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A 87.9 kg clock initially at rest on a horizontal floor requires a 659.2 N horizontal force to set it in motion.

After the clock is in motion, a horizontal force of 556.0 N keeps it moving with a constant speed.

Find the coefficient of static friction of the clock on the ground, µs

 May 9, 2014

Best Answer 

 #1
avatar+33657 
+5

The resistive force is given by μs*W, where W is the weight.  The resistance is overcome a force of 659.2N is applied, so 

μs*87.9*9.81 = 659.2

μs = 659.2/(87.9*9.81)

659.2(87.9×9.81)=0.7644680093563833

or μs ≈ 0.764

 May 9, 2014

1+0 Answers

 #1
avatar+33657 
+5
Best Answer

The resistive force is given by μs*W, where W is the weight.  The resistance is overcome a force of 659.2N is applied, so 

μs*87.9*9.81 = 659.2

μs = 659.2/(87.9*9.81)

659.2(87.9×9.81)=0.7644680093563833

or μs ≈ 0.764

Alan May 9, 2014

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