force and Momentum Example • Initial velocity (vo) and final velocity (v) are unknown • Do know change in velocity (v-vo) - 35 m/sec • Equation used can be used to find the solution ∆𝑣 𝐹=𝑚 ∆𝑡 𝑚 3 𝑠 𝐹 = (5.00 × 10 𝑘𝑔) 9𝑠 4 𝑘𝑔– 𝑚 𝐹 = 1.94 × 10 2 𝑠𝑒𝑐 35 4 𝐹 = 1.94 × 10 𝑁 © Copyright 2016 – Rev 2 ELO 6.1 Operator Generic Fundamentals 183 Conservation of Momentum • One of most useful properties of momentum is that it is conserved – Means that if no net external force acts upon an object, momentum of object remains constant – Using 𝐹 = ∆𝑃 = 0 ∆𝑝 , ∆𝑡 it can be seen that if force (F) is equal to zero, then • Important when solving problems involving collisions, explosions, etc., where external force is negligible, and momentum before event (collision, explosion) equals momentum following event © Copyright 2016 – Rev 2 ELO 6.1 Operator Generic Fundamentals 184 Conservation of Momentum Examples • Applies when a bullet is fired from a gun – Prior to firing the gun, both the gun and the bullet are at rest (i.e., VG and VB are zero) ⇒ total momentum is zero: 𝑚𝐺 𝑣 𝐺 + 𝑚 𝐵 𝑣 𝐵 = 0 or 𝑚𝐺𝑣𝐺 = −𝑚𝐵𝑣𝐵 • When gun is fired, momentum of recoiling gun is equal and opposite to momentum of bullet – In other words, momentum of bullet (mBvB) is equal to momentum of gun (mGvG), but of opposite direction © Copyright 2016 – Rev 2 ELO 6.1 Operator Generic Fundamentals 185 Law of Conservation of Momentum • Sum of a system's initial momentum is equal to sum of a system's final momentum: Σ 𝑃𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = Σ 𝑃𝑓𝑖𝑛𝑎𝑙 • Where a collision of two objects occurs, conservation of momentum can be stated as: 𝑃1 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑃2 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑃1 𝑓𝑖𝑛𝑎𝑙 + 𝑃2 𝑓𝑖𝑛𝑎𝑙 or (𝑚1𝑣1)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + (𝑚2𝑣2)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑚1𝑣1)𝑓𝑖𝑛𝑎𝑙 + (𝑚2𝑣2)𝑓𝑖𝑛𝑎𝑙 • In a case where two bodies collide and have identical final velocities, this equation applies: 𝑚1𝑣1 + 𝑚2𝑣2 = (𝑚1 + 𝑚2)𝑣𝑓 © Copyright 2016 – Rev 2 ELO 6.1 Operator Generic Fundamentals 186 Law of Conservation of Momentum • Consider two railroad cars rolling on a level, frictionless track • Cars collide, become coupled, and roll together at a final velocity (vf) •Momentum before and after the collision is expressed with (𝑚1𝑣1)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + (𝑚2𝑣2)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑚1𝑣1)𝑓𝑖𝑛𝑎𝑙 + (𝑚2𝑣2)𝑓𝑖𝑛𝑎𝑙 • If initial velocities of two objects (v1 and v2 ) are known, then final velocity (vf ) can be calculated by rearranging the equation 𝑚1 𝑣1 + 𝑚2 𝑣2 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑓𝑖𝑛𝑎𝑙 = 𝑚1 + 𝑚2 Figure: Momentum © Copyright 2016 – Rev 2 ELO 6.1 Operator Generic Fundamentals 187 Law of Conservation of Momentum Example • Railroad cars in figure below have masses of m1 = 1,000 kg and m2 = 1,300 kg. First car (m1) is moving at a velocity of 9 m/sec and second car (m2) is moving at a velocity of 4 m/sec. First car overtakes second car and couples with it. • Calculate final velocity of two cars. Figure: Momentum © Copyright 2016 – Rev 2 ELO 6.1 Operator Generic Fundamentals 188 Law of Conservation of Momentum Solution • Final