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Question:

Consider the diagram shown below in which two masses of m and 2m are placed on a fixed triangular wedge. The coefficient of friction between block A and the wedge is 2/3, while that for block B and the wedge is 1/3. If the whole system is released from rest, then acceleration of block A is:

4m23g

m2√2g

Zero

2m23g

Solution:

Given, μa=2/3 and μb=1/3
Let the acceleration of the whole system be a then,
As per the diagram,
For block B, 2mgcos45° + 2ma = T + μbNb —————(1)
Nb = 2mgsin45°
And, For block A, T = mgcos45° + μaNa
Na = mgsin45°
So, T = mgcos45° + μa mgsin45° —————(2)
From eq(1) and (2)
2mgcos45° + 2ma = mgcos45° + μa mgsin45° + μb 2mgsin45°
After solving,
We get a = 0
So the acceleration of the block A will be zero.