PROB 10.17 A 12.0-kg box resting on a horizontal, frictionless surface is attach

Question

PROB 10.17

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 1.80 kg and diameter 0.520 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.

Explanation / Answer

moment of inertia of disk is I = 0.5*m*R^2 = 0.5*1.8*(0.52/2)^2 = 0.06084 kg*m^2

Tension in the horizontal wire Th = net force acting on 12 kg box = m*a = 12*a

Th = 12*a.....................(1)

Tension in the vertical wire is Tv

weght suspended is w = 5*9.8 = 49 N

Net force on the suspended weight is = 49-Tv = 5*a

Then Tv = 49-5a.......(2)

angular accelaration alpha = a/r = a/(0.52/2) = a/0.28

Net torque is (Tv-Th)*r = (Tv-Th)*0.28

(49-5a-12a)*0.26 = I*alpha = 0.06084 *(a/0.26)

0.234*a+(17*0.26*a) = 49*0.26

4.654*a = 12.74

accelaration a = 12.74/4.654 = 2.73 m/s^2

Part A)

Th = 12*2.73 = 32.76 N

Part B) Tv = 49-(5*2.73) = 35.35 N

Part C) accelaration a = 2.73 m/s^2

Part D) Horizontal force is Th = 32.76 N

Vertical force is Tv+(1.8*9.8) = 35.35+(1.8*9.8) = 52.99 = 53 N

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