Question

Friction of one block on top of another. Full explanation please.

A 4.0 kg block is put on top of a 5.0 kg block. The coefficient of static friction between the two blocks Ls mus = 0.33. The assembly of blocks is placed on a horizontal, frictionless table as on the figure below. What is the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together without any slippage? 20.25 N 29.11 N 11.27 N 9.00 N 16.08 N

Answer 1

Here there is no relative motion between the blocks till their accelerations are equal.

Draw the free-body diagram of the 5 kg block
mg acts downwards
N acts upwards
Applied force F acts towards right
Frictional force f acts towards left

F - f = Ma
The frictional force is maximum = umg
So
F = Ma + umg = 5a + 0.33*4*9.8 = 5a + 12.93........................(1)

Now draw the free-body diagram for the 4 kg block.
mg acts downwards
N acts upwards
f acts towards right

So f = ma
or umg = ma
or a = ug = 0.33 * 9.8 = 3.234

Substitute in (1) :
F = 5(3.234) + 12.93 = 29.1 N

F= 29.1 N