Question

A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.00-kg weight by a thin, light wire that passes without slippage over a frictionless pulley (the figure (Figure 1) ). The pulley has the shape of a uniform solid disk of mass 2.00 kg and diameter 0.520 m .

Part A

After the system is released, find the horizontal tension in the wire.

Part B

After the system is released, find the vertical tension in the wire.

Part C

After the system is released, find the acceleration of the box.

Part D

After the system is released, find magnitude of the horizontal and vertical components of the force that the axle exerts on the pulley.

Answer 1

Equation for vertical motion. Let the vertical tension is T1 and hanging mass m1 (5kg) moves downard with acceleration a then

If the horizontal tension is T2 the equation of motion for second mass (12kg) will be

Pully will be under the influence of vertical and horizontal tension hence its equation of motion will be

By adding all the three equation we get

(a) Horizontal tension

(b) Vertical tension

(c) acceleration is

(d) Since the pully is not in linear motion hence net force on it will be zero hence horizontal force will be equal to T2 but opposite of it and vertical force is equal to T1 and it will also be opposite of it i.e.