One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and its magnitude is slowly increased. When the force reaches 44.5 N, the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block?

Question

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and its magnitude is slowly increased. When the force reaches 44.5 N, the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block?

Answer 1

To find the magnitude of the horizontal force that should be applied to the lower block, we need to understand the forces acting on the system.

Let's denote the coefficient of static friction between the blocks as μ, and the weight of each block as W. Since the blocks are identical, the weight of each block is the same.

When the upper block just begins to slide, the maximum static friction force between the two blocks is equal to the force applied to the upper block. Therefore, the force of static friction between the blocks is also 44.5 N.

The force diagram for the system is as follows:

-----------------
| Block 2 |
-----------------
| Block 1 |
-----------------
| Horizontal |
| Surface (Friction) |
-----------------

We know that the force of static friction between the lower block and the horizontal surface is given by the equation:

Friction force = μ * Normal force

Since the blocks are resting on each other, the normal force on the lower block is equal to the weight of both blocks combined (2W).

Therefore, the force of static friction between the lower block and the horizontal surface can be written as:

μ * 2W = 44.5 N

To find the magnitude of the horizontal force that should be applied to the lower block, we need to consider that this force has to overcome the force of static friction between the blocks. Therefore, the horizontal force should be slightly greater than the force of static friction between the blocks.

So, the magnitude of the horizontal force that should be applied to the lower block is slightly greater than 44.5 N.