Question

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

Answer 1

Explanation:

where is ur figure I can't see

Answer 2

Explanation:

Step 1:

(a) 5 a + T - 5 g = 0

From Body Free Diagram-1

$T=5 g-5 a \quad \ldots \ldots e q^{n}(i)$

$\left(\frac{1}{2}\right) T-4 g-8 a=0$

T - 8 g - 16 a = 0

Step 2:

From Body Free Diagram-2

$T=8 g+16 a \quad \ldots \ldots e q^{n}(i i)$

From (i) and (ii) equations we obtain:

5 g - 5 a = 8 g + 16 a

$21 a=-3 g-a=-\frac{9}{7}$

$So, the 5 kg mass acceleration is \frac{9}{7} ms^2 (upward) 97 and That of a mass of 4 \mathrm{kg}, 2 a=\frac{2 g}{7} (downward)$

Step 3:

(b) From Body Free Diagram-3

$4 a-\frac{T}{2}=0$

8 a - T = 0

T = 8 a

Again,

T + 5 a - 5 g = 0

Step 4:

From Body Free Diagram

8 a + 5 a - 5 g = 0

13 a - 5 g = 0

$a=\frac{5 g}{13} \quad(\text { downward })$

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

Step 5:

(c)

T + 1 a - 1 g = 0

From Body Free Diagram

T = 1 g - 1 a                        .....(i)

From free body diagram again,

$\frac{T}{2}-2 g-4 a=0$

T - 4 g - 8 a = 0                   .....(ii)

Step 6:

From I Equation

1 g - 1 a - 4 g - 8 a = 0

$a=\frac{g}{3} \quad \text { (downward) }$

mass acceleration 1 kg is  $\frac{g}{3}$  (upwards)

2 kg mass acceleration is  $\frac{2g}{3}$ (downwards)