A man stacks a small block of wood on top of a larger block and then moves both by pushing the

Question

bottom block. The large block has a mass of 25 kg, and the smaller block has a mass of 15 kg. The
coefficient of static friction between the two blocks is 0.45. How fast can the man accelerate the blocks
without having the smaller block slide off the top?

Answer 1

To solve this problem, we will consider the forces acting on the smaller block and the larger block separately.

Let's denote the acceleration of the blocks as a, and let's assume the direction of acceleration is to the right.

For the smaller block:
The only force acting on the smaller block is the static friction force (Fs) between the two blocks. The maximum value of static friction force can be calculated using the equation Fs ≤ μsN, where μs is the coefficient of static friction and N is the normal force between the two blocks. In this case, the normal force is equal to the weight of the smaller block, which is N = mg = 15 kg * 9.8 m/s^2 = 147 N. Plugging in these values, we have Fs ≤ 0.45 * 147 N = 66.15 N.

For the larger block:
The only force acting on the larger block is the force of gravity (weight). The weight of the larger block is W = mg = 25 kg * 9.8 m/s^2 = 245 N.

Since the smaller block can slide off the top if the static friction force is less than the force of gravity, we can set up the inequality Fs ≥ W, which becomes 66.15 N ≥ 245 N.

Therefore, the maximum acceleration the man can apply without the smaller block sliding off is given by Newton's second law: Fs = ma, which becomes 66.15 N = (15 kg + 25 kg) * a. Solving for acceleration, we have a = 66.15 N / 40 kg = 1.65 m/s^2.

Therefore, the man can accelerate the blocks up to a speed of 1.65 m/s^2 without having the smaller block slide off the top.