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 2.20kg and diameter 0.600m

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 magnitude of the horizontal and vertical components of the force that the axle exerts on the pulley.

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 2.20kg and diameter 0.600m 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 magnitude of the horizontal and vertical components of the force that the axle exerts on the pulley.

Explanation / Answer

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 ). The pulley has the shape of a uniform solid disk of mass 2.30kg and diameter 0.580m.



(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 magnitude of the horizontal and vertical components of the force that the axle exerts on the pulley.

Answer

horizotal tension =T1

vertical tension =T2

5g-T2 =5a

hence T2 =5g-5a .......1

(T2-T1)r =I?

? =a/r

ie(T2-T1)*0.29 =[(2.3)0.29^2]/2 *(a/0.29)

hence T2-T1 =1.15a ......2

also T1 =12a .....3

from 1 and 2 and 3

we get 5g -5a -12a =1.15a

hence a =2.7 m/sec^2

hence T1 =12*2.7 =32.4 N

T2 =5g-5a =35.5 N

d) horizontal component of force exerted by axel is same as the horizontal component of tension=32.4

vertical component oforce exerted by axel is same as the vertical component of tension =T2 +2.3g = 58.04 N

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