Question

A weight of 30 N is supported by two strings AB and AC of lengths 0.5 m and 1.2 m respectively. If BC is horizontal and of length 1.3 m, calculate the tension in AC and the angle BCA. A diagram is required with the solution.

Answer 1

The angle BCA:

$$\angle BCA=\arccos\dfrac{BC^2+AC^2-AB^2}{2BC·AC}=22.6°.$$

The tension in AC can be found from the equilibrium:

$$-W+T_1\cos\theta_1+T_2\cos\theta_2=0,\\ T_1\sin\theta_1=T_2\sin\theta_2:\\ T_1=27.7\text{ N},\\ T_2=11.5\text{ N}.$$