Finding the Coefficient of Static Friction for a Concrete Block and Tongs

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SUMMARY

The discussion focuses on calculating the coefficient of static friction for a 500-N concrete block being lifted with tongs. The user initially applies the equations of equilibrium, specifically \(\Sigma F = 0\) and \(\Sigma M = 0\), to derive forces at points A and B. However, the user encounters discrepancies in their calculations, ultimately arriving at an incorrect coefficient of static friction value of 3. The forum contributors emphasize the importance of clearly defining the body in equilibrium and accurately identifying the point about which moments are taken to avoid errors in the calculations.

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Homework Statement



A 500-N concrete block is to be lifted by the pair of tongs shown. Determine the smallest allowable value of the coefficient of static friction between the block and the tongs at F and G.

See Diagram: [PLAIN]http://img192.imageshack.us/img192/1315/sprob826.jpg


Homework Equations


Fm=\musN
\SigmaF=0
\SigmaM=0

The Attempt at a Solution



Well,

deframing everything
\SigmaFy=0, so 500N+Ay+By=0
\SigmaFx=0, Ax+Bx=0

Then I take the moments about A and B in the frame AB and I get Ay = 250 and By =250 . Then I say, Ff=250=\muNf. Nf= 250/\mu.

Then I take the moment about one of the frames because they're symmetrical. So \SigmaMa=0,(250)(.180)-(Nf)(.540)=0. Then I get Nf=83.33, so then \mus= 250/83.33 = 3. But the thing is, that isn't the answer. I have the answer only but I can't seem to arrive at it. What am I doing wrong?

Thanks!
 
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You didn't clearly identify the body for which you are making equilibrium statements, and whatever that body is, you need to draw its diagram, and put on it every force that exists. Your phrase "taking moments about one of the frames" is sloppy, and probably led to your error. About which axis (or POINT in two dimensions) are you taking moments?
 

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