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 A, B, D

Step 1: Understanding the System
We have a horizontal uniform beam of weight \( W \) and length \( L \), hinged at point \( O \). The other end of the beam is supported by a light, inextensible rope that is fixed at point \( Q \), at height \( L \) above the hinge at point \( O \). A block of weight \( \alpha W \) is attached at point \( P \) of the beam.
The rope can sustain a maximum tension of \( (2\sqrt{2}) W \). Our goal is to determine the reaction forces at point \( O \) and the conditions under which the rope will break.

Step 2: Vertical Component of Reaction Force at \( O \)
We will first analyze the forces acting on the system. Consider the vertical forces acting on the beam. The forces acting on the beam are:
- The weight of the beam, \( W \), acting vertically downward at the center of mass of the beam, i.e., at \( L/2 \) from the hinge \( O \). - The weight of the block, \( \alpha W \), acting vertically downward at point \( P \), which is a distance \( L \) from the hinge \( O \). - The vertical component of the tension in the rope, denoted by \( T_{\text{vertical}} \), which acts vertically upward.
For vertical force equilibrium, we sum the forces acting vertically and set them equal to zero: \[ T_{\text{vertical}} = W + \alpha W \] Thus, the vertical reaction at point \( O \) is independent of \( \alpha \), as it only depends on the weight of the beam and the block.
Therefore, the correct statement is:
(A) The vertical component of reaction force at \( O \) does not depend on \( \alpha \).

Step 3: Horizontal Component of Reaction Force at \( O \)
Now let's analyze the horizontal forces. The horizontal force acting at point \( O \) is due to the tension in the rope, specifically the horizontal component of the tension force, denoted as \( T_{\text{horizontal}} \).
The horizontal equilibrium condition gives: \[ T_{\text{horizontal}} = W \] This condition holds true when \( \alpha = 0.5 \), as the forces are balanced when the weight of the beam equals the horizontal tension component.
Therefore, the correct statement is:
(B) The horizontal component of reaction force at \( O \) is equal to \( W \) for \( \alpha = 0.5 \).

Step 4: Condition for Rope Breaking
The maximum tension the rope can withstand is given as \( T_{\text{max}} = (2\sqrt{2}) W \). The total tension in the rope depends on the forces acting on the beam and the geometry of the system.
We calculate the total tension in the rope by summing the forces and taking into account the angles involved in the system. Using the torque equilibrium condition around point \( O \), we find that the rope will break when the tension exceeds the maximum allowed value.
The rope will break when: \[ \alpha > 1.5 \] This is the condition when the total tension in the rope exceeds the maximum allowed value.
Therefore, the correct statement is:
(D) The rope breaks if \( \alpha > 1.5 \).

Final Answer:
The correct answers are:
(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 \( \alpha = 0.5 \)
(D) The rope breaks if \( \alpha > 1.5 \)