# A 1 kg mass on a frictionless inclined plane is connected by a pulley to a hanging 0.5kg mass,as in the diagram below.At what angle will the system be in equilibrium?

$\cos30^{\large\circ}=\sin 60^{\large\circ}=\large\frac{\sqrt 3}{2}$$,\cos 60^{\large\circ}=\sin 30^{\large\circ}=\large\frac{1}{2}$$,\cos 45^{\large\circ}=\sin 45^{\large\circ}=\large\frac{1}{\sqrt 2}$

$\begin{array}{1 1}0^{\large\circ}\\-30^{\large\circ}\\30^{\large\circ}\\45^{\large\circ}\end{array}$

Answer : $30^{\large\circ}$
The system will be in equilibrium when the net force acting on the 1 kg mass is equal to zero. A free-body diagram of the forces acting on the 1 kg mass shows that it is in equilibrium when the force of tension in the pulley rope is equal to $mg \sin\theta$ , where m = 1 kg and $\theta$ is the angle of the inclined plane.
Since the system is in equilibrium, the force of tension in the rope must be equal and opposite to the force of gravity acting on the 0.5 kg mass. The force of gravity on the 0.5 kg mass, and hence the force of tension in the rope, has a magnitude of 0.5 g. Knowing that the force of tension is equal to $mg \sin \theta$ , we can now solve for $\theta$ :
$mg\sin \theta=0.5g$
(1kg)$\sin \theta=0.5$
$\sin \theta=0.5$
$\theta=30^{\large\circ}$