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 pulley has the shape of a uniform solid disk of mass 2.40 kg and diameter 0.420 m.

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

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

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

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

Answer 1

Given data:

Box m = 12kg

pulley m = 2.40 kg

2nd weight = 5 kg

r = 0.120 m

for 12kg

F_net = ma

F_net = 12*a = T_h

T_h = 12a

for 5kg

F_net = ma

F_net = 5*a

F_net = mg - T_v

mg - T_v = ma
T_v = 49.05 - 5a

for pulley

net torque = (T_v - T_h)*r = I

(T_v - T_h)*r = ½*m*r²*(a/r)

(T_v - T_h)*r = ½*m*r*a

T_v - T_h = ½*m*a

T_v - T_h = ½*2.4*a

49.05 - 5a - 12a = 1.2a

49.05 = 18.2a

a = 2.69 m/s²

A)

T_h = 12* 2.69 = 32.34 N

B)

T_v = 49.05N - 5kg*2.69m/s²

T_v = 35.57 N

C)

a = 2.69 m/s²

D)

in this case

F_X = T_h = 32.34 N

F_Y = T_v + m_p*g

F_Y = 35.57N + (9.81)(2.40)

F_Y = 59.11N