velocity (vf ) can be easily calculated using the equation below 𝑚1 𝑣1 + 𝑚2 𝑣2 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑓𝑖𝑛𝑎𝑙 = 𝑚1 + 𝑚2 𝑚 𝑚 1,000 𝑘𝑔 9.0 + 1,300 𝑘𝑔 4 𝑠 𝑠 𝑣𝑓𝑖𝑛𝑎𝑙 = 1,000 𝑘𝑔 + 1,300 𝑘𝑔 𝑚 𝑣𝑓𝑖𝑛𝑎𝑙 = 6.17 𝑠𝑒𝑐 Figure: Momentum © Copyright 2016 – Rev 2 ELO 6.1 Operator Generic Fundamentals 189 Momentum Knowledge Check Select all statements about momentum that are true. A. Momentum is conserved in all collisions. B. Momentum is conserved in elastic collisions, but not in inelastic collisions. C. Momentum of an object may be changed by applying force to the object. D. Momentum is a vector quantity. Correct answers are A, C, and D. © Copyright 2016 – Rev 2 ELO 6.1 Operator Generic Fundamentals 190 Elastic and Inelastic Collisions ELO 6.2 – Using the conservation of momentum, calculate the velocity of an object (or objects) following a collision of two objects. • Law of conservation of momentum does not consider whether collision is elastic or inelastic • Elastic collision - both momentum and kinetic energy are conserved – Example - head-on collision of two billiard balls of equal mass • Inelastic collision - momentum is conserved, but system kinetic energy is not conserved – Example - head-on collision of two automobiles where part of initial kinetic energy is lost as metal crumples during the impact © Copyright 2016 – Rev 2 ELO 6.2 Operator Generic Fundamentals 191 Collisions and Velocity • To calculate the velocity of objects after elastic collisions: – Draw the interaction, including known masses and velocities of all objects, before and after the collision – Determine which directions (e.g. up and to the right) are positive in this solution. – Organize the known information, including masses and velocities of all objects, before and after the collision – Determine the unknown that you must solve for – Use the formula given below to determine the solution. – Formula to use if objects are together after collision: 𝑚1𝑣1 + 𝑚2𝑣2 𝑣𝑓 = 𝑚1 + 𝑚2 – Formula to use if objects are separate after collision: (𝑚1𝑣1)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + (𝑚2𝑣2)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑚1𝑣1)𝑓𝑖𝑛𝑎𝑙 + (𝑚2𝑣2)𝑓𝑖𝑛𝑎𝑙 © Copyright 2016 – Rev 2 ELO 6.2 Operator Generic Fundamentals 192 Collisions and Velocity Example • Consider two railroad cars that have masses of m1 = 1,000 kg and m2 = 1,300 kg. The first car (m1) is moving at a velocity of 9 m/sec and the second car (m2) is moving at a velocity of 4 m/sec. The first car overtakes the second car and couples with it. • Calculate the final velocity of the two cars. • The final velocity (vf ) can be calculated as follows: 𝑚1𝑣1 + 𝑚2𝑣2 𝑣𝑓 = 𝑚1 + 𝑚2 𝑚 𝑚 𝑣𝑓 = ( 1,000𝑘𝑔 9 + 1,300 𝑘𝑔 4 )/(1,000𝑘𝑔 + 1,300 𝑘𝑔) sec sec 𝑣𝑓 = 6.17 𝑚/sec © Copyright 2016 – Rev 2 ELO 6.2 Operator Generic Fundamentals 193 Collisions and Velocity Knowledge Check An object with mass of 1.0 x 102 grams is traveling at 50 m/s when it strikes an object with mass of 10 kg, which is at rest. Assuming that the first object is at rest after the elastic collision, what is the velocity and direction of the second object? A. 0.50 m/s, in the same direction that the first object was moving. B. 0.50 m/s, in the opposite direction that the first object was moving. C. 5.0 m/s, in the same direction that the first object was moving. D. 5.0 m/s in the opposite direction that the first object was moving. Correct answer is A. © Copyright 2016 – Rev 2 ELO 6.2 Operator Generic Fundamentals 194 Energy, Work and Power TLO 7 – Describe energy, work, and power in mechanical systems. 