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CHAPTER 10 Force and Force-Related Parameters © 2011 Cengage Learning Engineering. All Rights Reserved. 10-1 Outline In this chapter we will • explain what we mean by force • discuss Newton’s laws in mechanics • explain tendencies of unbalanced mechanical forces • quantify the tendencies of a force acting at a distance (moment), over a distance (work), and over a period of time (impulse) • discuss mechanical properties of materials © 2011 Cengage Learning Engineering. All Rights Reserved. 10-2 What Is Force? Force represents the interaction between 2 objects Objects: car and person Interaction: person pushing the car © 2011 Cengage Learning Engineering. All Rights Reserved. 10-3 What Is Force? Force exerted by your hand on a lawn mower. © 2011 Cengage Learning Engineering. All Rights Reserved. Force exerted by bumper hitch on the trailer. 10-4 Definition of Force All forces are defined by their magnitudes, their directions, and the point of applications Examples to demonstrate the effect of magnitude, direction, and the point of application of a force on the same object © 2011 Cengage Learning Engineering. All Rights Reserved. 10-5 Tendencies of a Force Examples to demonstrate the tendencies of a force © 2011 Cengage Learning Engineering. All Rights Reserved. 10-6 Units of Force Force is a derived parameter force mass accelerati on ma Units of force kg m m 1 N 1 kg 1 2 N 2 s s slug ft ft 1 lb f 1 slug 1 2 lb f lb s2 s 1 lb f 4.448 N © 2011 Cengage Learning Engineering. All Rights Reserved. 10-7 Types of Forces Spring forces and Hooke’s Law F kx © 2011 Cengage Learning Engineering. All Rights Reserved. x 10-8 Example 10.1 – Spring Constant Given: a spring as shown; dead weight is attached to the end of the spring and the corresponding deflection is measured. Find: the spring constant Results of the Experiments © 2011 Cengage Learning Engineering. All Rights Reserved. 10-9 Example 10.1 – Spring Constant Solution: Spring constant k is determined by calculating the slope of a forcedeflection line change in force slope change in deflection 19.6 4.9N k 0.54 N/mm 36 9mm © 2011 Cengage Learning Engineering. All Rights Reserved. 10-10 Types of Forces • Friction forces dry friction Fmax N weight applied force viscous (fluid) friction normal force © 2011 Cengage Learning Engineering. All Rights Reserved. 10-11 friction force Example 10.2 – Friction Force Given: the coefficient of static friction between a book and a desk surface is 0.6. The book weighs 20 N. A horizontal force of 10 N is applied to the book. Find: would the book move? If not, find the friction force and magnitude of the horizontal force F to set the book in motion. weight Solution: applied force Fmax N 0.620 12 newtons normal force since Fmax > 10 N, the book would not move. It would take a 12 N horizontal force to set the book in motion. © 2011 Cengage Learning Engineering. All Rights Reserved. 10-12 friction force Newton’s Laws in Mechanics • First law an object at rest remains at rest if no unbalanced forces acting on it an object in motion with a constant velocity, and if there are no unbalanced forces acting on it, the object will continue to move with the same velocity © 2011 Cengage Learning Engineering. All Rights Reserved. 10-13 Newton’s Laws in Mechanics Second law F ma © 2011 Cengage Learning Engineering. All Rights Reserved. 10-14 Newton’s Laws in Mechanics • Third law For every action there is a reaction The forces of action and reaction have the same magnitude and act along the same line, but they have opposite directions © 2011 Cengage Learning Engineering. All Rights Reserved. 10-15 Newton’s Laws in Mechanics • Universal law of gravitation attraction Two masses attract each other with a force that is equal in magnitude and opposite in direction Gm1m2 F r2 G = 6.673 x10-11 m3/kg.s2 weight of an object is the force that is exerted on the mass of the object by earth’s gravity © 2011 Cengage Learning Engineering. All Rights Reserved. 10-16 Newton’s Laws in Mechanics Weight of an object having a mass m on the earth is GM earthm GM earth W , let g 2 2 Rearth Rearth W mg Mearth = 5.97 x 1024 kg Rearth = 6378 x 103 m g = 9.8 m/s2 © 2011 Cengage Learning Engineering. All Rights Reserved. 