Shown in the above figure, a rigid and uniform 1 m long rod AB held in horizontal position by two st

Q: Why does the weight of the rod act exactly at the 50 cm mark?

Because the problem states the rod is 'uniform'. This means its mass is distributed evenly along its length, so its center of mass (where gravity effectively acts) is exactly at its geometric center.

Q: Why did we choose point B as the pivot for calculating torque?

Torque is force multiplied by perpendicular distance. By choosing point B as the pivot, the distance for the force $T_B$ becomes zero. This makes its torque zero, effectively eliminating the unknown $T_B$ from our equation and allowing us to solve directly for $T_A$.

Q: Could I have chosen point A as the pivot instead?

Absolutely! If you choose point A, you will find the tension $T_B$ first. Once you have $T_B$, you can use translational equilibrium ($T_A + T_B = \text{Total Weight}$) to find $T_A$. It's just a two-step process instead of one.

Q: Do I need to convert the distances from centimeters to meters?

In this specific case, no. Because every term in the torque equation is multiplied by a distance in centimeters, the units will cancel out when you divide both sides. However, it is generally a good habit to work in standard SI units (meters) to avoid mistakes in more complex problems.

Q: What if the rod wasn't uniform?

If the rod wasn't uniform, its center of mass would not be at the 50 cm mark. The problem would have to provide the specific location of the center of mass, or give you a linear mass density function to calculate it using integration.

Question

Shown in the above figure, a rigid and uniform $1	ext{ m}$ long rod $AB$ held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is of mass $m$ and has another weight of mass $2m$ hung at a distance of $75	ext{ cm}$ from $A$. The tension in the string at $A$ is

Options:

(A) $1	ext{ mg}$

(B) $2	ext{ mg}$

(C) $0.75	ext{ mg}$

(D) $0.5	ext{ mg}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a perfectly rigid and uniform rod suspended horizontally from the ceiling by two strings attached to its ends, $A$ and $B$. The rod itself has a mass $m$. Because it is uniform, we can confidently say that its entire weight acts exactly at its geometric center. Since the rod is $100	ext{ cm}$ long, this weight vector $mg$ points straight down from the $50	ext{ cm}$ mark.

But that's not all! There is an additional mass of $2m$ hung at a distance of $75	ext{ cm}$ from end $A$. This creates another downward force of $2mg$ at that specific location. Our mission is to find the tension in the string at end $A$, which we will call $T_A$.

### The Master Equation: Rotational Equilibrium
Since the rod is perfectly horizontal and completely at rest, it is in a state of **rotational equilibrium**. This is a powerful physical constraint: it means the net torque (the turning effect of forces) about *any* point on the rod must be exactly zero.

$$\sum 	au = 0$$

To find $T_A$ directly and elegantly, we need to choose our pivot point wisely. If we choose point $B$ as our pivot, the torque produced by the tension $T_B$ becomes zero (since its distance from the pivot is zero). This brilliant choice eliminates $T_B$ from our equation entirely, leaving $T_A$ as the only unknown!

### Balancing the Torques
Let's set up our torque equation about point $B$. The tension $T_A$ is pulling up on the left end, trying to rotate the rod clockwise around $B$. The two downward weights, $mg$ and $2mg$, are trying to rotate the rod counter-clockwise around $B$.

For equilibrium, the clockwise torque must perfectly balance the counter-clockwise torque:

$$	au_{	ext{clockwise}} = 	au_{	ext{counter-clockwise}}$$

Now, let's calculate the lever arms (perpendicular distances) from our pivot $B$ (which is at the $100	ext{ cm}$ mark):
- Distance of $T_A$ from $B$ = $100	ext{ cm} - 0	ext{ cm} = 100	ext{ cm}$
- Distance of $mg$ from $B$ = $100	ext{ cm} - 50	ext{ cm} = 50	ext{ cm}$
- Distance of $2mg$ from $B$ = $100	ext{ cm} - 75	ext{ cm} = 25	ext{ cm}$

Plugging these into our torque balance equation:

$$T_A 	imes 100 = (mg 	imes 50) + (2mg 	imes 25)$$

### Final Calculation
Let's simplify the right side of our equation. The torque from the rod's weight is $50mg$. The torque from the suspended mass is $2mg 	imes 25 = 50mg$.

Adding them together gives us the total counter-clockwise torque:

$$T_A 	imes 100 = 50mg + 50mg$$

$$T_A 	imes 100 = 100mg$$

Dividing both sides by $100$, we arrive at our final, elegant answer:

**$$T_A = 1	ext{ mg}$$**

The tension in the string at $A$ is exactly equal to $1	ext{ mg}$.

Select Answer:

Visualized Solution

$Free Body Diagram$

$\sum τ_{B} = 0$

$Balancing Torques$

$Torque Equation$

$Simplification$

$Final Answer$

$Translational Equilibrium$

The Sigma Insight: Torque and Angular Momentum

Analyzing the Setup Imagine a perfectly rigid and uniform rod suspended horizontally from the ceiling by two strings attached to its ends, $A$ and $B$

The Master Equation

Balancing the Torques Let's set up our torque equation about point $B$

Final Calculation Let's simplify the right side of our equation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Free Body Diagram

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑τB​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Balancing Torques

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Torque Equation

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Translational Equilibrium