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Newton’s Laws of Motion

This homework assignment focuses on Newton's Third Law of Motion and applications of friction in various scenarios. It includes problems and examples to reinforce understanding of these concepts. Due date: Monday, September 30th.

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Newton’s Laws of Motion

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  1. Newton’s Laws of Motion Homework 4 Due Monday September 30 Chap 4:Pb. 32, Pb. 35, Pb.48, Pb 49, Pb 52,Pb 54

  2. Newton’s Third Law of Motion Any time a forceis exerted on an object, that force is caused by another object. Newton’s third law: Whenever one object exerts a forceon a second object, the second exerts an equal force in the opposite directionon the first.

  3. Newton’s Third Law:interaction pairs Newton’s third Law:for every action, there is an equalandoppositereaction. We call these two equal and opposite forces interaction partners. Note: they operate on different objects Note: they involve the same interaction: eg. Gravity, contact…

  4. Newton’s Third Law of Motion A key to the correct application of the third law is that the forces are exerted on different objects.Make sure you don’t use them as if they were acting on the same object.

  5. Newton’s Third Law of Motion Rocket propulsioncan also be explained using Newton’s third law: hot gases from combustion spew out of the tail of the rocket at high speeds. Thereaction force is what propels the rocket. Note that the rocket does not need anything to “push” against.

  6. Newton’s Third Law of Motion Conceptual Example 4-5: Third law clarification. Michelangelo’s assistant has been assigned the task of moving a block of marble using a sled. He says to his boss, “When I exert a forward force on the sled, the sled exerts an equal and opposite force backward. So how can I ever start it moving? No matter how hard I pull, the backward reaction force always equals my forward force, so the net force must be zero. I’ll never be able to move this load.” Is he correct?

  7. Problem Solving Strategy 1) Draw a FBD for the object (or objects). Labeling all forces with simple vector symbols. 2) Pick a coordinate system 3) Add up all of the forces in the x-direction. 4) Decide whether or not this sum is equal to zero or ma 5) Add up all of the forces in the y-direction 6) Decide whether or not this sum is equal to zero or ma 7) Solve these two simultaneous equations for your unknown(s).

  8. Problem 27 Problem 27: A box weighing 77.0 N rests on a table. A rope tied to the box runs vertically upward over a pulley and a weight is hung from the other end. Determine the force that the table exerts on the box if the weight hanging on the other side of the pulley weighs (a) 30.0 N, (b) 60.0 N, and (c) 90.0 N.

  9. Problem 51 Problem 51:The Figure below shows a block of mass mA on a smooth horizontal surface, connected by a thin cord that passes over a pulley to a second block of mass mB which hangs vertically. (a) Draw a free-body diagram for each block, showing the force of gravity on each, the force (tension) exerted by the cord, and any normal force. (b) Apply Newton’s second law to find formulas for the acceleration of the system and for the tension in the cord. Ignore friction and the masses of the pulley and cord.

  10. Chapter 5:Using Newton’s Laws: Friction, Circular Motion, Drag Forces

  11. Applications of Newton’s Laws Involving Friction Friction is always present when two solid rough surfaces slide along each other.

  12. Applications of Newton’s Laws Involving Friction Sliding friction is called kinetic friction. Approximation of the frictional force: Ffr = μkFN Here, FN is the normal force, and μk is the coefficient of kinetic friction,which is different for each pair of surfaces. This equation is not a vector equation.

  13. Applications of Newton’s Laws Involving Friction Static friction applies when two surfaces are at rest with respect to each other (such as a book sitting on a table). The static frictional force is as big as it needs to be to prevent slipping, up to a maximum value. Ffr ≤ μsFN Usually it is easier to keep an object sliding than it is to get it started. and μk< μs

  14. Question Its more difficult to start moving a heavy carton from rest than it is to keep pushing it with constant velocity, because A) The normal force is greater when the carton is at rest. B) μs<μk . C) Initially, the normal force is not perpendicular to the applied force. D) μk< μs . E) μs=μk

  15. Applications of Newton’s Laws Involving Friction Example 5-3: Pulling against friction. A 10.0-kg box is pulled along a horizontal surface by a force of 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction is 0.30. Calculate the acceleration.

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