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 2.20 kg and diameter 0.520 m .After the system is released, find the horizontal tension in the wire.After the system is released, find the vertical tension in the wire.After the system is released, find the acceleration of the box.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

Let's call the horizontal tension Th and the vertical tension Tv.

fbd for the 12 kg mass gives us

Fnet = ma = 12kg * a = Th

Dropping units, we have Th = 12a

fbd for 5 kg mass gives us

Fnet = ma = 5kg * a = mg - Tv = 5kg * 9.8m/s² - Tv

Tv = 49 - 5a

fbd for pulley gives us

net torque tau = (Tv - Th)*r = I*alpha = ½mr²(a/r) = ½mra ? r cancels

Tv - Th = ½ma = ½ * 2.2kg * a = 1.1kg * a

Plug in for Tv and Th:

49 - 5a - 12a = 1.1a

49 = 18.1a

a = 2.707 m/s² part C

A) Th = 12kg * 2.707m/s² = 32.49 N

B) Tv = 49N - 5kg*2.707m/s² = 35.46 N

D) The horizontal and vertical components are simply Th and Tv.

Horizontal = Th = 32.49N

Vertical = Tv = 35.46N