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posted by  bookworm2 on 11/2/2007 5:50:27 PM  |  status: Closed  

torque?

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Question Details:

The drawing shows a bicycle wheel resting against a small step whose

height is h = 0.120 m. The weight and radius of the wheel are W = 25.0 N and

r = 0.340 m. A horizontal force is applied to the axle of the wheel. As the

magnitude of increases, there comes a time when the wheel just begins to

rise up and loses contact with the ground. What is the magnitude of the force

when this happens?

 
 
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posted by Jimbo84 on 11/2/2007 6:14:48 PM  |  status: Live
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bookworm2's comment:
"Thanks so much!"
Response Details:
The force applied horizontally must cause a torque that will be equal to that caused by the wheel, first we must find the horizontal distance from the step to the center of the wheel (where the force of gravity is drawn)
 
By assembling a triangle we can find this value
0.34-0.12=0.22
from pythagorean theorem
L2=0.342-0.222=0.0672
L=0.26 m
 
Then we match the torques so that torque from gravity= torque from applied force
The horizontal distance from the applied force will be 0.34-0.12=0.22m
L1Fg=L2Fapplied
0.26m(25N)=0.22(F)
 
solve for F and you see that
 
F=
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