7.1 Describe the following terms: a. Energy b. Potential energy c. Kinetic energy d. Work 7.2 Calculate energy and work for a mechanical system. 7.3 State the First Law of Thermodynamics, "Conservation of Energy.“ 7.4 State the mathematical expression for power. 7.5 Calculate power in a mechanical system. © Copyright 2016 – Rev 2 TLO 7 Operator Generic Fundamentals 195 Energy ELO 7.1 – Describe the following terms: energy, potential energy, kinetic energy, and work. • Energy - measure of the ability to do work • Energy determines capacity of a system to perform work and may be stored in various forms • Basic mechanical systems involve concepts of potential and kinetic energy • Both thermal and mechanical energy can be separated into two categories: – Transient Energy - energy in motion; energy being transferred from one place to another – Stored Energy - energy contained within a substance or object © Copyright 2016 – Rev 2 ELO 7.1 Operator Generic Fundamentals 196 Potential Energy • Potential Energy - energy stored in an object because of its position – Example: potential energy of an object above surface of earth in gravitational field • PE also applies to energy due to separation of electrical charge and to energy stored in a spring – Energy due to position of any force field • Example: Energy stored in hydrogen and oxygen as PE to be released upon burning – Relative separation distance associated with elemental forms of hydrogen and oxygen represents PE – Compound water is formed during burning, resulting in a release of PE of separation © Copyright 2016 – Rev 2 ELO 7.1 Operator Generic Fundamentals 197 Potential Energy • When discussing mechanical PE, we look at the position of an object, relative to other objects or to a reference point – Measure of an object's position is its vertical distance above a reference point – Reference point is normally earth's surface, but can be any point © Copyright 2016 – Rev 2 ELO 7.1 Operator Generic Fundamentals 198 Potential Energy • PE of object represents work required to elevate object to that position from reference point 𝑃𝐸 = 𝑤𝑜𝑟𝑘 𝑡𝑜 𝑒𝑙𝑒𝑣𝑎𝑡𝑒 = 𝑤𝑒𝑖𝑔ℎ𝑡 × ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑔𝑧 𝑔𝑐 OR = mgz (if using SI units) • Where: 𝑃𝐸 = potential energy in N-m (Joules) or (ft-lbf) 𝑚 = mass in kg (lbm) 𝑔 = 9.8 m/sec2 (32.17 ft/sec2) 𝑔𝑐 = 32.17 (lbm-ft)/(lbf-sec2) 𝑧 = height above a reference in m (ft) • Note: 𝑔𝑐 is used only when using the English system of measurement © Copyright 2016 – Rev 2 ELO 7.1 Operator Generic Fundamentals 199 Kinetic Energy • Kinetic Energy - energy stored in object due to its motion • KE of an object represents amount of energy required to increase velocity of object from rest (v = 0) to its final velocity • Can also represent work an object it can do as it pushes against something in slowing down (waterwheel or turbine) © Copyright 2016 – Rev 2 ELO 7.1 Operator