Topic: Engineering Mechanics
A pin jointed uniform rigid rod of weight 981 N and length 1 m is kept horizontally by an external force F as shown in the figure below. The force F is removed suddenly. At the instant of force removal, the magnitude of vertical reaction (in N) developed at the support is close to
- A. Zero
- B. 245
- C. 490
- D. 981
Answer: B. 245
Explanation: When force F is removed, the rod begins to rotate about the pin. Taking moments about the pin: $W \times \frac{L}{2} = I \alpha$. The moment of inertia of the rod about one end is $I = \frac{mL^{2}}{3}$. So, $W \times \frac{L}{2} = \frac{mL^{2}}{3} \alpha = \frac{W}{g}\frac{L^{2}}{3} \alpha$. Solving for $\alpha$, we get $\alpha = \frac{3g}{2L}$. The linear acceleration of the center of mass is $a = \frac{L}{2} \alpha = \frac{3g}{4}$. Using Newton's second law for vertical translation: $W - R = ma = \frac{W}{g} \left(\frac{3g}{4}\right) = \frac{3W}{4}$. Thus, the reaction $R = W - \frac{3W}{4} = \frac{W}{4} = \frac{981}{4} = 245.25~N$.