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 1.90 kg and diameter 0 460 m After the system is released, find the acceleration of the box a = 2.73 m/s^2 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) = 1.90kg

radius= diameter/2 = 0.460 / 2 = 0.230 m

moment of inertia=(1/2mr^2) = 0.5*2*(0.230)^2 = 0.0529 kg-m^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 = 12a ......................(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/0.230

Net torque = [ Tv -Th]*r = [Tv - Th ]0.230

Net torque = [49 - 5a - 12a ]0.23

Net torque = [49 - 17a]0.230

but net torque = I alpha= Ia/r = 0.0529a / 0.230 = 0.230 a

0.230a = [49 - 17a]0.230 --------(3)

a = 2.7222 m/s^2

Th = 12a = 32.666 N

horizontal tension is 32.666 N

vertical tension = Tv = 49 - 5a = 35.388 N

vertical tension is 35.388 N

acceleration of box is 2.7222 m/s^2

magnitude of the horizontal component of the force F that the axle exerts on the pulley =Th = 32.666 N

magnitude of the vertical components of the force that the axle exerts on the pulley,

= Tv + weight of wheel = 35.388 + 18.62 = 54 N

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