1. 1 Some application & Forces of Friction

2. 2 Example: When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass, as in Figure, the arrangement is called an Atwood machine. The device is sometimes used in the laboratory to measure the free-fall acceleration. Determine the magnitude of the acceleration of the two objects and the tension in the lightweight cord. When Newton’s second law is applied to object m 1, we obtain Solution:

3. 3 Similarly, for object m 2 we, find When (2) is added to (1), T cancels and we have Example: find the acceleration and the tension of Atwood's machine in which m 1 =2kg, m 2 =4kg. Then we have

4. 4 example: A ball of mass m 1 and a block of mass m 2 are attached by a lightweight cord that passes over a frictionless pulley of negligible mass, as in Figure. The block lies on frictionless incline of angle Find the magnitude of the acceleration of the two objects and the tension in the cord. Equations of motion for m 1 (1) (2) Solution:

5. 5 Equations of motion for m 2 (4) (3) Now, solve (2), (3) simultaneously. The acceleration When this is substituted into (2) we find the normal force Example: if m 1 =10kg, m 2 =5kg and =45 find a and T.

6. 6 problem: Two blocks of mass 3.50 kg and 8.00 kg are connected by amassless string that passes over a frictionless pulley. The inclines are frictionless. Find (a) the magnitude of the acceleration of each block and (b) the tension in the string M1=1kg / M2=3kg / F=5N Find (a) the magnitude of the acceleration of each block and (b) the tension in the string problem:

7. 7 Example: A 3.0 kg mass hangs at one end of a rope that is attached to a support on a railroad car. When the car accelerates to the right, the rope makes an angle of 4.0° with the vertical. Find the acceleration of the car. In the vertical direction: Solution:

8. 8 In the horizontal direction:

9. 9 Problem: A train is given an acceleration of 5 m/s 2. What is the tension between the two cars with a mass of 2000 kg.

10. 10 Forces of Friction In most cases we need to include friction. For many surfaces the force of sliding friction is proportional to the normal force. That is, as the normal force increases, the force of friction increases linearly. When this is true, we write with the proportionality constant called, the coefficient of kinetic friction. This is the friction when the object is moving.

11. 11 If I push a table and it doesn't move, there must also be a force pushing back. That force comes from friction, as well. It is the friction when the object is not moving, or static, called the coefficient of static friction. It is always greater than or equal to the coefficient of kinetic friction. It is often harder to start something sliding than to keep it sliding, due to the difference in these coefficients.

12. 12 The Static force of friction ( f s ) is the force of friction between two objects when there is no motion. The Kinetic force of friction ( f k ) is the force of friction between two objects when there is motion.

13. 13 Example : An 18.0 kg box is released on a 37.0° incline and accelerates down the incline at 0.27 m/s 2. Find the coefficient of friction and the frictional force. (Draw diagram) Forces in y-direction are: Solution :

14. 14 Forces in x-direction are: Now, we can relate these to the coefficient of kinetic friction,

15. 15 Problem: Forces are being applied to a box sitting on a surface with friction. Will the box move horizontally (along the surface)? F1=50N, F2=50N, Mass of the block 10kg, and ms=0.4. Hint: must be greater or equal to

16. 16 Problem: A 20 kg sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 80.0 N and is directed at an angle or 30.0° above the horizontal. Determine the coefficient of kinetic friction.