00:01
This problem, we have to consider two blocks that are resting on top of one top of other.
00:09
The lower block has a given mass and is resting on a frictionless table.
00:14
The upper block has a different mass and the coefficient of static friction between the two blocks is given by mu.
00:24
A force of f is applied as shown in the figure and we have to find the maximum value of force f and for which the upper block can be pushed horizontally so that the two blocks move together without slipping so the force is applied to the upper block only now the lower block also moves due to the applied force because the upper block applies friction on the lower block so friction on the upper block by the lower block will be f so, friction on the lower block by the upper block will be in the opposite direction, which is f prime.
01:32
So we consider the friction force.
01:39
For the upper block, the vertical force is normal force in the upward direction and weight m g of the upper block is in the downward direction.
01:51
So the forces on the upper block are normal force.
02:01
This has mass m and the lower block has mass capital m.
02:06
So for the upper block, the forces are normal force in the upper direction, weight, mg in the downward direction, friction force to the left and the applied force capital f to the right.
02:33
Now we have to find the maximum value of applied force.
02:38
So when the applied force is maximum, the static friction force between the two blocks will also be maximum and the bar...