Generic Fundamentals 200 Kinetic Energy • Because KE is due to motion of object, and motion is measured by velocity, KE for an object can be calculated in terms of its velocity: 2 1 𝑚𝑣 𝐾𝐸 = 𝑚𝑣 2 𝑜𝑟 𝐾𝐸 = 2 2𝑔𝑐 • Where: 𝐾𝐸 = kinetic energy in N-m (Joules) or (ft-lbf) 𝑚 = mass in kg (lbm) 𝑣 = velocity in m/sec (ft/sec) 𝑔𝑐 = (32.17 lbm-ft)/(lbf-sec2) • Note: 𝑔𝑐 is used only when using the English system of measurement © Copyright 2016 – Rev 2 ELO 7.1 Operator Generic Fundamentals 201 Thermal and Internal Energy • Thermal Energy - energy related to temperature (higher temperature, greater molecular movement, and greater energy) – If an object has more thermal energy than adjacent substance, substance at higher temperature will transfer thermal energy (at a molecular level) to cooler substance – Note: This energy is moving from one place to another (energy in motion) referred to as transient energy, more commonly – heat • Internal Energy - energy stored in a substance because due to motion and position of particles of substance ‒ Only stored energy in a solid material ‒ Important to heat transfer and fluid flow © Copyright 2016 – Rev 2 ELO 7.1 Operator Generic Fundamentals 202 Mechanical Energy and Work • Mechanical Energy - energy related to motion or position – Transient mechanical energy - commonly referred to as work – Stored mechanical energy exists in two forms: KE or PE – Both KE and PE can be found in fluids and solid objects • Work - commonly thought of as activity requiring exertion – Physics Definition of Work - work done by a force acting on a moving object if the object has component of motion in the direction of the force – Work is done by a person, a machine, or an object by applying a force and causing something to move © Copyright 2016 – Rev 2 ELO 7.1 Operator Generic Fundamentals 203 Work • If you push on a box (apply a force) and it moves three feet, work has been performed by you to the box, while work has been performed on the box – If you push on the box and it does not move, then work, by our definition, has not been accomplished • Work is defined mathematically as: 𝑊 = 𝐹 × 𝑑 – 𝑊 = work done in J (ft-lbf) – 𝐹 = force applied to the object in N (lbf) – 𝑑 = distance the object is moved with the force applied in m (ft) © Copyright 2016 – Rev 2 ELO 7.1 Operator Generic Fundamentals 204 Energy Definitions Knowledge Check Match the following terms to the appropriate definitions A. Energy stored in an object because of its motion 1. Work B. The energy stored in an object because of its position 2. Thermal energy C. Force applied through a distance 3. Kinetic energy D. That energy related to temperature 4. Potential energy Correct answers are: 1-C, 2-D, 3-A, 4-B. © Copyright 2016 – Rev 2 ELO 7.1 Operator Generic Fundamentals 205 Calculating Energy and Work ELO 7.2 – Calculate energy and work for a mechanical system. • To calculate potential energy, use the following steps: – Determine whether the English or metric system is being used – Determine the height above reference point and mass of the object in question. – Ensure all units are consistent. Convert to appropriate units if necessary. – Apply the appropriate formula 𝑃𝐸 = 𝑚𝑔𝑧 (SI system) 𝑃𝐸 = 𝑚𝑔𝑧 𝑔𝑐 (English system) – Calculate the potential energy. © Copyright 2016 – Rev 2 ELO 7.2 Operator Generic Fundamentals 206 PE Example • What is the potential energy of a 50 kg object suspended 10 meters above the ground? 