One block rests upon a horizontal surface. A second identical block rests upon t

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 48.5 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?

Explanation / Answer

The force should be 3 times of 48.5

When the applied force is slughtly greaetr than static friction the block strts sliding.
Fnet = 0

F_applied - Friction due to surface - Frcition due to upper block

F_applied - 2*48.5 - 48.5 = 0

F_applied = 3*48.5

= 145.5 N

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