A 10 kg body P is in equilibrium under the action of its weight, a tension T in a string at 60 degrees to the horizontal, and a horizontal force F. Calculate the tension T.

The body P is in equilibrium, meaning the net force acting on it is zero. The forces acting on body P are:

• The weight of the body (W) acting vertically downwards.
• The tension (T) in the string, acting upwards and to the right at an angle of $60^\circ$ to the horizontal.
• A horizontal force (F) acting to the left.

Given mass $m = 10 \, \text{kg}$ and acceleration due to gravity $g = 10 \, \text{m/s}^2$.

$$W = mg$$ $$W = 10 \, \text{kg} \times 10 \, \text{m/s}^2$$ $$W = 100 \, \text{N}$$

The tension T acts at an angle of $60^\circ$ to the horizontal.

• Vertical component of tension: $T_y = T \sin(60^\circ)$ (upwards)
• Horizontal component of tension: $T_x = T \cos(60^\circ)$ (to the right)

Since the body is in equilibrium, the sum of vertical forces must be zero. The upward vertical component of tension must balance the downward weight.

$$\sum F_y = 0$$ $$T \sin(60^\circ) - W = 0$$ $$T \sin(60^\circ) = W$$

Substitute the value of W and $\sin(60^\circ)$.

$$\sin(60^\circ) = \frac{\sqrt{3}}{2}$$ $$T \left(\frac{\sqrt{3}}{2}\right) = 100 \, \text{N}$$ $$T = \frac{100 \, \text{N}}{\frac{\sqrt{3}}{2}}$$ $$T = \frac{200}{\sqrt{3}} \, \text{N}$$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$:

$$T = \frac{200 \sqrt{3}}{3} \, \text{N}$$

Using $\sqrt{3} \approx 1.732$:

$$T \approx \frac{200 \times 1.732}{3} \, \text{N}$$ $$T \approx \frac{346.4}{3} \, \text{N}$$ $$T \approx 115.47 \, \text{N}$$

The tension in the string is $\text{115.47 N}$.