𝑚 𝑃𝐸 = (50 𝑘𝑔)(9.81 2 )(10.0 𝑚) 𝑠 2 2 𝑃𝐸 = 4.90 × 10 𝑁– 𝑚 or 4.90 × 10 𝐽 © Copyright 2016 – Rev 2 ELO 7.2 Operator Generic Fundamentals 207 Calculating Kinetic Energy • To calculate kinetic energy, use the following steps: – Determine the velocity and mass of the object in question. – Ensure all units are consistent. Convert to appropriate units if necessary. – Use the formula to calculate the kinetic energy. © Copyright 2016 – Rev 2 ELO 7.2 Operator Generic Fundamentals 208 KE Example • What is the kinetic energy of a 10 kg object that has a velocity of 8 m/sec? 2 1 𝑘𝑔 𝑚 𝐾𝐸 = 𝑚𝑣 2 = 10 8.0 2 2 𝑠𝑒𝑐 𝐾𝐸 = 5 𝑘𝑔 64 𝑚2 sec 2 2 𝐾𝐸 = 3.2 × 10 𝐽 © Copyright 2016 – Rev 2 ELO 7.2 Operator Generic Fundamentals 209 Calculating Work • To calculate work use the following steps: – Determine the force applied and the distance through which the force acts on the object – Ensure all units are consistent. Convert to appropriate units if necessary. – Use the formula (𝑊 = 𝐹 × 𝑑) to calculate the work done. © Copyright 2016 – Rev 2 ELO 7.2 Operator Generic Fundamentals 210 Work Examples • An individual pushes a large box for three minutes. During that time, a constant force of 200 N is exerted to the box, but it does not move. How much work has been accomplished? 𝑊 = 𝐹 × 𝑑 𝑊 = 200 𝑁 𝑥 0 𝑚 𝑊 = 0 𝐽 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 • The same box is pushed again. A horizontal force of 200 N is applied to the box, and the box moves five meters horizontally. How much work has been done? 𝑊 = 𝐹𝑥𝑑 𝑊 = 200 𝑁 𝑥 5 𝑚 𝑊 = 1,000 𝐽 © Copyright 2016 – Rev 2 ELO 7.2 Operator Generic Fundamentals 211 Calculating Energy and Work Knowledge Check Match the terms below to the appropriate formula. A. 1 mv2 2 1. Work B. Mgz 2. Momentum C. Fd 3. Kinetic energy D. mv 4. Potential energy Correct answers are: 1-C, 2-D, 3-A, 4-B. © Copyright 2016 – Rev 2 ELO 7.2 Operator Generic Fundamentals 212 Law of Conservation of Energy ELO 7.3 – State the First Law of Thermodynamics, "Conservation of Energy." • The First Law of Thermodynamics - "energy cannot be created or destroyed, only altered in form." – PE - measure of force applied to an object, in order to raise it from the point of origin to some height – Energy expended in raising object is equivalent to PE gained by object because of its height – Example of transfer of energy as well as alteration of type of energy © Copyright 2016 – Rev 2 ELO 7.3 Operator Generic Fundamentals 213 Law of Conservation of Energy • Example: Person throwing a baseball – While ball is in person’s hand, it contains no KE – Force must be applied to ball in order to throw it – Ball leaves hand of person throwing it with a velocity, giving KE equal to work applied by person’s hand • Mathematically transfer of energy can be described by following simplified equation: 𝐸𝑛𝑒𝑟𝑔𝑦𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝐸𝑛𝑒𝑟𝑔𝑦𝑎𝑑𝑑𝑒𝑑 – 𝐸𝑛𝑒𝑟𝑔𝑦𝑟𝑒𝑚𝑜𝑣𝑒𝑑 = 𝐸𝑛𝑒𝑟𝑔𝑦𝑓𝑖𝑛𝑎𝑙 © Copyright 2016 – Rev 2 ELO 7.3 Operator Generic Fundamentals 214 Law of Conservation of Energy • Energy initial - energy