A body weighing 13 kg is suspended by two strings 5m and 12m long, their other ends being fastened t

Q: Why is the median to the hypotenuse equal to half the hypotenuse?

In any right-angled triangle, the midpoint of the hypotenuse is the circumcenter. This means the distance from the midpoint to all three vertices is equal. Thus, the median is equal to half the hypotenuse.

Q: Why do we use $\sin$ for horizontal and $\cos$ for vertical?

It depends on how you define your angles. In this solution, we defined angles $A$ and $B$ with respect to the vertical line $MC$. Therefore, the horizontal components involve the sine of these angles, and the vertical components involve the cosine.

Q: Can we use Lami's Theorem?

Absolutely! Lami's theorem is a powerful tool for three concurrent forces in equilibrium. You would just need to find the angles between the forces. However, resolving components is often less prone to calculation errors in these types of problems.

Question

A body weighing 13 kg is suspended by two strings 5m and 12m long, their other ends being fastened to the extremities of a rod 13m long. If the rod be so held that the body hangs immediately below the middle point, then tensions in the strings are

Options:

(A) 5 kg and 12 kg

(B) 5 kg and 13 kg

(C) 12 kg and 13 kg

(D) 5 kg and 5 kg

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a lab, looking at a simple setup: a rod of length $13$ m, with two strings of $5$ m and $12$ m attached to its ends, supporting a $13$ kg weight. It seems like a standard mechanics problem, but there is a hidden elegance here.

The numbers $5$, $12$, and $13$ are not random; they are a Pythagorean triplet. This means the strings and the rod form a perfect right-angled triangle.

By recognizing that $5^2 + 12^2 = 13^2$, we immediately know that the angle between the two strings is $90^\circ$. This geometric insight is the anchor for everything that follows.

### The Magic of the Midpoint

The problem states the body hangs directly below the midpoint of the rod. Let's call the midpoint $M$. Because the body is in equilibrium, the string $MC$ must be perfectly vertical.

Recall a beautiful property of right-angled triangles: the median to the hypotenuse is exactly half the length of the hypotenuse. Since the hypotenuse is $13$ m, the median $MC$ must be $6.5$ m.

This creates two isosceles triangles, $	riangle AMC$ and $	riangle BMC$, where $MA = MC = MB = 6.5$ m. This symmetry is the key to unlocking the angles.

### Resolving the Forces

At point $C$, we have three forces: the weight $W = 13$ kg acting downwards, and the tensions $T_1$ and $T_2$ acting along the strings. Since the system is in equilibrium, the net force must be zero.

We resolve these forces into horizontal and vertical components. For horizontal equilibrium, the horizontal pull of $T_1$ must balance the horizontal pull of $T_2$:

$$T_1 \sin A = T_2 \sin B$$

For vertical equilibrium, the sum of the vertical components of the tensions must support the weight:

$$T_1 \cos A + T_2 \cos B = W$$

### The Final Calculation

We derive the trigonometric ratios from our right-angled triangle: $\sin A = \frac{12}{13}$, $\cos A = \frac{5}{13}$, $\sin B = \frac{5}{13}$, and $\cos B = \frac{12}{13}$. Substituting these into our equilibrium equations, we get:

$$T_1 \left( \frac{12}{13} ight) = T_2 \left( \frac{5}{13} ight) \Rightarrow 12 T_1 = 5 T_2$$

Using the vertical equilibrium equation $T_1 \left( \frac{5}{13} ight) + T_2 \left( \frac{12}{13} ight) = 13$, we multiply by $13$ to obtain:

$$5 T_1 + 12 T_2 = 169$$

Substituting $T_2 = \frac{12}{5} T_1$ into this equation, we find:

$$5 T_1 + 12 \left( \frac{12}{5} T_1 ight) = 169$$

$$25 T_1 + 144 T_1 = 845$$

$$169 T_1 = 845 \Rightarrow T_1 = 5 	ext{ kg}$$

Consequently, $T_2 = 12$ kg. The elegance of the result—**$5$ kg and $12$ kg**—is a testament to the beauty of physics and geometry working in harmony.

Animated Solution for Mathematics - Vector Algebra: A body weighing 13 kg is suspended by two strings 5m and 12m long, their other ends being fastened to the extremities of a rod 13m long. If the rod be so held that the body hangs immediately below the middle point, then tensions in the strings are

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Visualized Solution

The Geometric Setup: Right-Angled Triangle $△ A B C$

The Midpoint $M$ and Median $M C$

Isosceles Triangles $△ A M C$ and $△ B M C$

Finding $sin$ and $cos$ of Angles $A$ and $B$

Horizontal Force Equilibrium at Point $C$

Relating $T_{1}$ and $T_{2}$

Vertical Force Equilibrium at Point $C$

Calculating the Final Tensions

The Sigma Insight: Components of a Vector

Analyzing the Setup

The Magic of the Midpoint

Resolving the Forces

The Final Calculation

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Animated Solution for Mathematics - Vector Algebra: A body weighing 13 kg is suspended by two strings 5m and 12m long, their other ends being fastened to the extremities of a rod 13m long. If the rod be so held that the body hangs immediately below the middle point, then tensions in the strings are

Select Answer:

Visualized Solution

The Geometric Setup: Right-Angled Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ABC

The Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M and Median .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MC

Isosceles Triangles .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △AMC and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △BMC

Horizontal Force Equilibrium at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​

Vertical Force Equilibrium at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C