As shown in the figure, a block of mass M1 = 0.470 kg is initially at rest on a slab of mass M2 = 0.850 kg, and the slab is initially at rest on a level table. A string of negligible mass is connected to the slab, runs over a frictionless pulley on the edge of the table, and is attached to a hanging mass M3. The block rests on the slab but is not tied to the string, so friction provides the only horizontal force on the block. The slab has a coefficient of kinetic friction ìk = 0.330 and a coefficient of static friction ìs = 0.570 with both the table and the block. When released, M3 pulls on the string and accelerates the slab, which accelerates the block. Find the maximum mass of M3 that allows the block to accelerate with the slab, without sliding on top of the slab.

Question

As shown in the figure, a block of mass M1 = 0.470 kg is initially at rest on a slab of mass M2 = 0.850 kg, and the slab is initially at rest on a level table. A string of negligible mass is connected to the slab, runs over a frictionless pulley on the edge of the table, and is attached to a hanging mass M3. The block rests on the slab but is not tied to the string, so friction provides the only horizontal force on the block. The slab has a coefficient of kinetic friction ìk = 0.330 and a coefficient of static friction ìs = 0.570 with both the table and the block. When released, M3 pulls on the string and accelerates the slab, which accelerates the block. Find the maximum mass of M3 that allows the block to accelerate with the slab, without sliding on top of the slab.

Answer 1

To find the maximum mass of M3 that allows the block to accelerate with the slab without sliding on top of the slab, we need to analyze the forces and apply Newton's second law.

1. Draw a free body diagram for each object:
- Block (M1): The forces acting on the block are the friction force and the tension force from the string.
- Slab (M2): The forces acting on the slab are the friction force from the table and the tension force from the string.
- Hanging mass (M3): The only force acting on it is its weight.

2. Determine the acceleration of the system:
- Since the block is not sliding on top of the slab, the acceleration of the block is the same as the acceleration of the slab.
- We can use Newton's second law for the slab: ΣF = M2 * a, where ΣF is the net force on the slab.
- The net force on the slab is the tension force from the string minus the friction force: ΣF = T - μk*M2*g, where μk is the coefficient of kinetic friction, and g is the acceleration due to gravity.
- The net force on the block is just the friction force: ΣF = μs*M1*g, where μs is the coefficient of static friction.
- Since the block and the slab have the same acceleration, we can set the above two equations equal to each other.
- T - μk*M2*g = μs*M1*g

3. Solve for the maximum mass of M3:
- Substitute the given values into the equation and solve for T (tension force).
- Once you have T, you can find the maximum mass of M3 by setting T equal to the weight of M3: T = M3 * g.
- Rearrange the equation to solve for M3: M3 = T / g.

4. Substitute the calculated values and solve for M3:
- Plug in the values for μk, μs, M1, and M2. Also, use the known value of g (acceleration due to gravity).

By following these steps, you can determine the maximum mass of M3 that allows the block to accelerate with the slab without sliding on top of the slab.