A Moment Please: Help MiYu with the Test Question

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The discussion revolves around a test question concerning forces and moments related to a winch system. The user MiYu seeks clarification on how to approach the problem, which involves applying Newton's third law and balancing forces and torques. The analysis shows that the net force is zero, leading to the equations 200 + R2 - R1 = 0 and 200*40 - R1*2 - R2*2 = 0. After solving these equations, the results indicate R1 = 2100 and R2 = 1900. The explanation provided successfully clarifies MiYu's doubts.
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A "moment" please...

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.
 
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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+R2-R1 = 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*40-R1*2-R2*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
 
Gee thanks! I've understood! ^^
 
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