First slide

Rotational motion

Question

A uniform rod of mass 15 kg and of length 5m is held stationary with the help of a light string as shown in the figure. The tension in the string is

Moderate

A

150 N
B

225 N
C

100 N
D

none of the above

Solution

The free-body diagram of the rod is as shown in the figure. (Force exerted by the hinge on the rod are not shown.)
$G i v e n : O B \sin θ = 3 \Rightarrow \sin θ = \frac{3}{5} O A = \frac{O B}{2} = \frac{5}{2}$
Now, take torque about the hinge,
$\begin{array}{l} T \times (O B \sin θ) - 15 g \times (O A \cos θ) = 0 \\ \Rightarrow T \times 3 - 15 g \times (\frac{5}{2} \times \frac{4}{5}) = 0 \\ \Rightarrow T = 100 N \end{array}$

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