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 2.40 kg and diameter 0.520 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.Express your answers separated by a comma.

Explanation / Answer

mass of pulley ( solid disk) =2.4kg

radius= diameter/2 = 0.520 / 2 =0.26 m

moment of inertia=(1/2mr^2) = 0.5*2.4*(0.26)^2=0.08112 kgm^2

mass of box = M = 12 kg

acceleration of box= a

net force on box=ma

but net force on box =tension in horizontal portion of wire =Th

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

but net force on suspended weight =ma=5a

49 - Tv=5a

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

If alpha is angular acceleration of pulley,

alpha=linear acceleration/ radius =a/0.26

Net torque=[ Tv -Th]*r=[Tv-Th ]0.26

Net torque=[49 - 5a- 12 a ]0.26

Net torque=[49-17a]0.26

but net torque= I alpha= Ia /r=0.08112 a/0.26=0.312 a

0.312a=[49-17a]0.26--------(3)

a=2.692 m/s^2

Th=12a=32.307 N

(A) horizontal tension is 32.307 N

(B)vertical tension=Tv=49 -5a=35.54 N

vertical tension is 35.54 N

(C)acceleration of box is 2.692 m/s^2

(D)magnitude of the horizontal component of the force F that the axle exerts on the pulley=Th =32.307 N

magnitude of the vertical components of the force that the axle exerts on the pulley= Tv + weight of wheel= 35.54+19.6=54.14 N

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