For the situation shown in the figure, pulleys are light and smooth. Pulleys C and D are fixed, but A and B are movable. Find the value of m (in kg) if m is in equilibrium. (take, m1 = 2 kg and m2 =1 kg)

Solution:

Let tension of $m_{2}$ is T and tension of $m_{1}$ is 2T

Let acceleration of $m_{1}$ is a and acceleration of $m_{2}$ is 2a

$F o r m_{2}$

$T - m_{2} g = m_{2} (2 a) \Rightarrow T - 1 (10) = 1 (2 a) . . . . . . (i)$

Free body diagram (FBD) of m₂

$For m_{1} : m_{1} g - 2 T = m_{1} (a) \Rightarrow 2 (10) - 2 T = 2 a . . . . . . (i i)$

Free body diagram (FBD) of m₁

From Eqs. (i)2 and (ii), we get

$\begin{array}{l} 2 T - 20 = 4 a \\ \Rightarrow 20 - 2 T = 2 a \\ \Rightarrow a = 0 m / s^{2} \\ Or substituting in (1) \\ \Rightarrow T - 10 = 2 (0) \\ \Rightarrow T = 10 N \end{array}$

Free body diagram of m

$For m, 2 T = m g \Rightarrow 2 (10) = m (10) \Rightarrow m = 2 kg$

For the situation shown in the figure, pulleys are light and smooth. Pulleys C and D are fixed, but A and B are movable. Find the value of m (in kg) if m is in equilibrium. (take, m₁ = 2 kg and m₂ =1 kg)

Solution:

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