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 2.30kg and diameter 0.440m .

Part A

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

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Part B

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

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Part C

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

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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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Figure 1 of 1

Explanation / Answer

Applying equation of motion for 12 kg block

Th = M1 a = 12 a

For 5 kg block

M2 g - Tv = M2 a

Tv = M2 g - M2 a

= 5 * 9.8 - 5 * a

= 49 - 5 a

The torque acting on the pully is

(Tv - Th) R = 0.5 M R a

  Tv - Th = 0.5 M a

49 - 5a - 12a = 0.5 M a

49 - 17a = 0.5Ma

49 = a (0.5M + 17)

a = 49 / (0.5M + 17)

= 49 / (0.5M + 17)

= 49 / (0.5 *2.30 + 17)

= 2.699 m/s2

Th = 12 * 2.699 = 32.39 N

Tv = 49 - 5 * 2.699 = 35.505 N

The horzontal for acting on the pully is equivalent to its horzontal tension

Fx = Th = 32.39 N

The vertical force acting on the pully is

Fy = Tv + Mg

= 35.505 + 2.30 * 9.8

= 57.905 N

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