Crate Friction on Horizontal Floor

AI Thread Summary
A user is calculating the maximum static friction force for a 35 kg crate being pushed with a 110 N force, using a coefficient of static friction of 0.37. They initially calculated the maximum static friction force as 127 N, but the book states the answer is 130 N. The discussion clarifies that the discrepancy is due to significant figures; the book rounded the answer to two significant digits. The user confirms their understanding and appreciates the reassurance regarding their calculations. This highlights the importance of significant figures in physics problems.
VinnyCee
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I have a problem with an answer that I don't know how to arrive at.

There is a 35 kg crate on a horizontal floor that a person is pushing with a 110 N force. The coefficient of static friction is 0.37. What is the maximum value f of s, max for which the crate will not move?

I thought that the answer is f of s, max = (coefficient of static frction)(35 kg)(9.8 m / s^2) = 127 N. However, the book lists an answer of 130 N. Am I going about his problem correctly?
 
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You're absolutely correct! f_{s, max} = \mu_s N = \mu_s mg Because there are only 2 significant digits in the values for m, \mu_s etc., the answer should be expressed with only 2 significant digits. You got an answer of 127 N which contains 3 significant digits. The book simply rounded up to 130 N to express the answer in 2 significant digits. Wasn't that simple? You were doing everything correctly!
 
Thank you for the reassurance! I don't think I would have gotten any further in the assignment stressing over that one little answer! :redface:
 
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