Two blocks A and B of masses mA = 1 kg and mB = 3 kg are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontal, so that the block A does not slide over the block B is:
->The answer is 16N
But considering the answer as 16N if we find the acceleration of both the blocks, then they are coming to be different.....
Answers
To apply a force F on the block B such that the block A does not move over B, both the blocks should move together.
In the figure the FBD of the block B is when the blocks move separate to each other or when there is a relative motion between them. But here no relative motion occurs.
Here f_A is the friction between the blocks A and B and f_B is the friction between the block B and the surface, which are equal to 2 and 8 Newtons respectively.
Since the force F is acting only on B, the block A moves only by the friction f_A. Hence, maximum acceleration of the system is given by,
a_(max) = f_A / m_A = 2 / 1 = 2 m s^(-2).
In the FBD of the whole system,
F_(max) - f_B = 2 (1 + 3)
F_(max) - 8 = 2 × 4
F_(max) = 8 + 8
F_(max) = 16 N.
So this is the maximum force. By this force, both the blocks move together with an acceleration of 2 m s^(-2).
Here the individual accelerations of the blocks have no significance since the blocks are moving together so that there is no relative motion between the blocks. That's why we don't have to find out acceleration for each of the block. We don't have to be confused too!
If we find accelerations of both the blocks each, we may get them different, but remember, those accelerations are possible if the force F exceeds F_(max) and if there is relative motion between them.
Answer:
answer 16 N
Explanation: