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A 10.0-kg box resting on a horizontal, frictionless surface is attached to a 6.00-kg weight by a thin, light wire that passes over a frictionless pulley. The pulley has the shape of a uniform solid disk of mass 4.00kg and diameter 0.200m.

a) Find the acceleration of the box after the system is released.

b) Find the tension in the wire on both sides of the pulley.

c) Find the horizntal and vertical components of the force that axle exerts on the pulley.
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Added Mon, 29 Jun '15
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mass of pulley ( solid disk) =4.00kg

radius= diameter/2 = 0.20 / 2 =0.1 m

moment of inertia=(1/2mr^2) = 0.5*4*(0.1)^2=0.02 kgm^2

mass of box = M = 10 kg

acceleration of box= a

net force on box=ma

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

Th= 10 a..........................(1)

tension in vertical portion of wire = Tv

weight suspended =mg=6*9.8 =58.8 N

net force on suspended weight = 58.8 - Tv

but net force on suspended weight =ma=6a

58.8 - Tv=6a

Tv=58.8 -6a .........................(2)

If alpha is angular acceleration of pulley,

alpha=linear acceleration/ radius =a/0.1

Net torque=[ Tv -Th]*r=[Tv-Th ]0.1

Net torque=[58.8 - 6a- 10 a ]0.1

Net torque=[58.8-16a]0.1

but net torque= I alpha= Ia /r=0.02 a/0.1=0.2 a

0.2a=[58.8-16a]0.1--------(3)

a=3.2667 m/s^2

Th=10a=32.667 N

horizontal tension is 32.667 N

B) vertical tension=Tv=58.8-6a=39.1998 N

vertical tension is 39.1998 N

A) acceleration of box is 3.2667m/s^2

C) magnitude of the horizontal component of the force F that the axle exerts on the pulley=Th =32.667 N

magnitude of the vertical components of the force that the axle exerts on the pulley= Tv + weight of wheel= 39.1998+ 39.2=78.3998? N
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Added Mon, 29 Jun '15
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