Chapter 4 Dynamics: Newton s Laws of Motion

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1 Chapter 4 Dynamics: Newton s Laws of Motion

2 Force Newton s First Law of Motion Mass Newton s Second Law of Motion Newton s Third Law of Motion Weight the Force of Gravity; and the Normal Force Applications Involving Friction, Inclines

3 Recalling Last Lecture

4 Tension in a Flexible Cord When a cord, string or rope pulls on an object, it is said to be under tension, and the force it exerts is called a tension force. The figure shows you pulling a SYSTEM consisting of two boxes connected by a cord. The force you apply on the system is. Note: If you consider each individual component of the system, the force is actually only applied on box A and NOT on box B. The force will be transmitted to box B via the tension in the cord.

5 Tension in a Flexible Cord In the figures: Fig. (b) is a diagram of all forces applied on box A. Note that box A exerts a force on the cord connecting the two boxes. According to Newton s third law, the cord responds with a force (tension force),, of same magnitude but opposite direction. Fig. (c) is a diagram of forces applied on B. Now the cord pulls box B. If we neglect the mass of the cord, this pull will correspond to a force equal to the tension found on the other end of the cord (connected to A). Note that this is NOT true if a mass is assigned to the cord.

6 Tension in a Flexible Cord Problem 4.78 (textbook: (a) What minimum force F is needed to lift the piano (mass M) using the pulley apparatus shown in Fig. 4 60? (b) Determine the tension in each section of rope: F F, F, and, T1, T2 T3 F T4

7 Problem 4.78 (textbook): (a) To find the minimum force, assume that the piano is moving with a constant velocity. Since the piano is not accelerating, F = Mg T 4 For the lower pulley, since the tension in a rope is the same throughout, and since the pulley is not accelerating, it is seen that F + F = 2 F = Mg F = F = Mg 2 T1 T 2 T1 T1 T 2 Also, since, then F = F T 2 F = Mg 2 F r T2 F r T1 Upper Pulley F r T3 (b) Draw a free-body diagram for the upper pulley. From that diagram, we see that F = F + F + F = T 3 T1 T 2 3Mg 2 F r T4 Lower Pulley F r T2 F r T1 F r F = F = Mg 2 F = 3Mg 2 F = Mg T1 T 2 T 3 T 4

8 Friction and Inclined Planes On a microscopic scale, most surfaces are rough. When we try to SLIDE an object across a table, the roughness of both surfaces (object s and table s surfaces) tends to oppose the object s motion. This effect is called sliding friction, or kinetic friction. The opposition to the body s motion is due to a force called friction force.

9 Friction and Inclined Planes The friction force acts in the opposite direction to the object s motion. In the figure, the forces applied on the box are : the applied force; : the normal force; : the gravitational force; : the friction force. The reason for the friction force is not very well understood, though it has been very well modeled. Experiments have shown that the magnitude of the friction force is approximately proportional to the magnitude of the normal force between two surfaces, and does not depend on the total surface area of contact between the two surfaces. (4.5)

10 Friction and Inclined Planes (4.5) The term is called coefficient of kinetic friction, and its value depends on the nature of the two contacting surfaces. Sliding friction (or kinetic friction) is not present when the object is not SLIDING. However, you might have experienced the fact that it takes different (magnitude) forces to slide a same object across different surfaces. This is due to the so called static friction. The force of static friction is parallel to the two contacting surfaces. Note: Let me stress again there is no sliding force if the object moves (or is at rest) without sliding.

11 Friction and Inclined Planes The force of static friction varies with the applied force on the object The more force you apply, the bigger is the force of static friction. There will be a critical applied force that will ultimately set the object in motion. This point will correspond to the maximum force of static friction: (4.6) The tem is called the coefficient of static friction. Note that, at any point, it is true that force of static friction will be such that:

12 Friction and Inclined Planes The magnitude of the friction forces depends on the nature of the two surfaces in contact.

13 Tension in a Flexible Cord Problem 4-36 (textbook: If the coefficient of kinetic friction between a 35-kg crate and the floor is 0.30, what horizontal force is required to move the crate at a steady speed across the floor? What horizontal force is required if is zero?

14 Problem 4.36 (textbook): A free-body diagram for the crate is shown. The crate does not accelerate vertically, and so F = mg N Also, the crate does not accelerate horizontally (steady speed), and so F = F P fr F r fr F r N F r mg r P Putting this together, we have F F µ F µ mg ( )( )( ) 2 2 = = = = kg 9.8 m s = 103= N P fr k N k If the coefficient of kinetic friction is zero, then the horizontal force required is 0 N, since there is no friction to counteract. Of course, it would take a force to START the crate moving, but once it was moving, no further horizontal force would be necessary to maintain the motion.

