
#1
Dec2803, 11:15 PM

P: 6

Hello everyone!!!
Well, I found this moment question in a test paper...so if you know how to answer it please do help me clarify my doubts! Pls visit: http://sg.geocities.com/be_do_get/Qn3a.doc It takes a while to load, so be patient! Thank you! Love, MiYu. 



#2
Dec2903, 12:25 AM

P: 125

The pegs are fixed relative to the winch. When a force (effort) is applied, the handle bar pushes against peg A to the left, so peg A exerts an opposite and equal force to the right (Newton's third law). Similarly, the handle bar pushes against peg B to the right, so peg B exerts an opposite and equal force to the left. R1 and R2 are in opposite directions because one is above the center of the winch while the other is below the center of the winch.
The center of the winch does not undergo any translational motion. Hence the net force is 0. 200+R2R1 = 0. Also balance the torques (i.e., the "moments") since there is no rotation about the center of the winch unless the exerted force > 200. Let torques that point out of the page be +, and those that point into the page be . 200*40R1*2R2*2 = 0. Solve[{200 + R2  R1 == 0, 200*40  R1*2  R2*2 == 0}, {R1, R2}]{{R1 > 2100, R2 > 1900}} Solving these two equations (on Mathematica), I get R1 = 2100 and R2 = 1900. Hope this helps! Ying 



#3
Dec2903, 08:24 AM

P: 6

Gee thanks! I've understood! ^^



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