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Question

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

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Solution

## (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 $⇒21a=-3g-a=-\frac{9}{7}$ 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 $⇒a=\frac{5g}{13}\left(\mathrm{downward}\right)$ 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 $⇒a=\frac{g}{3}\left(\mathrm{downward}\right)$ 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)

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