A bar of uniform cross section and weighing 100 N is held horizontally using two massless and inextensible strings S1 and S2 as shown in the figure.

The tensions in the strings are

Correct Answer:
T₁ = 0 N and T₂ = 100 N

Step-by-Step Explanation:

1. Analysis of the Diagram:
Based on the provided diagram, we can identify the following components and details:

• A horizontal Bar of total length $L$ supported by a Rigid support at the top.
• Two vertical, massless, and inextensible strings, labeled S1 and S2, exert upward tensions $T_1$ and $T_2$ respectively.
• String S1 is attached to the leftmost end of the bar.
• String S2 is attached at the midpoint of the bar, at a distance of $\frac{L}{2}$ from the left end. The remaining segment to the right end is also labeled $\frac{L}{2}$.

2. Center of Gravity and Weight:
The bar is uniform and has a weight of $W=100\text{N}$. For a uniform bar, the weight acts vertically downwards through its center of gravity, which is located at its geometric center (midpoint), i.e., at a distance of $\frac{L}{2}$ from either end.

Notably, both the downward gravitational force ($W$) and the upward tension force from string S2 ($T_2$) act at this exact same midpoint.

3. Rotational Equilibrium:
For the bar to remain in static equilibrium, the net torque acting on it about any point must be zero. Let us calculate the torque about the midpoint of the bar:

$$\sum \tau =0$$

Taking torque about the midpoint (center of gravity):

• The weight $W$ acts at the midpoint, so its torque is zero: ${\tau }_W=0$.
• The tension $T_2$ acts at the midpoint, so its torque is zero: ${\tau }_{T_2}=0$.
• The tension $T_1$ acts at the leftmost end, at a perpendicular distance of $\frac{L}{2}$ from the midpoint. Its torque is: ${\tau }_{T_1}=T_1\times \frac{L}{2}$.

Setting the sum of torques about the midpoint to zero:
$$T_1\times \frac{L}{2}=0$$
Since the length $L\ne 0$, this directly gives:
$$T_1=0\text{N}$$

4. Translational Equilibrium:
For the bar to remain in translational equilibrium, the net vertical force acting on the bar must be zero:
$$\sum F_y=0$$
$$T_1+T_2-W=0$$
Substituting $T_1=0\text{N}$ and $W=100\text{N}$ into the equation:
$$0+T_2-100=0$$
$$T_2=100\text{N}$$

Therefore, the tension in string S1 is 0 N and the tension in string S2 is 100 N.