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Physics 111 Lecture 04 Force and Motion I: The Laws of Motion SJ 8th Ed.: Ch. 5.1 – 5.7 • • • • • • • • • 5.1 5.2 5.3 5.4 5.5 5.6 5.7 Dynamics - Some history Force Causes Acceleration Newton’s First Law: zero net force Mass Newton’s Second Law Free Body Diagrams Gravitation Newton’s Third Law Application to Sample Problems The Concept of a Force Newton’s First Law and Inertial Frames Mass Newton’s Second Law Gravitational Force and Weight Newton’s Third Law Using Newton’s Second Law Copyright R. Janow – Spring 2012 Dynamics - Newton’s Laws of Motion Kinematics described motion only – no real Physics. Why does a particle have a certain acceleration? New concepts (in 17th century): • Forces - pushes or pulls - cause acceleration • Inertia (mass) measures how much matter is being accelerated – resistance to acceleration Newton’s 3 Laws of Motion: • Codified kinematics work by Galileo and other early experimenters • Introduced mathematics (calculus) as the language of Physics • Allowed detailed, quantitative prediction and control (engineering). • Ushered in the “Enlightenment” & “Clockwork Universe”. • Are accurate enough (with gravity) to predict all common motions plus those of celestial bodies. • Fail only for v ~ c and quantum scale (very small). Sir Isaac Newton 1642 – 1727 • Formulated basic laws of mechanics • Invented calculus in parallel with Liebnitz • Discovered Law of Universal Gravitation • Many discoveries dealing with light and optics • Ran the Royal Mint for many years during old age • Many rivalries & conflicts, few friends, no spouse or children, prototype “geek” Copyright R. Janow – Spring 2012 Force: causes any change in the velocities of particles Newton’s A force is that which definition: causes an acceleration Contact forces involve “physical contact” between objects F = - kx Field forces act through empty space without physical contact Action at a distance through intervening space? How? Given atomic physics, Is there any such thing as a real contact force? The four fundamental forces of nature are: Gravitation, Electromagnetic, Nuclear, and Weak force Copyright R. Janow – Spring 2012 Force: what causes any change in the velocities of particles Units: 1 Newton = force that causes 1 kg to accelerate at 1 m/s2 1 Pound = force that causes 1 slug to accelerate at 1 ft/s2 Forces are VECTORS Operate with vector rules F Replace a force acting at a point by its components at the same point. Fy ĵ Fx î F1 Superposition: A set of forces at a point have the same effect that their vector resultant force would F3 F2 Fnet Notation: Fnet Fi i Definition: A body is in EQUILIBRIUM if the net force applied to it equals zero Copyright R. Janow – Spring 2012 Relative motion - Inertial Reference Frames in 1 Dimension A frame of reference amounts to selecting a coordinate system. • Describe point P in both frames using x1, v1, a1 and x2, v2, a2. • Origins coincide at t = 0 • Constant v12 = relative velocity of origin of 1 in frame 2 • x12 is the distance between origins at some time t. 2 y1 y Transform the coordinates: x2 (t) x1(t) x12(t) v12 Transform the velocities: dx 2 dt dx1 dt dx12 v2 v1 v12 dt a2 dt dv1 dt Then a12 = 0 and dv12 dt a1 a2 x2 P x2 Find the accelerations: dv2 x1 x12 dv1 dt a1 dv12 dt a12 0 The acceleration of a moving object is the same for a pair of inertial frames. Example: An accelerating car viewed from a train and the ground, with the train itself moving at constant velocity Inertial frames can not be rotating or accelerating relative to one another or to the Copyright R. Janow – Spring 2012 fixed stars. Non-inertial frames fictitious forces. x1 Newton’s First Law (1686) Don’t moving objects come to a stop if you stop pushing? • Stopping implies negative acceleration, due to friction or other forces opposing motion. • What effect does inertia have on a curve on an icy road? “The Law of Inertia”: A body’s velocity is constant (i.e., a = 0) if the net force acting on it equals zero Alternate statement: A body remains in uniform motion along a straight line at constant speed (or remains at rest) unless it is acted on by a net external force. Above assume an “inertial reference frame”: • Equations of physics look simplest in inertial systems. • Non-inertial