Crate Friction on Horizontal Floor

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SUMMARY

The discussion centers on calculating the maximum static friction force for a 35 kg crate on a horizontal floor being pushed with a 110 N force. The correct formula for maximum static friction is f_{s, max} = μ_s N = μ_s mg, where the coefficient of static friction (μ_s) is 0.37. The calculated maximum static friction force is 127 N, but the book rounds this to 130 N due to significant figures, as the values used only contain two significant digits. The participant confirms their method was correct but needed reassurance regarding the rounding convention.

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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! [tex]f_{s, max} = \mu_s N = \mu_s mg[/tex] Because there are only 2 significant digits in the values for [tex]m, \mu_s[/tex] 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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