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Question
Find the acceleration of the block of mass M in the situation of figure in the following. The coefficient of friction between the two blocks is μ1 and that between the bigger block and the ground is μ2.
Solution
Let the acceleration of block of mass M is ‘a’ towards right. So, the block ‘m’ must go down with an acceleration ‘2a’.
As the block ‘m’ is in contact with the block ‘M’, it will also have acceleration ‘a’ towards right. So, it will experience two inertial forces as shown in figure (b). From free body diagram
R1 – ma = 0
⇒ R1 = ma (i)
Also,
2ma + T – Mg + μ1R1 = 0
⇒ T = Mg – (2 + μ1) ma (ii)
Using the free body diagram (c), we can write,
T + μ1R1 + Mg – R2 = 0
Substituting the value of R1 from (i), we get,
R2 = T + μ1 ma + mg
Substituting the value of T from (ii), we get,
R2 = (Mg – 2ma – μ1ma) = μ1 ma + Mg + ma
∴ R2 = Mg + Ma – 2ma ….(iii)
Again, using the free body diagrams (c),
T + T – R – Ma – μ2R2 = 0
⇒ 2T – Ma – ma – μ2(Mg + mg – 2ma) = 0
Substituting the values of R1 and R2 from (i) and (iii), we get:
2T = (M + m)a + μ2 (Mg + mg – 2ma) ….(iv)
From equations (ii) and (iv), we have:
2T = 2mg – 2(2 + μ1) ma
= (M + m) a + μ1(Mg + mg – 2ma)
⇒ 2mg – μ2 (M + m)g = a[M + m – 2μ2m + 4m + 2μ1m]
Therefore, the acceleration of the block of mass M in the given situation is given by `a = (2m + μ2(M + m)g)/(M + m(5 + 2(μ1 - μ2))`
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