A uniform rod of length $1$ m and mass of $2$ kg is attached to a side support at $O$ as shown in the figure. The rod is at equilibrium due to upward force $T$ acting at $P$. Assume the acceleration due to gravity as $10$ m/s$^2$. The value of $T$ is
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For rods, weight acts at center; always take torque about pivot to simplify.
Concept:
For rotational equilibrium:
\[
\sum \tau = 0
\]
Step 1: Weight of rod.
\[
W = mg = 2 \times 10 = 20 \text{ N}
\]
Acts at center (0.5 m from O).
Step 2: Take moments about O.
\[
T \times 1 = 20 \times 0.5
\]
Step 3: Solve.
\[
T = 10 \text{ N}
\]