A wedge M of mass 35 ~kg has a vertical left face and a right face inclined at an angle of 37^ to the horizontal. A block m₂ of mass 5 ~kg is placed against the vertical left face, and a block m₁ of mass 10 ~kg is placed on the inclined right face. The two blocks are connected by a light, inextensible string that passes over a smooth, light pulley fixed at the top of the wedge. A horizontal force F is applied to the

Step-by-step solution

Let a be the acceleration of the wedge to the left. Since m₁ moves down the incline with relative acceleration a_r = 2 ~m/s^2 , m₂ must move up the vertical face with the same relative acceleration a_r . For block m₂ in the vertical direction: T - m₂ g = m₂ a_r T - 50 = 5(2) T = 60 ~N For block m₁ along the incline: In the frame of the wedge, a pseudo force m₁ a acts on m₁ towards the right. Its component down the incline is m₁ a 37^ . Equation of motion for m₁ down the incline: m₁ g 37^ + m₁ a 37^ - T = m₁ a_r 10(