frames (e.g., rotating) require fictitious forces & accelerations in physics equations (e.g., centrifugal and Coriolis forces). First Law (our text): An object that does not interact with other objects (no net force, isolated) can be put into a reference frame in which the object has zero acceleration (i.e., it’s motion can be transformed to an inertial frame if it is not in one already). Copyright R. Janow – Spring 2012 Mass: the Measure of Inertia Apply the same force to different objects. Different accelerations result. Why? Example: Apply same force to baseball, bowling ball, automobile, RR train Mass measures inertia: • the amount of “matter” in a body (i.e., how many atoms of each type) • resistance to changes in velocity (i.e., acceleration) when a force acts For a given force applied to m1 and m2: m1 m2 a2 a1 Mass is a scalar: A mass and resulting acceleration on it are inversely proportional Attach m 1 to m2 The result behaveswith m m1 m2 Mass is intrinsic to an object: • it doesn’t depend on the environment or state of motion (for v << c), or time. Don’t confuse mass with weight (a force): W mg If m 2 kg then Wearth 20 N. but Wmoon 3.3 N. The same “inertial mass” value also measures “gravitational mass” – - a particle’s effect in producing gravitational pull on other Copyright R. Janowmasses – Spring 2012 Newton’s Second Law SIMPLE PROPORTIONALITY WHEN IN AN INERTIAL FRAME Fnet Summarizing: Fi ma i Vector sum of forces acting ON a particle Inertia of particle Acceleration resulting from Fnet DIRECTION OF ACCELERATION AND NET FORCE ARE THE SAME Cartesian component equations: Fnet,x Fix ma x Fnet,y i Other ways to Write 2nd Law: Fnet Units for Force: [F] SYSTEM SI CGS British 1 dyne = 10-5 N Fiy d2r Fi m dt 2 i [m] [a] FORCE Newton (N) Dyne Pound (lb) 1 gm = 10-3 kg ma y Fnet,z Fiz ma z i i dp Fnet where p mv dt M L / T2 MASS Kg gm slug ACCELERATION m/s2 cm/s2 ft/s2 1 lb = 4.45 kg 1 kg WEIGHS 2.2 lb 1 slug = 14. 59 kg Copyright R. Janow – Spring 2012 Tug of war 4-1: Three students can all pull on the ring (see sketch) with identical forces of magnitude F, but in different directions with respect to the +x axis. One of them pulls along the +x axis with force F1 as shown. What should the other two angles be to minimize the magnitude of the ring’s acceleration? a) q2 = 0, q3 = 0 b) q2 = 180, q3 = -180 c) q2 = 60, q3 = -60 d) q2 = 120, q3 = -120 e) q2 = 150, q3 = -150 F2 F3 q3 q2 F1 x 4-2: What should the other two angles be to maximize the magnitude of the ring’s acceleration? a) q2 = 0, q3 = 0 b) q2 = 180, q3 = -180 c) q2 = 60, q3 = -60 d) q2 = 120, q3 = -120 e) q2 = 150, q3 = -150 Copyright R. Janow – Spring 2012 Example: A hockey puck whose mass is 0.30 kg is sliding on a frictionless ice surface. Two forces act horizontally on it as shown in the sketch. Find the magnitude and direction of the puck’s acceleration. m = 0.3 kg Apply 2nd Law to x and y directions Fnet,x F1x F2x ma x F1 cos(20) F2 cos(60) Fnet,y F1y F2y ma y F1 sin(20) F2 sin(60) Evaluate: Fnet,x 8.7 N Fnet,y 5.2 N ax ay Fnet,x m Fnet,y m 8.7 29 m / s2 0.3 5.2 17 m / s2 0.3 FBD Convert to polar coordinates: a [a2x a2y ]1 / 2 34 m/s2 q tan-1 (ay / ax ) 31o The net force and acceleration vectors have the same direction A unit vector in that direction is: ay ax a 29 17 â î ĵ î ĵ 0.85 î 0.50 ĵ |a| |a| |a| 34 34 Copyright R. Janow – Spring 2012 Measuring Force and Mass On frictionless surface (e.g., air track): Apply enough horizontal force F0 to give the standard mass m0 the standard acceleration a0 m0 = 1 kg a0 = 1 m/s2 The (standard) force unit thereby defined = 1N. F0 m0a0 Note: x-components only above, F & a are in same direction Measuring another mass: • Apply standard force, record resulting acceleration a F0 / m m F0 / a What about forces in the y – direction, left out above? Copyright R. Janow – Spring 2012 Gravitational Force, Weight, “Normal” Force Mass in free fall on Earth’s surface accelerates at g: FBD m g has the same direction (toward center of Earth) and magnitude for all masses (in a volume large compared to a human). Fg mg ĵ g = 9.8 m/s2 = 32.2 ft/s2 at the Earth’s surface Fg " the weight" • “Action at a distance” • The weight is independent of how a mass is moving (perhaps other forces also act). Mass in contact with a “horizontal” surface (table, air track, …) may be in equilibrium for y: " Equilibrium" for y Fy 0 a y FBD ay = 0 N = Fg (m does not accelerate) N " normal force" perpendicular to surface m Fg mg ĵ N N is a contact force that adjusts to Fg Fg pushes on the surface – - the surface pushes back with N If ay not = 0, N does not equal Fg Copyright R. Janow – Spring 2012 Newton’s Third Law Bodies interact by pushing or pulling on each other 3rd Law (antique version): Each action has an equal and opposite reaction More modern version: When two bodies interact the forces that each exerts on the other are always equal in magnitude and opposite in direction Example: gravity acting at a distance F12 force on object 1 due to object 2 F21 force on object 2 due to object 1 F12 F21 Example: box on a level surface m Fg mg ĵ N Fs Fe • • • • • • • Fg is the pull of Earth on the box (weight) The 3rd law reaction is Fe - the box’s pull on the Earth N is the surface’s push on the box Fs is the box’s push on the surface ONLY forces on the box (Fg & N) affect it’s motion Fg & N are NOT a 3rd law pair (Fsb & N are a pair) Why then does Fg = N ??? If you ever find a force w/o the 3rd law reaction, Janow – Spring 2012 you can build a perpetualCopyright motionR.machine Free Body Diagrams (FBDs) Drawing the FBDs is the most important step in analyzing motion. • DRAW FBDs FIRST – before you start writing down equations. • Pictorial sketches are not the same as FBDs. • Model bodies in your system as point particles. There may be several. Sometimes you can treat the system as one object. • Choose coordinates. • Include in FBDs only forces that act ON your system. • Exclude forces exerted BY bodies in your system on other bodies. • Neglect internal forces. When you break up a system for analysis, INCLUDE formerly internal forces that become external. • Don’t forget action-at-a-distance forces (fields) such a gravity M N m Fg N m Fg N Fg Mg ĵ Is this a FBD? m ' Fg mg ĵ Copyright R. Janow – Spring 2012 FBDs: show only forces on bodies Friction Frictionless m N F Fg No friction parallel to surface ax F / m Sliding or static friction f (later) m f N F Fg Friction is a resistive force parallel to surface ax (F f ) / m f N Contact force - always opposed to motion Cords Tension only, no compression Pulling creates tension T – the force transmitted by the cord T is the same everywhere in a zero-mass, unstretchable cord Support FBD for body FBD for cord FBD for support T’ = T T T F 3rd law pair m rd law pair 3 m T’ Fg T Fg T T' F Equilibrium Copyright R. Janow – Spring 2012 Newton’s Laws - Summary Newton’s First Law A body’s velocity is constant ( a 0 ) if the net external force on it is zero - Motion is along a straight line - Find and use an inertial frame of reference Newton’s Second Law Fnet F net force vector m a i i In Cartesian components: Fx max Fy ma y Fz maz Newton’s Third Law If body A exerts a foce on body B, then body B exerts and equal and Opposite force on body A. Copyright R. Janow – Spring 2012 Method for solving Newton’s Second Law problems • Draw or sketch system. Adopt coordinates. Name the variables, • Draw free body diagrams. Show forces acting on particles. Include gravity (weights), contact forces, normal forces, friction. • Apply Second Law to each part Fnet Fi ma • Make sure there are enough (N) equations; Extra conditions connecting unknowns (constraint equations) may be applicable • Simplify and solve the set of (simultaneous) equations. • Interpret resulting formulas. Do they make intuitive sense? Are the units correct? Refer back to the sketches and original problem • Calculate numerical results, and sanity check anwers (e.g., right order of magnitude?) Systems with several components may have several unknowns…. …and need an equal number of independent equations Copyright R. Janow – Spring 2012 Example: Traffic Light in Equilibrium Conceptualize: cables are massless and don’t break no motion Categorize: equilibrium problem – accelerations = 0 Model as particles in equilibrium 2 FBDs FBD of Knot FBD of Light T3 Fg = 122 N FBD of light yields T3 = Fg ( alight = 0) FBD of knot yields: Fx Fy 0 T2 cos(53o ) T1 cos(37o ) 0 T2 sin(53o ) T1 sin(37o ) T3 Solve upper for T2: Fg T2 cos( 37o ) T1 o cos( 53 ) o 1.33T1 T1 [sin( 37 ) 1.33 sin(53o )] 1.22 N T1 73.4 N