Exercise 10.16 Part A A 12.0-kg box resting on a horizontal, frictionless surfac
ID: 1452884 • Letter: E
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Exercise 10.16 Part A 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 2.50 kg and diameter 0.440 m After the system is released, find the horizontal tension in the wire. Submit My Answers Give Up Part B After the system is released, find the vertical tension in the wire. Submit My Answers Give Up Part C After the system is released, find the acceleration of the box. m/s2 Submit My Answers Give Up Figure 1 01 Part D 12.0 kg 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. 5.00 kg Submit My Answers Give UpExplanation / Answer
mass of pulley ( solid disk) =2.50kg
radius= diameter/2 = 0.440 / 2 =0.22 m
moment of inertia=(1/2mr^2) = 0.5*2.5*(0.22)^2=0.0605 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/0.22
Net torque=[ Tv -Th]*r=[Tv-Th ]0.22
Net torque=[49 - 5a- 12 a ]0.22
Net torque=[49-17a]0.22
but net torque= I alpha= Ia /r=0.0605 a/0.22=0.275 a
0.275a=[49-17a]0.24-----------(3)
a=2.7 m/s^2
Th=12a=32.4 N
(A) horizontal tension is 32.4 N
(B)vertical tension=Tv=49 -5a=35.5 N
vertical tension is 35.5 N
(C)acceleration of box is 2.7 m/s^2
(D)magnitude of the horizontal component of the force F that the axle exerts on the pulley=Th =32.4 N
magnitude of the vertical components of the force that the axle exerts on the pulley= Tv + weight of wheel= 35.5+ 14.7=50.2 N
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