A bead of weight w can slide on smooth circular wire in a vertical plane. The bead is attached by a

Q: Wait, why is the reaction force $R$ acting radially outwards? Shouldn't it be perpendicular to the tangent?

That is a brilliant observation! The reaction force is indeed perpendicular to the surface of the wire. Since the wire is circular, the normal to the tangent at any point always passes through the center of the circle. Thus, the normal reaction force must act along the radius, pointing outwards from the center. You are thinking exactly like a physicist!

Q: How did we get $2\theta$ as the angle between the weight $w$ and the reaction $R$?

Let's look at the geometry. The weight $w$ acts vertically downwards. The radius $OB$ makes an angle $\theta$ with the vertical. The reaction force $R$ acts along the radius $OB$. If you extend the vertical line downwards, the angle between the vertical and the radius $OB$ is $\theta$. However, the angle between the vertical and the reaction $R$ is the angle between the vertical and the radius $OB$. Since the radius $OB$ makes an angle $\theta$ with the vertical, the angle between the vertical and the reaction $R$ is $\theta$. Wait, let's re-verify. The angle between the vertical and the radius $OB$ is $\theta$. The angle between the vertical and the reaction $R$ is $\theta$. Actually, the angle between the vertical and the radius $OB$ is $\theta$. The angle between the vertical and the reaction $R$ is $\theta$. Let's look at the triangle. The angle at the center $O$ is $2\theta$. The angle between the vertical and $OB$ is $\theta$. The angle between the vertical and $OA$ is $0$. The angle between $OA$ and $OB$ is $2\theta$. Yes, the angle between the reaction $R$ (along $OB$) and the weight $w$ (vertical) is indeed $2\theta$.

Q: Can I solve this using components instead of Lami's Theorem?

Absolutely! You can resolve the forces into horizontal and vertical components. You would set $\sum F_x = 0$ and $\sum F_y = 0$. It is a perfectly valid method and will yield the exact same result. Lami's Theorem is just a shortcut for concurrent forces, but mastering components is essential for more complex problems where forces aren't concurrent.

Question

A bead of weight w can slide on smooth circular wire in a vertical plane. The bead is attached by a light thread to the highest point of the wire and in equilibrium, the thread is taut and make an angle $	heta$ with the vertical then tension of the thread and reaction of the wire on the bead are

Options:

(A) $T = w \cos 	heta, R = w 	an 	heta$

(B) $T = 2w \cos 	heta, R = w$

(C) $T = w, R = w \sin 	heta$

(D) $T = w \sin 	heta, R = w \cot 	heta$

Sigma Percentile · Accepted Answer

### The Dance of Equilibrium: A Journey into Mechanics

Welcome, future engineers. Today, we are not just solving a problem; we are stepping into a world of balance.

Imagine you are standing in a laboratory, looking at a smooth, circular wire fixed in a vertical plane. A small bead, weighted by gravity, rests on this wire. It is held in place by a light, inextensible thread attached to the highest point of the wire.

This is a classic JEE Advanced scenario—a test of your ability to visualize geometry and apply the laws of statics. Let us break this down together.

### Phase 1: The Geometric Blueprint

Before we touch a single equation, we must see the system. Let $O$ be the center of our circular wire.

We have point $A$ at the very top and point $B$ where our bead resides. If we draw lines from the center $O$ to $A$ and $B$, we create a triangle, $	riangle OAB$.

Because $OA$ and $OB$ are both radii of the same circle, they are equal. This makes $	riangle OAB$ an isosceles triangle.

Since the thread makes an angle $	heta$ with the vertical, and the vertical line passes through $O$ and $A$, the angle $\angle OAB$ is $	heta$. By the properties of isosceles triangles, the base angles are equal, so $\angle OBA$ is also $	heta$.

This simple geometric insight is the foundation of our entire solution.

### Phase 2: The Forces at Play

Now, let us identify the forces acting on the bead at point $B$. We have three distinct forces keeping the bead in perfect equilibrium.

First, the weight $w$, pulling the bead vertically downwards. Second, the tension $T$ in the thread, pulling the bead along the line $BA$.

Third, the normal reaction $R$ from the smooth wire. Because the wire is circular, the normal force must act along the radius, pointing outwards from the center $O$ through $B$.

We have three concurrent forces: $w$, $T$, and $R$. When you see three forces in equilibrium, your mind should immediately jump to Lami's Theorem.

### Phase 3: The Elegance of Lami's Theorem

Lami's Theorem is a beautiful tool. It states that for three concurrent forces in equilibrium, each force is proportional to the sine of the angle between the other two.

Let us calculate these angles. The angle between the reaction $R$ (along $OB$) and the weight $w$ (vertical) is $2	heta$.

The angle between the tension $T$ and the weight $w$ is $180^{\circ} - 	heta$. Similarly, the angle between the tension $T$ and the reaction $R$ is $180^{\circ} - 	heta$.

Now, we write our equation:

$$
\frac{T}{\sin(2	heta)} = \frac{R}{\sin(180^{\circ} - 	heta)} = \frac{w}{\sin(180^{\circ} - 	heta)}
$$

This is the heart of the problem.

### Phase 4: The Algebraic Resolution

Let us solve for the reaction $R$ first. Using the equality $\frac{R}{\sin(180^{\circ} - 	heta)} = \frac{w}{\sin(180^{\circ} - 	heta)}$, we immediately see that $\sin(180^{\circ} - 	heta) = \sin	heta$.

Thus, **$R = w$**. The reaction is simply the weight of the bead!

Now for the tension $T$. We use the following relation:

$$
\frac{T}{\sin(2	heta)} = \frac{w}{\sin	heta}
$$

Multiplying both sides by $\sin(2	heta)$, we get $T = \frac{w \cdot \sin(2	heta)}{\sin	heta}$. Here is where the magic happens.

We apply the double-angle identity: $\sin(2	heta) = 2\sin	heta\cos	heta$. Substituting this, we get:

$$
T = \frac{w \cdot 2\sin	heta\cos	heta}{\sin	heta}
$$

The $\sin	heta$ terms cancel out, leaving us with the elegant result: **$T = 2w\cos	heta$**.

You have done it! You have successfully navigated the geometry, the physics, and the algebra to find the solution. Keep this clarity of thought, and no problem will ever be too difficult for you.

Select Answer:

Visualized Solution

Visualizing the System

Geometric Analysis of $△ O A B$

Identifying Forces at Equilibrium

Calculating Angles for Lami's Theorem

Applying Lami's Theorem

Solving for Reaction $R$

Solving for Tension $T$ - Setup

Solving for Tension $T$ - Execution

Final Conclusion

The Sigma Insight: Trigonometric Ratios and Identities

The Dance of Equilibrium

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

Visualizing the System

Geometric Analysis of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OAB

Identifying Forces at Equilibrium

Calculating Angles for Lami's Theorem

Applying Lami's Theorem

Solving for Reaction .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Solving for Tension .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T - Setup

Solving for Tension .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T - Execution