Question

Find the acceleration of the block of mass M in the situation of figure (6−E10). The coefficient of friction between the two blocks is μ₁ and that between the bigger block and the ground is μ₂.

Answer 1

Let us the acceleration of the block of mass M be a and let it be towards right. Therefore, the block of mass m must go down with acceleration 2a. As both the blocks are in contact, it (block of mass m) will also have acceleration a towards right. Hence, it will experience two inertial forces as shown in the free body diagram given below.



(Free body diagram 1)
From the free body diagram 1, we have:


(Free body diagram-2)

R₁ − ma = 0
⇒ R₁ = ma ....(i)
Again,
2ma + T − Mg + μ₁R₁ = 0
⇒ T = Mg − (2 + μ₁) ma ....(ii)

Using the free body diagram 2, we have:

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 −2,
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
$a=\frac{\left[2m+m_2\left(M+m\right)\right]g}{M+m\left[5+2\left({\mathrm{\mu }}_1-{\mathrm{\mu }}_2\right)\right]}$