10-17 Example 10.3 – Newton’s Laws Given: an exploration vehicle with a mass of 250 kg on Earth Find: the mass of the vehicle on Moon (gMoon = 1.6 m/s2) and on planet Mars (gMars = 3.7 m/s2); weight of vehicle on Moon and Mars Solution: The mass of the vehicle is the same on Earth, Moon and Mars. The weight of the vehicle is determined from W = mg, On Moon : W 250 kg 1.6 m s 400 N On Mars : W 250 kg 3.7 m s 925N On Earth : W 250 kg 9.8 m s 2 2450 N 2 2 © 2011 Cengage Learning Engineering. All Rights Reserved. 10-18 Example 10.4 – Newton’s Laws Given: The space shuttle orbits the Earth at altitudes ranging from 250 km to 965 km; mass of an astronaut is 70 kg on Earth Find: The g value and the weight of the astronaut in orbit Solution: Space shuttle orbiting at an altitude of 250 km above Earth’s surface. GM Earth g R2 3 m 11 24 6.673 10 5 . 97 10 kg 2 kg s 2 9 . 07 m s 2 6378 103 250 103 m W 70 kg 9.07 m s 2 635 N © 2011 Cengage Learning Engineering. All Rights Reserved. 10-19 Example 10.4 – Newton’s Laws Solution (continued): Space shuttle orbiting at an altitude of 965 km above the Earth’s surface. GM Earth g R2 3 11 m 24 6.673 10 5 . 97 10 kg 2 kg s 2 7 . 38 m s 2 6378 103 965 103 m W 70 kg 7.38 m s 2 517 N The near weightless conditions that you see on TV are created by the orbital speed of the shuttle. For example, when the shuttle is orbiting at an altitude of 935 km at a speed of 7744 m/s, it creates a normal acceleration of 8.2 m/s2. It is the difference between g and normal acceleration [(9.8 – 8.2) m/s2] that creates the condition of weightlessness. © 2011 Cengage Learning Engineering. All Rights Reserved. 10-20 Moment, Torque – Force Acting at a Distance Moment is the measure of the tendency of a force acting about an axis or a point hinge Moment is a vector Example – opening and closing a door To open or close a door, we apply a pulling or pushing force on the doorknob. This force will make the door rotate about its hinge. In mechanics, this tendency of force is measured in terms of a moment of a force about an axis or a point. © 2011 Cengage Learning Engineering. All Rights Reserved. 10-21 Moment, Torque – Force Acting at a Distance Calculating moment of a force Line of action C A M A d1 F M B d2 F B F © 2011 Cengage Learning Engineering. All Rights Reserved. MC 0 10-22 Example 10.5 – Moment, Torque 0.07 cos 35o Given: the locations and forces applied as shown Find: the resulting moment of the forces about point O Solution: 0.1 cos 35o M O 50 N 0.05 m 50 N 0.07 cos 35 m 100 N 0.1cos 35 m 13.55 N m © 2011 Cengage Learning Engineering. All Rights Reserved. 10-23 Example 10.6 – Moment, Torque Given: two forces are applied as shown Find: moment about points A, B, C, and D Solution: Note the two forces have equal magnitude and act in opposite directions. M M M M A B C D 100 N 0 100 N 0.1 m 10 N m 100 N 0.1 m 100 N 0 10 N m 100 N 0.25 m 100 N 0.15 m 10 N m 100 N 0.35 m 100 N 0.25 m 10 N m This pair of forces is called a couple. © 2011 Cengage Learning Engineering. All Rights Reserved. 10-24 Internal Force When an object is subjected to an external force, internal forces are created inside the material to hold the material and the components together. An example of internal force. When you try to pull apart the bar, inside the bar material, internal forces develop that keep the bar together as one piece © 2011 Cengage Learning Engineering. All Rights Reserved. 10-25 Reaction Force Reaction forces are developed at the supporting boundaries to keep the object held in position as planned © 2011 Cengage Learning Engineering. All Rights Reserved. 10-26 Reaction Force Examples to demonstrate various support conditions and how they influence the behavior of an object. © 2011 Cengage Learning Engineering. All Rights Reserved. 10-27 Work – Force Acting Over a Distance Mechanical work is defined as the component of the force – that moves the object – times the distance the object moves work = F x d units: N.m, J, lb.ft…. distance, d force © 2011 Cengage Learning Engineering. All Rights Reserved. 10-28 Work Done on a Car d Work done on the car by the force F is equal to W1-2 = (F cos θ)(d) © 2011 Cengage Learning Engineering. All Rights Reserved. 10-29 Mechanical Power Power is the time rate of doing work. It represents how fast you want to do the work. work power time units: J/s (watts), N.m/s lb.ft/s © 2011 Cengage Learning Engineering. All Rights Reserved. 10-30 Example 10.7 – Work Given: a box with weight of 100 N on the ground as shown Find: work required to lift the box 1.5 m above the ground Solution: W 100 N1.5 m 150 N m © 2011 Cengage Learning Engineering. All Rights Reserved. 10-31 Example 10.7 – Power Given: a box with weight of 100 N on the ground as shown. We want to lift the box in 3 seconds Find: power required Solution: work 150 J power 50 watts time 3s © 2011 Cengage Learning Engineering. All Rights Reserved. 10-32 Pressure – Force Acting Over an Area Pressure provides a measure of intensity of a force acting over an area. force pressure area weight contact area Units: Pa (N/m2), kPa, MPa, GPa psi, ksi © 2011 Cengage Learning Engineering. All Rights Reserved. 10-33 Pressure An experiment demonstrating the concept of pressure using bricks (a) and (b) © 2011 Cengage Learning Engineering. All Rights Reserved. 10-34 Common Units of Pressure 1N 1 Pa 2 m 1 lb 1 psi 1 in 2 1 lb 1 psf 2 1 ft © 2011 Cengage Learning Engineering. All Rights Reserved. 1 kPa 1000 Pa 1 ksi 1000 psi 1 ksf 1000 psf 10-35 Pressure – Pascal’s Law Pascal’s law states that for a fluid at rest, pressure at a point is the same in all directions © 2011 Cengage Learning Engineering. All Rights Reserved. 10-36 Pressure – Pascal’s Law For a fluid at rest, the pressure increases with the depth of fluid. P gh P = fluid pressure at a point (Pa) = density of fluid (kg/m3) g = acceleration due to gravity (m/s2) h = height of fluid column (m) © 2011 Cengage Learning Engineering. All Rights Reserved. 10-37 Pressure – Pascal’s Law Buoyancy is the force that a fluid exerts on a submerged object. buoyancy force FB ρVg (N) V volume of the object (m 3 ) density of fluid kg m 3 g accelerati on due to gravity 9.8 m s 2 © 2011 Cengage Learning Engineering. All Rights Reserved. 10-38 Example 10.8 – Pressure Given: a water of height h in a water tower Find: develop a table showing the water pressure in a pipeline located at the base of the water tower relating to h; use the table to determine h if a pressure of 50 psi is desired Solution: We will first develop the equation for pressure P as a function of height h P gh 1 ft 2 ft lb slug P 2 1.94 3 32.2 2 h ft 2 in ft s 144 in 0.4338h ft © 2011 Cengage Learning Engineering. All Rights Reserved. 10-39 Example 10.8 – Pressure Solution (continued): Substituting values of h from 10 ft to 240 ft yields the pressure in the following table: For a pressure of 50 psi, the water level in the tower should be approximately 120 ft © 2011 Cengage Learning Engineering. All Rights Reserved. 10-40 Pressure in Hydraulic Systems In an enclosed fluid system, the pressure is nearly constant throughout the system. The relationship between the forces F1 and F2 could be established. P1 F1 A1 since P1 P2 , P2 P1 P2 F2 A2 F1 F2 A1 A2 © 2011 Cengage Learning Engineering. All Rights Reserved. F2 10-41 A2 F1 A1 Pressure in Hydraulic System Push to raise load To raise the load – push the arm of the hand pump down – fluid will be pushed into the load cylinder, which in turn creates a pressure that is transmitted to the load piston, and consequently the load is raised. To lower the load – open the release valve. The fluid will return to the reservoir and the load is lowered. Open to lower load reservoir © 2011 Cengage Learning Engineering. All Rights Reserved. 10-42 Pressure in Hydraulic System To raise the load – move the control lever UP – fluid will be pushed into the load cylinder, which in turn creates a pressure that is transmitted to the load piston, and consequently the load is raised. Gear or rotary pump © 2011 Cengage Learning Engineering. All Rights Reserved. To lower the load – push the control lever down. The fluid will return to the reservoir and the load is lowered. 10-43 Example 10.12 – Hydraulic System Given: the hydraulic system shown, g = 9.81 m/s2 Find: the load that can be lifted by the hydraulic system. Solution: F1 m1 g 100 kg 9.81 m/s 2 981 N A2 0.15 m 981 N 8829 N F2 F1 2 A1 0.05 m 2 F2 8829 N m2 kg 9.81 m/s 2 © 2011 Cengage Learning Engineering. All Rights Reserved. m2 900 kg 10-44 Example 10.12 – Hydraulic System Solution (continued): Alternate approach A R2 m1 g F2 2 F1 m2 g 2 A1 R1 2 2 2 R2 15 cm 100 kg 900 kg m2 m1 2 2 R1 5 cm The second approach is preferred. We can see clearly the relationship between m1 and m2. © 2011 Cengage Learning Engineering. All Rights Reserved. 10-45 Stress Stress provides a measure of the intensity of a force acting over an area. force stress area units: Pa (N/m2), kPa, MPa, GPa psi, ksi force component normal to area normal stress area force component parallel to area shear stress area Normal stress is often called pressure © 2011 Cengage Learning Engineering. All Rights Reserved. 