A $2.5 kg$ body is suspended by two inextensible strings, XP and YP, inclined at $40^\circ$ and $50^\circ$ respectively to the downward vertical. We need to find the tension in both strings. We are given $g = 10 m/s^2$.

1. Problem Description

A $2.5 kg$ body is suspended by two inextensible strings, XP and YP, inclined at $40^\circ$ and $50^\circ$ respectively to the downward vertical. We need to find the tension in both strings. We are given $g = 10 m/s^2$.

2. Solution Steps

Let $T_1$ be the tension in string XP and $T_2$ be the tension in string YP. The angles they make with the vertical are $40^\circ$ and $50^\circ$ respectively.

The weight of the body is $W = mg = 2.5 \times 10 = 25 N$.

Since the system is in equilibrium, the sum of forces in the horizontal and vertical directions must be zero.

Horizontal components:

$T_1 \sin(40^\circ) = T_2 \sin(50^\circ)$

Vertical components:

$T_1 \cos(40^\circ) + T_2 \cos(50^\circ) = W = 25$

From the horizontal equation, we can write:

$T_1 = T_2 \frac{\sin(50^\circ)}{\sin(40^\circ)}$

Substitute this into the vertical equation:

$T_2 \frac{\sin(50^\circ)}{\sin(40^\circ)} \cos(40^\circ) + T_2 \cos(50^\circ) = 25$

$T_2 (\sin(50^\circ) \cot(40^\circ) + \cos(50^\circ)) = 25$

$T_2 (\sin(50^\circ) \frac{\cos(40^\circ)}{\sin(40^\circ)} + \cos(50^\circ)) = 25$

$T_2 = \frac{25}{\sin(50^\circ) \cot(40^\circ) + \cos(50^\circ)}$

Using a calculator, $\sin(50^\circ) \approx 0.766$ and $\cos(50^\circ) \approx 0.643$ and $\sin(40^\circ) \approx 0.643$ and $\cos(40^\circ) \approx 0.766$.

Therefore, $\cot(40^\circ) = \frac{\cos(40^\circ)}{\sin(40^\circ)} \approx \frac{0.766}{0.643} \approx 1.191$

$T_2 = \frac{25}{0.766 \times 1.191 + 0.643} = \frac{25}{0.912 + 0.643} = \frac{25}{1.555} \approx 16.077 N$

Now we can find $T_1$:

$T_1 = T_2 \frac{\sin(50^\circ)}{\sin(40^\circ)} = 16.077 \times \frac{0.766}{0.643} \approx 16.077 \times 1.191 \approx 19.148 N$

3. Final Answer

The tension in string XP is approximately $19.15 N$ and the tension in string YP is approximately $16.08 N$.