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 2.30 kg and diameter 0.460 m

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

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

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

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

Explanation / Answer

let the acceleration of the blocks is a

Now, using the second law of motion and effective mass

acceleration , a = net force/effective mass

a = 5*9.8 /( 5 + 12 + 0.5 * 2.30 * r^2 /r^2)

a = 2.7 m/s^2

1)

now, let the horizontal tension is Th

for 12 kg block

Th = 12 * 2.7

Th = 32.4 N

horizontal tension is 32.4 N

2)

Now, for 5 kg block

5 * 9.8 - Tv = 5 * 2.7

Tv = 35.5 N

vertical tension is 35.5 N

3)
acceleration of the box = 2.7 m/s^2

4)

for the pulley , as the pulley is in rest ,

magnitude of the horizontal = Th

magnitude of the horizontal = 32.4 N

magnitude of the vertical = Tv

magnitude of the vertical = 35.5 N

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