A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.0

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 0.400 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.

Express your answers separated by a comma.

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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 0.400 m .

Part A

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

Explanation / Answer

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 = (Tv - Th)*r = I = 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 = 2.70 m/s² part C

A) Th = 12kg * 2.70m/s² = 32.48 N

B) Tv = 49N - 5kg*2.70m/s² = 35.47 N

C)acceleration of box is 2.70m/s^2

D)magnitude of the horizontal component of the force F that the axle exerts on the pulley=Th =32.48 N

magnitude of the vertical components of the force that the axle exerts on the pulley=

Tv + weight of wheel= 35.47 + 21.56=57.03 N

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