initially stored in an object/substance – Energy can exist in various combinations of KE and PE • Energy added - energy added to object/substance: – Heat can be added o Heat is energy gained or lost at a microscopic level – Energy can be added in form of stored energy in any mass added, such as water to a fluid system – Work can be done on a system o Energy gained at macroscopic level © Copyright 2016 – Rev 2 ELO 7.3 Operator Generic Fundamentals 215 Law of Conservation of Energy • Energy removed - energy removed from an object/substance – Heat can be rejected – Work can be done by the system – Energy can be in form of energy stored in any mass removed • Energy final - energy remaining within object/substance after all energy transfers and transformations occur – Energy can exist in various combinations of KE, PE, flow, and internal energy © Copyright 2016 – Rev 2 ELO 7.3 Operator Generic Fundamentals 216 Simplified Energy Balance • Each component of Energy Balance can be broken down: – 𝐸𝑛𝑒𝑟𝑔𝑦𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐾𝐸1 + 𝑃𝐸1 – 𝐸𝑛𝑒𝑟𝑔𝑦𝑎𝑑𝑑𝑒𝑑 = work done on and heat added to system – 𝐸𝑛𝑒𝑟𝑔𝑦𝑟𝑒𝑚𝑜𝑣𝑒𝑑 = work done by and heat removed from system – 𝐸𝑛𝑒𝑟𝑔𝑦𝑓𝑖𝑛𝑎𝑙 = 𝐾𝐸2 + 𝑃𝐸2 • Resulting energy balance: 𝐾𝐸1 + 𝑃𝐸1 + 𝐸𝑎𝑑𝑑𝑒𝑑 − 𝐸𝑟𝑒𝑚𝑜𝑣𝑒𝑑 = 𝐾𝐸2 + 𝑃𝐸2 • Neglecting any heat removed or added to system, can replace Eadded and Eremoved with associated work terms: 𝐾𝐸1 + 𝑃𝐸1 + 𝑊𝑜𝑛 = 𝐾𝐸2 + 𝑃𝐸2 + 𝑊𝑏𝑦 © Copyright 2016 – Rev 2 ELO 7.3 Operator Generic Fundamentals 217 Conservation of Energy Knowledge Check Match the terms below to the appropriate definitions. A. Energy cannot be created or destroyed, only altered in form. 1. Kinetic energy B. 𝐾𝐸1 + 𝑃𝐸1 + 𝐸𝑎𝑑𝑑𝑒𝑑 = 𝐾𝐸2 + 𝑃𝐸2 + 𝐸𝑟𝑒𝑚𝑜𝑣𝑒𝑑 2. Conservation of Energy C. Energy due to position 3. Potential Energy D. Energy due to motion 4. Energy Balance Correct answers are: 1-B, 2-D, 3-C, 4-A. © Copyright 2016 – Rev 2 ELO 7.3 Operator Generic Fundamentals 218 Power ELO 7.4 – State the mathematical expression for power. • Power: a measure of the rate at which energy is used. • Thermal power: term used to refer to the transfer of heat. • Mechanical power: term used to describe when work is being done. • Power - amount of energy used per unit time or rate of doing work – Measured in units of: – Joules/sec (also known as watts) – British Thermal Units (BTUs) – Horsepower – ft-lbf/sec © Copyright 2016 – Rev 2 ELO 7.4 Operator Generic Fundamentals 219 Thermal Power • Thermal Power - measure of thermal energy used per unit time – Rate of heat transfer or heat flow rate – Examples of thermal power units: o British Thermal Units (BTU) o Kilowatts (kW) • Thermal power is calculated by: ℎ𝑒𝑎𝑡 𝑢𝑠𝑒𝑑 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 = 𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 © Copyright 2016 – Rev 2 ELO 7.4 Operator Generic Fundamentals 220 Mechanical Power • Mechanical Power - Mechanical energy used per unit time ⇒ rate at which work is done • Expressed in units of: – Joules/sec (joules/s) or watt (W) in mks – Feet-pounds force per second (ft-lbf/s) or horsepower (hp) in English system • Mechanical power can be calculated using the following mathematical expression: 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑃𝑜𝑤𝑒𝑟 = 𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 © Copyright 2016 – Rev 2 ELO 7.4 Operator Generic Fundamentals 221 Mechanical Power • Because work can be defined as force multiplied by distance, can also use: 𝐹𝑑 𝑃= 𝑡 • Where: 𝑃 = Power (W or ft-lbf/s) 𝐹 = Force (N or lbf) 𝑑 = distance (m or ft) 𝑡 = time (sec) • In English system of measurement, horsepower is commonly used for describing power ratings of equipment such as pumps, motors and engines • One horsepower is equivalent to 550 ft-lbf/s and 745.7 watts © Copyright 2016 – Rev 2 ELO 7.4 Operator Generic Fundamentals 222 Mechanical Power 𝐹𝑑 • In the equation, 𝑃 = , d divided by t (distance per unit time) is 𝑡 same as velocity ⇒ alternate description of power is: 𝐹𝑣 𝑃= 550 • Where: 𝑃 = power (hp) 𝐹 = force (lbf) 𝑣 = velocity (ft/s) © Copyright 2016 – Rev 2 ELO 7.4 Operator Generic Fundamentals 223 Power Knowledge Check Match the terms below to the appropriate definitions. A. Energy used per unit time 1. One Horsepower B. 1 J/s 2. Thermal power C. 550 ft-lbf/s 3. D. Heat used per unit time 4. Power One Watt Correct answers are: 1-D, 2-A, 3-C, 4-B. © Copyright 2016 – Rev 2 ELO 7.4 Operator Generic Fundamentals 224 Calculating Power ELO 7.5 – Calculate power in a mechanical system. • To calculate power use the following steps: – Determine the information known and the information needed – Ensure that units are consistent, convert units as necessary – Choose the appropriate formula based on the information available and the terms being used • Formulae for calculating power: – 𝑃 = – 𝑃 = 𝐹𝑣 550 𝐹𝑑 𝑡 ℎ𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟 – 𝑃𝑜𝑤𝑒𝑟 = © Copyright 2016 – Rev 2 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 ELO 7.5 Operator Generic Fundamentals 225 Power Example 1 A pump provides a flow rate of 10,000 lpm. The pump does 1.5 x 108 Joules of work every 100 minutes. What is the power of the pump? Answer: 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑃𝑜𝑤𝑒𝑟 = 𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 1.5 × 108 𝐽𝑜𝑢𝑙𝑒𝑠 1 𝑚𝑖𝑛 𝑃= 100 𝑚𝑖𝑛 60 𝑠𝑒𝑐 𝑃 = 2.5 × 104 𝑊𝑎𝑡𝑡𝑠 © Copyright 2016 – Rev 2 ELO 7.5 Operator Generic Fundamentals 226 Power Example 2 A boy rolls a ball with a steady force of 1 lbf, giving the ball a constant velocity of 5 ft/s. What is the power used by the boy in rolling the ball? Answer: 𝑃= 𝐹𝑣 550 1 𝑙𝑏𝑓 𝑃= 5 𝑓𝑡 𝑠𝑒𝑐 550 𝑃 = 9.0 × 10−3 ℎ𝑝 © Copyright 2016 – Rev 2 ELO 7.5 Operator Generic Fundamentals 227 Power Example 3 A race car traveling at constant velocity can go one quarter mile (1,320 ft) in 5 seconds. If the motor is generating a force of 1,890 lbf pushing the car, what is the power of the motor in hp? Assume the car is already at full speed at t=0. Answer: 𝐹𝑑 𝑃= 𝑡 1,890 𝑙𝑏𝑓 1,320 𝑓𝑡 𝑃= 5 𝑠𝑒𝑐 1 ℎ𝑝 𝑓𝑡– 𝑙𝑏𝑓 550 𝑠𝑒𝑐 𝑃 = 907 ℎ𝑝 © Copyright 2016 – Rev 2 ELO 7.5 Operator Generic Fundamentals 228 Calculating Power Knowledge Check A man constantly exerts 50.0 lbf to move a crate 50 feet across a level floor. The move requires 10 seconds. What is the power expended by the man pushing the crate? A. 0.450 horsepower B. 2.5 x 103 foot-pounds force C. 4.50 horsepower D. 2.5 x 103 watts Correct answer is A. © Copyright 2016 – Rev 2 ELO 7.5 Operator Generic Fundamentals
