Question

One end of a horizontal uniform beam of weight $W$ and length $L$ is hinged on a vertical wall at point $O$ and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point $Q$, at a height $L$ above the hinge at point $O$ a block of weight $\alpha W$ is attached at the point $P$ of the beam, as shown in the figure (not to scale). The rope can sustain a maximum tension of $(2 \sqrt{2})$ W Which of the following statement(s) is(are) correct?

• A. The vertical component of reaction force at $O$ does not depend on $\alpha$
• B. The horizontal component of reaction force at $O$ is equal to $W$ for $\alpha=0.5$
• C. The tension in the rope is $2 \,W$ for $\alpha=0.5$
• D. The rope breaks if $\alpha > 1.5$

Answer 1

The Correct Option is D

Step 3: Maximum tension condition

The maximum tension the rope can sustain is \( T_{\text{max}} = (2\sqrt{2})W \). Using this, we can set up the equation for the system's torque balance:

Torque due to beam's weight + Torque due to block's weight = Torque due to rope's tension

This gives the equation:

\( W \times \frac{L}{2} + \alpha W \times L = T_{\text{max}} \times L \)

Substituting \( T_{\text{max}} = 2\sqrt{2}W \), we get:

\( W \times \frac{L}{2} + \alpha W \times L = (2\sqrt{2}W) \times L \)

Canceling out the common factors of \( W \) and \( L \), we get:

\( \frac{1}{2} + \alpha = 2\sqrt{2} \)

Simplifying this, we find:

\( \alpha = 2\sqrt{2} - \frac{1}{2} \approx 2.828 - 0.5 = 2.328 \)

Step 4: Conclusion

The rope will break when \( \alpha > 1.5 \). This is because when \( \alpha \) exceeds this value, the torque generated by the block becomes too large for the rope to handle, and it reaches its maximum tension threshold.

Final Answer:

The rope breaks if \( \alpha > 1.5 \).