1. Aim: More Atwood Machines Answer Key HW 6 Do Now: Draw a free-body diagram for the following frictionless inclined plane: m2m2 m1m1 M θ Mg m2m2 m1m1 M m 1 gsinθ m 2 gsinθ T 1,2 T T

2. m2m2 m1m1 M θ Block 1 Block 2 v 1. Blocks 1 and 2 of masses m 1 and m 2, respectively, are connected by a light string as shown above. These blocks are further connected to a block of mass M by a light string that passes over a pulley of negligible mass and friction. Blocks 1 and 2 move with a constant velocity v down the inclined plane, which makes an angle θ with the horizontal. The kinetic friction force on block 1 is f and that on block 2 is 2f. Calculator Allowed **15 min**

3. a. On the figure below, draw and label all the forces on block m 1 m1m1 θ T f N m1gm1g m1m1

4. Express your answers to each of the following in terms of m 1, m 2, g, θ, and f. b. Determine the coefficient of kinetic friction between the inclined plane and block 1.

5. Mg c. Determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the plane. m2m2 m1m1 M m 1 gsinθ m 2 gsinθ f 2f T 1,2 T T

6. d. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the inclined plane.

7. m1gm1g T a. On the diagram below, draw and identify all the forces acting on the block of mass M1. 2. In the system shown above, the block of mass M1 is on a rough horizontal table. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction  k between M1 and the table is less than the coefficient of static friction  s FNFN F m1m1 m1m1 Calculator Allowed **10 min**

8. b. In terms of M 1 and M 2 determine the minimum value of  s that will prevent the blocks from moving. ΣF = 0T = M 2 g T = F F T = μ s M 1 g M 2 g = μ s M 1 g M 2 = μ s M 1 M2M2 T M2gM2g

9. The blocks are set in motion by giving M2 a momentary downward push. In terms of M 1, M 2,  k, and g, determine each of the following: c. The magnitude of the acceleration of M 1ΣF = ma T – F F = M 1 aM 2 g – T = M 2 a T - μ k M 1 g = M 1 a M 2 g – T = M 2 a M 2 g - μ k M 1 g = M 1 a + M 2 a M 2 g - μ k M 1 g = a(M 1 + M 2 )

10. d. The tension in the string. T - μ k M 1 g = M 1 a