10-46 Strain • Strain is a description of the deformation of a body in terms of the relative displacement of the particles in the body. That is, how much the body stretches. • Deformation, or strain, is typically caused by the application of an external force. Engineering Fundamentals, By Saeed Moaveni, Fourth Edition, Copyrighted 2011 10-47 Mechanical Properties of Materials • Modulus of elasticity a measure of how easily material stretches • Modulus of Rigidity (also called shear modulus) a measure of how easily material twists • Tensile strength • Compressive strength © 2011 Cengage Learning Engineering. All Rights Reserved. 10-48 Mechanical Properties of Materials Modulus of elasticity, E – measures how easily a material will stretch when pulled (subject to a tensile force) or how well a material will shorten when pushed (subject to compressive force). When subjected to the same force, which piece of material will stretch more? Esteel > Ealuminum > Erubber L L © 2011 Cengage Learning Engineering. All Rights Reserved. 10-49 L Tensile Test of Metal Specimen Tensile test set up Original specimen © 2011 Cengage Learning Engineering. All Rights Reserved. Final specimen 10-50 Stress-Strain Diagram σe = elastic stress (σY)u = upper-yield stress (σY)l = lower-yield stress σY = ultimate stress σf = fracture stress Stress-strain diagram for a mild steel sample © 2011 Cengage Learning Engineering. All Rights Reserved. 10-51 Modulus of Elasticity Hooke' s law F x E E A L solve for E yields, FL E Ax © 2011 Cengage Learning Engineering. All Rights Reserved. x 10-52 Other Mechanical Properties of Materials • Modulus of rigidity or shear modulus measures how easily a material can be twisted or sheared is used to select materials for shafts or rods subjected to twisting torques value is determined using a torsional test machine © 2011 Cengage Learning Engineering. All Rights Reserved. 10-53 Other Mechanical Properties of Materials • Tensile strength or ultimate strength maximum tensile force per unit of original cross-sectional area of specimen • Compressive strength maximum compressive force per unit of cross-sectional area of specimen © 2011 Cengage Learning Engineering. All Rights Reserved. 10-54 Modulus of Elasticity and Shear Modulus of Selected Materials © 2011 Cengage Learning Engineering. All Rights Reserved. 10-55 Strength of Selected Materials © 2011 Cengage Learning Engineering. All Rights Reserved. 10-56 Strength of Selected Materials © 2011 Cengage Learning Engineering. All Rights Reserved. 10-57 Example 10.13 – Material Properties Given: a structural member and a load of 4000 N distributed uniformly over the crosssectional area of the member Find: select a material to carry the load safely Solution: We will consider the strength of the material as the design factor in this example. 4000 N 16 MPa 0.05 m0.005 m Aluminum alloy or structural steel material with yield strength of 50 MPa and 200 MPa, respectively, could carry the load safely. © 2011 Cengage Learning Engineering. All Rights Reserved. 10-58 Summary • You should have a good understanding of force and its common units. • You should know the types of forces. • You should understand the tendency of an unbalanced force, to rotate or to translate. • You should know that the application of forces can lengthen, shorten, bend, and twist objects. • You should be familiar with Newton’s laws in mechanics. © 2011 Cengage Learning Engineering. All Rights Reserved. 10-59 Summary • You should understand pressure and should know Pascal’s law for static fluids relationship between fluid pressure and depth of fluid how hydraulic systems work fluid properties: viscosity, bulk modulus of compressibility • You should know the mechanical properties of materials: modulus of elasticity, shear modulus, tensile and compressive strength © 2011 Cengage Learning Engineering. All Rights Reserved. 10-60 Summary • You should know how a material behaves under applied forces • You should understand the effect of an applied force on an object stress: intensity of a force acting within the material Internal forces: forces created inside the material to hold the material and the components together moment: force acting at a distance work: force acting over a distance linear impulse: force acting over a period of time © 2011 Cengage Learning Engineering. All Rights Reserved. 10-61