T 97.4 N 2 R. Janow – Spring 2012 Copyright Example: Particle Motion Under a Net Force y Mass m is moving on a frictionless horizontal surface, acted on by an external force of magnitude F, making an angle q with the x-axis. F m Find expressions for the acceleration along the horizontal and vertical directions Along x: Fx F cos(q) max ax N q x F cos(q) m W W = weight Along y: Is ay zero, or does particle accelerate upward? Fy F sin(q) N W may Set ay = 0. Crossover to ay > 0 when N also = 0, i.e, when 0 F sin(q) N W Fsin(q) W When Fy is > the weight, the particle accelerates upward ay When Fy is < the weight, the particle does not accelerate ay 0 F sin(q) W m N0 N W F sin(q) Copyright R. Janow – Spring 2012 Bootstraps 4-3: The man and the platform weigh a total of 500 N. He pulls upward on the rope with force F. What force would he need to exert in order to accelerate upward with a = 0.1 g? Is this possible? a) F = 50 N b) F = 1000 N c) F = 550 N d) F = 500 N e) He cannot lift himself by his own bootstraps at all. Copyright R. Janow – Spring 2012 Example: Sliding and Hanging Blocks Block S, mass M is sliding on a frictionless horizontal surface. Block H, mass m hangs from a massless, unstretchable cord wrapped over a massless pulley. Find expressions for the accelerations of the blocks and the tension in the cord. N T No friction, No mass W Mg T’ Apply 2nd Law to blocks S & H separately for x & y FBD for Block S FBD for Block H N T’ T a W = Mg W’ = mg a’ choose positive down Fxs T Ma Fys N W Mays 0 Fxh maxh 0 Fyh W' T' ma' Constraints: T’ = T, a’ = a Find formula for T Mm T Ma g mM W’ = mg Eliminate T,T’,a’ W' Ma ma mg ma Ma could also use m a g system approach m M To find this Copyright R. Janow – Spring 2012 Example: Block Sliding on a Ramp [“Inclined Plane”] Mass m is accelerating along a frictionless inclined surface as shown, making an angle q with the horizontal. m N a W = mg q Find expressions for the acceleration and the normal force Forces acting ON the block are N and W • N is normal to the surface • W is vertical as usual Choose x-y axes aligned to ramp, for which: Assume a y 0, a x a (positive down & right) Wy W cos(q) W x Wsin(q) Apply 2nd Law to x and y Fy N Wy N mgcos(q) 0 N mgcos(q) Fx Wx max mg sin(q) ax a gsin(q) Does not depend on m Why does the block accelerate? Do you expect N to equal W? y FBD N Wx 90q Wy Check: What happens as q 90o as q 0o q q W Copyright R. Janow – Spring 2012 x Example: “Atwood’s Machine” with Massless Pulley No forces or motion along x Both masses have the same acceleration a (constraint). The tension T in the unstretchable cord is the same on both sides of the massless pulley (another constraint). Find expressions for the acceleration and the tension In the cord. Find the force Wtot supporting the pulley FBD for m1 a T FBD for m2 Fy1 T m1g m1a m1g T a Fy2 m2g T m2a m2g Add the equations T m1g m2g T m2a m1a a = 0 for m1 = m2 Wtot T T a a m2 m1 m2 m1 a g a is clockwise for m2 > m1 a = g for m1 = 0 or a = - g for m2 = 0 Subtract the equations T m1g m2g T m2a m1a FBD for pulley Wtot Wtot 2T 0 T 2m1m2 m2 m1 Wtot T = 0 for m1 or m2 equals zero. T = m1g for m2 = m1 g 4m1m2 m2 m1 g Wtot = 2m1g if m2 = m1 Otherwise not so Copyright R. Janow – Spring 2012 Example: Your Weight in an Elevator FBD for A passenger whose mass m = 72.2 kg is standing on a passenger platform scale in an elevator. What weight does the scale read for him as the elevator (and + himself) accelerates up and down? • Fg = mg is the real weight, which doesn’t change • The building is an inertial frame, as is the elevator when traveling at constant speed. • When the elevator and passenger are accelerating their non-inertial frame fictitious forces N is the scale reading = apparent weight • Apply Second Law in the building’s reference frame Fy N mg ma No forces or motion along x N Fg mg N m (a g) Interpretations: N N N N = > < = mg for a mg for a mg for a 0 for a = = 0 > 0 < 0 -g Normal weight for elevator at rest or moving with constant v Increased apparent weight for elevator accelerating up Decreased apparent weight for elevator accelerating down Free fall - weightless mg = force exerted by gravity velocity has no effect on N – apparent weight Fictitious force appears in non-inertial frame Copyright R. Janow – Spring 2012