Part D Question 6 A g box resting on a horizontal, frictionless surface is attac

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Part D

Question 6 A g 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 1.70 kg and diameter 0.460 m. Figure 1 of 1 12.0 kg 5.00 kg Part B After the system is released, find the vertical tension in the wire 35.3 N My Answers Give Up Correct Part C After the system is released, find the acceleration of the box a 2.75 m/s My Answers Give Up Correct 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. AE Fe, F. 33,35.3 Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

mass of pulley ( solid disk) =1.70kg

radius= diameter/2 = 0.460 / 2 =0.23 m

moment of inertia=(1/2mr^2) = 0.5*1.7*(0.23)^2=0.0449kgm^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/0.23

Net torque=[ Tv -Th]*r=[Tv-Th ]0.23

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

Net torque=[49-17a]0.23

but net torque= I alpha= Ia /r=0.0449 a/0.23=0.19 a

0.19a=[49-17a]0.23-----------(3)

a=2.74 m/s^2

Th=12a=32.88 N

(A) horizontal tension is 32.88 N

(B)vertical tension=Tv=49 -5a=35.3 N

vertical tension is 35.3 N

(C)acceleration of box is 2.74 m/s^2

(D)magnitude of the horizontal component of the force F that the axle exerts on the pulley=Th = 32.88 N

magnitude of the vertical components of the force that the axle exerts on the pulley= Tv + weight of wheel= 35.3+ 33.3 = 68.6 N

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