Numerical Problems Based on Equilibrium of Concurrent Forces for Class 11 Physics

Let $T$ be the tension in the upper half of the rope, making an angle $\theta$ with the vertical. The lower half of the rope simply supports the 10 kg mass, so the downward force at the midpoint is the weight of the block.

Downward force, $W = mg = 10 \times 10 = 100\text{ N}$

Horizontal force, $F = 60\text{ N}$

For the midpoint to be in equilibrium, we resolve the tension $T$ into horizontal and vertical components:

1. Vertical Equilibrium:

2. Horizontal Equilibrium:

Dividing Equation 2 by Equation 1:

Note: The solution uses the force value of $200\sqrt{3}\text{ N}$ as provided in the figure to match the equilibrium conditions for a 20 kg mass.

Let the angle the upper rope makes with the horizontal be $\theta$.

Downward force at midpoint, $W = mg = 20 \times 10 = 200\text{ N}$

Horizontal applied force, $F = 200\sqrt{3}\text{ N}$

Resolving the tension $T$ of the upper rope into components:

1. Vertical Equilibrium: (Since $\theta$ is with the horizontal, the vertical component is sine)

2. Horizontal Equilibrium:

Dividing Equation 1 by Equation 2:

From the figure, string 1 makes an angle of $30^\circ$ with the horizontal, and string 2 makes an angle of $45^\circ$ with the horizontal. The system is in equilibrium at the junction.

1. Horizontal Equilibrium:

The horizontal components of the tensions must balance each other.

2. Vertical Equilibrium:

The sum of the upward vertical components must balance the downward weight ($200\text{ N}$).

Substitute Equation 1 into the vertical equilibrium equation:

Now, substitute the value of $T_1$ back into Equation 1 to find $T_2$: