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 2.50kg and diameter 0.420m .

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

mass of pulley ( solid disk) =2.50kg

radius= diameter/2 = 0.420 / 2 =0.210 m

moment of inertia=(1/2mr^2) = 0.5*2.5*(0.21)^2=0.0551 kgm^2

net force on box=ma

net force on box =tension in horizontal portion of wire

Th= 12* a                                 (1)

tension in vertical portion of wire = Tv

weight suspended =mg=5*9.8 =49 N

net force on suspended weight = 49 - Tv

net force on suspended weight =ma=5a

49 - Tv=5a

Tv=49 -5a                                      (2)

angular acceleration(alpha) =a/0.21

net torque= I *alpha= I*a/r=0.0551 a/0.21=0.262 a

also Net torque=[ Tv -Th]*r=[Tv-Th ]0.21 = [49 - 5*a- 12*a ]0.21

Net torque=[49-17*a]*0.21

0.262*a=[49-17a]*0.21

a=2.685 m/s^2

Th=12*a=32.22 N

vertical tension=Tv=49 -5a=35.575 N

D. magnitude of the horizontal component of the force F that the axle exerts on the pulley=

Th =32.22 N

magnitude of the vertical components of the force that the axle exerts on the pulley= Tv + weight of wheel= 35.575 + 2.5*9.8 = 60.075 N

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