15 Tension in a Flexible Cord Problem 4-38 (textbook Suppose that you are standing on a train accelerating at 0.20 g. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide?

16 Problem 4.38 (textbook): A free-body diagram for you as you stand on the train is shown. You do not accelerate vertically, and so F = mg N mg r F r fr F r N The maximum static frictional force is to the force needed to accelerate you. µ F s N, and that must be greater than or equal F ma µ F ma µ mg ma µ a g = 0.20g g = 0.20 fr s N s s The static coefficient of friction must be at least 0.20 for you to not slide.

17 Friction and Inclined Planes The problem with inclined plans are no different from the problems we have approached so far. However, remember: 1- The normal force is ALWAYS perpendicular to the plane where the object lies on. So, in an inclined plane its is not going to be in the vertical direction. (See next slide)

18 Friction and Inclined Planes Three forces can ALWAYS be identified acting on an object moving on an inclined place: Gravity (vertical); friend 2 Friction (along the surface); Normal (perpendicular to the surface). So, it is convenient to define your coordinate system such that the x direction coincides with the inclined line, and y in the direction perpendicular to the plane. friend 1 you You can surely have other applied forces: for instance, you (or you and a group of friends) can push or pull this object.

19 Tension in a Flexible Cord Problem 4-41 (textbook A 15.0-kg box is released on a 32º incline and accelerates down the incline at 0.30 m/s 2. Find the friction force impeding its motion. What is the coefficient of kinetic friction? F fr F r F r N y θ mg r θ x

20 Problem 4.41 (textbook): A free-body diagram for the box is shown. Write Newton s 2 nd law for each direction: F = mg sinθ F = ma x F = F mg cosθ = ma = 0 y N Notice that the sum in the y direction is 0, since there is no motion (and hence no acceleration) in the y direction. Solve for the force of friction. fr x y F r fr F r N θ mg r y θ x mg sin θ F = ma fr x ( ) ( )( ) F = mg = = fr 2 o 2 sinθ ma 15.0 kg 9.80 m s sin m s N 73 N x Now solve for the coefficient of kinetic friction. Note that the expression for the normal force comes from the y direction force equation above. F N fr F = µ F = µ mg cos θ µ = = = 0.59 fr k N k k 2 o mg cosθ 15.0 kg 9.80 m s cos 32 ( )( )( )

21 Tension in a Flexible Cord Problem 4-65 (textbook A bicyclist of mass 65 kg (including the bicycle) can coast down a 6.0º hill at a steady speed of 6.0 Km/h because of air resistance. How much force must be applied to climb the hill at the same speed and same air resistance?

22 Problem 4.41 (textbook): Consider a free-body diagram for the cyclist coasting downhill at a constant speed. Here we call F fr the friction due to air resistance (and not sliding or static friction). Since there is no acceleration, the net force in each direction must be zero. Write Newton s 2 nd law for the x direction. F = mg sinθ F = 0 F = mg sinθ x fr fr F r fr F r N y This establishes the size of the air friction force at 6.0 km/h, and so can be used in the next part. Now consider a free-body diagram for the cyclist climbing the hill. F p is the force pushing the cyclist uphill. Again, write Newton s 2 nd law for the x direction, with a net force of 0. F = F + mg sinθ F = 0 x fr F = F + mg sinθ = 2mg sinθ P fr ( )( )( ) = = 2 o kg 9.8 m s sin N P y F r P x θ θ θ mg r θ mg r F r N F r fr x

23 Assignment 5 Textbook (Giancoli, 6 th edition), Chapters 4 and 6: Due on Thursday, October 23, Problem 52 - page 102 of the textbook - Problem 87 - page 105 of the textbook - Problems 6 and 8 - page 162 of the textbook Tutorial next week, Oct. 15, CL-127 solving an old midterm.

24 Midterm Remind: Date: October 16, 1pm 2:15pm - There are four problems Full mark will be considered only when all problems have been completed - You ARE allowed to use a calculator - A equation sheet will be provided with some useful equations and identities. - You are NOT allowed to: - Use laptops; - Have cell phones (please, turn it off if you have one); - Use or consult any material other than those provided by me; -Make sure you bring your photo ID with you - Do not forget to write your name and UofR ID on each of the notebook used to solve your problems..

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