Question

Find the acceleration of the blocks A and B in the three situations shown in figure (5−E17).

Figure

Answer 1

(a) 5a + T − 5g = 0
From free-body diagram (1),
T = 5g − 5a .....(i)
Again, $\left(\frac{1}{2}\right)T-4g-8a=0$
⇒ T − 8g − 16a = 0


(1)

From free-body diagram (2),
T = 8g + 16a ......(ii)

From equations (i) and (ii), we get:
5g − 5a = 8g + 16a
$\Rightarrow 21a=-3g-a=-\frac{9}{7}$
So, the acceleration of the 5 kg mass is $\frac{9}{7}\mathrm{m}/{\mathrm{s}}^2\left(\mathrm{upward}\right)$ and that of the 4 kg mass is $\mathit{2}a\mathit{=}\frac{\mathit{2}\mathit{g}}{\mathit{7}}\mathit{}\left(\mathrm{downward}\right)$.

(b) From free body diagram-3,
$4a-\frac{T}{2}=0$


(2)


⇒ 8a − T = 0
⇒ T = 8a
Again, T + 5a − 5g = 0
From free body diagram-4,

8a + 5a − 5g = s0
⇒ 13a − 5g = 0
$\Rightarrow a=\frac{5g}{13}\left(\mathrm{downward}\right)$

Acceleration of mass 2 kg is $2a=\frac{10}{13}\left(g\right)$ and 5 kg is $\frac{5g}{13}$.

(c) T + 1a − 1g = 0
From free body diagram-5
T = 1g − 1a .....(i)
Again, from free body diagram-6,
$\frac{T}{2}-2g-4a=0$
⇒ T − 4g − 8a = 0 .....(ii)

From equation (i)
1g − 1a − 4g − 8a = 0
$\Rightarrow a=\frac{g}{3}\left(\mathrm{downward}\right)$
Acceleration of mass 1 kg is $\frac{g}{3}\left(\mathrm{upward}\right)$,
Acceleration of mass 2 kg is $\frac{2g}{3}\left(\mathrm{downward}\right)$.

(3)