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

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To determine the smallest allowable coefficient of static friction for lifting a 500-N concrete block with tongs, equilibrium equations and moments were applied. The calculations initially led to a frictional force of 250 N and a normal force of 83.33 N, suggesting a coefficient of static friction of 3. However, this result was incorrect, indicating a potential error in the equilibrium analysis or moment calculations. The discussion emphasizes the importance of clearly identifying the body in equilibrium and accurately drawing the force diagram. Properly defining the axis or point for moment calculations is crucial to avoid mistakes in the solution process.
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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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