One end of a horizontal uniform beam of weight W and length L is hinged on a vertical wall at point

Q: Why do we take the torque specifically about the hinge O?

Taking torque about the hinge O is a strategic move! Since the unknown reaction forces $R_x$ and $R_y$ pass directly through O, their lever arms are zero, meaning they create zero torque. This leaves the tension $T$ as the only unknown in our equation, making it super easy to solve.

Q: How did we figure out the angle of the rope is $45^\circ$?

Look at the geometry of the setup. The wall, the beam, and the rope form a right-angled triangle $\triangle OQP$. The height $OQ$ is given as $L$, and the base $OP$ (the beam's length) is also $L$. Since the perpendicular and base are equal, $\tan \theta = L/L = 1$, which means $\theta = 45^\circ$.

Q: Why doesn't the vertical reaction $R_y$ depend on the block's weight $\alpha W$?

This is a beautiful consequence of the geometry! The vertical component of the tension ($T \sin 45^\circ$) increases exactly by $\alpha W$ when you add the block. Since the rope perfectly compensates for the added weight, the hinge at O doesn't 'feel' any extra vertical load. It only supports half the beam's own weight.

Q: What exactly does it mean when the problem says "the rope breaks"?

Every material has a breaking point. Here, the rope can only sustain a maximum tension of $2\sqrt{2}W$. If the calculated tension $T$ for a given $\alpha$ exceeds this value, the physical rope will snap. We set up the inequality $T > T_{max}$ to find the critical value of $\alpha$.

Q: Could we have solved this by taking torque about point P instead?

Absolutely! If you take torque about P, the tension $T$ and the block's weight $\alpha W$ create zero torque. You would get an equation involving $R_y$ and $W$. It's a great way to double-check your answer for $R_y$, but you'd still need translational equilibrium to find $T$ and $R_x$.

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 ?

Options:

(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 2W for $\alpha = 0.5$

(D) The rope breaks if $\alpha > 1.5$

Sigma Percentile · Accepted Answer

### Decoding the Equilibrium of a Hinged Beam

Imagine you are an engineer tasked with designing a support structure. You have a horizontal beam hinged to a wall, held up by a single rope, and carrying a heavy block at its far end. This classic problem from rigid body dynamics tests your ability to systematically apply the laws of static equilibrium. Let's break it down step-by-step and uncover the elegant physics hidden within the math.

### Visualizing the Setup and the Free Body Diagram

The absolute most crucial step in any mechanics problem is drawing a flawless Free Body Diagram (FBD). Let's identify every force acting on our beam of length $L$ and weight $W$.

First, we need the angle of the rope. The rope is attached to point $Q$ on the wall, which is at a height $L$ above the hinge $O$. It connects to the end of the beam at point $P$, a horizontal distance $L$ away. This forms a perfect isosceles right triangle, meaning the rope makes an angle of $45^\circ$ with the beam.

Now, let's map the forces:
1. The tension $T$ pulling up and left at $45^\circ$ from point $P$.
2. The beam's own weight $W$ acting downwards exactly at its center of mass ($L/2$).
3. The block's weight $\alpha W$ acting downwards at the far end $P$.
4. The hinge at $O$ exerts a reaction force. Since we don't know its exact direction, we split it into a horizontal component $R_x$ and a vertical component $R_y$.

### The Master Move: Rotational Equilibrium

We have three unknowns: $T$, $R_x$, and $R_y$. If we try to balance forces first, we'll get stuck with multiple unknowns in our equations. The master move here is to apply rotational equilibrium by taking the torque about the hinge $O$.

Why $O$? Because the lines of action for both $R_x$ and $R_y$ pass directly through the hinge. Their lever arms are zero, meaning they create zero torque! This brilliant choice leaves the tension $T$ as our only unknown.

Setting the net torque to zero ($\sum 	au_O = 0$), the counter-clockwise torque from the tension must equal the clockwise torque from the weights:

$$T \sin 45^\circ \cdot L = W \cdot \frac{L}{2} + \alpha W \cdot L$$

Dividing out the length $L$ and substituting $\sin 45^\circ = \frac{1}{\sqrt{2}}$, we get:

$$\frac{T}{\sqrt{2}} = W(0.5 + \alpha)$$

$$T = \sqrt{2} W (0.5 + \alpha)$$

Let's test the options with this result. For $\alpha = 0.5$, the tension becomes $T = \sqrt{2} W (0.5 + 0.5) = \sqrt{2} W$. This proves that **Option C is incorrect**, as it claims the tension would be $2W$.

What about the breaking point? The problem states the rope can sustain a maximum tension of $2\sqrt{2} W$. For the rope to break, our calculated tension must exceed this limit:

$$\sqrt{2} W (0.5 + \alpha) > 2\sqrt{2} W$$

Canceling $\sqrt{2} W$ from both sides leaves $0.5 + \alpha > 2$, which simplifies to $\alpha > 1.5$. This perfectly matches **Option D**, making it a correct statement.

### Translational Equilibrium: Finding the Reactions

Now that we have the tension, finding the hinge reactions is a breeze. We simply balance the forces in the vertical and horizontal directions.

Let's start with the vertical forces ($\sum F_y = 0$). The upward forces must balance the downward forces:

$$R_y + T \sin 45^\circ = W + \alpha W$$

We already know from our torque equation that $T \sin 45^\circ = W(0.5 + \alpha)$. Substituting this in:

$$R_y + W(0.5 + \alpha) = W(1 + \alpha)$$

Watch what happens when we expand and solve for $R_y$:

$$R_y = W + \alpha W - 0.5 W - \alpha W$$

$$R_y = 0.5 W$$

This is a beautiful result! The $\alpha W$ terms completely cancel out. The vertical reaction at the hinge is exactly half the weight of the beam, completely independent of how heavy the block is. The vertical component of the rope's tension perfectly compensates for the added weight of the block. Therefore, **Option A is correct**.

Finally, let's balance the horizontal forces ($\sum F_x = 0$). The wall pushes right with $R_x$, and the rope pulls left with the horizontal component of tension:

$$R_x = T \cos 45^\circ$$

Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, we have:

$$R_x = \sqrt{2} W (0.5 + \alpha) \cdot \frac{1}{\sqrt{2}} = W(0.5 + \alpha)$$

Let's test this for $\alpha = 0.5$:

$$R_x = W(0.5 + 0.5) = W$$

This confirms that **Option B is also correct**.

### The Final Verdict

By systematically applying the conditions for static equilibrium, we've decoded the entire system. The correct statements are **A, B, and D**. This problem beautifully illustrates how a strategic choice of the pivot point for torque can drastically simplify the algebra, and how physical constraints (like a breaking tension) translate into mathematical inequalities.

Select Answer:

Visualized Solution

Visualizing the Forces

Rotational Equilibrium: $\sum τ_{O} = 0$

Calculating Tension $T$

Vertical Equilibrium: $\sum F_{y} = 0$

Calculating Vertical Reaction $R_{y}$

Horizontal Equilibrium: $\sum F_{x} = 0$

Final Conclusion

The Sigma Insight: Equilibrium of Concurrent Forces

Decoding the Equilibrium of a Hinged Beam

Visualizing the Setup and the Free Body Diagram

The Master Move

Translational Equilibrium

The Final Verdict

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