Question

Suppose the entire system of the previous questions is kept inside an elevator which is coming down with an acceleration a < g. Repeat parts (a) and (b).

Answer 1

From the free body diagram, we have:
R₁ + ma − mg = 0
⇒ R₁ = m (g − a)
= mg − ma ...(i)
Now,
F − T − μR₁ = 0 and
T − μR₁ = 0
⇒ F − [μ (mg − ma)] − μ(mg − ma) = 0
⇒ F − μ mg − μma − μmg + μma = 0
⇒ F = 2 μmg − 2 μma
= 2 μm (g − a)

(b) Let the acceleration of the blocks be a₁.

R₁ = mg − ma ....(i)
And,
2F −T − μR₁ = ma₁ ...(ii)
Now,
T = μR₁ + Ma₁
= μmg − μma + Ma₁

Substituting the value of F and T in equation (ii), we get:
2[2μm(g − a)] − (μmg − μma + Ma₁) − μmg + μma = ma₁
⇒ 4μmg − 4μma − 2μmg + 2μma= ma₁ + Ma₁
$\Rightarrow a_1=\frac{2\mathrm{\mu }m\left(g-a\right)}{M+m}$
Thus, both the blocks move with same acceleration a₁ but in opposite directions.