How Much Force to Overcome Static Friction and Move a Crate?

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To move the crate weighing 875 N, the minimum force required must overcome static friction, calculated using the coefficient of static friction (0.56). The formula for static friction is Fp = \mus * n, where n equals the weight of the crate. This results in a necessary force of approximately 490 N to initiate movement. Once the crate is in motion, the coefficient of kinetic friction (0.47) applies, allowing for a smaller force to maintain movement. Understanding the transition from static to kinetic friction is crucial for solving the problem correctly.
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Homework Statement



The weight of the crate is 875 N. You push it with a force of magnitude 300 N but it doesn't move

What is the magnitude Fp of the minimum force you need to exert on the crate to make it start sliding along the floor? Let the coefficent of the static friction \mus betwen the crate and the floor be 0.56 and that of kinetic friction, \muk, be 0.47



Homework Equations







The Attempt at a Solution



I tried solving it by using the formula fk = \mukn but it told me that I'm going about the problem wrong. It says that I need to study the condition for which the crate ceases to be at rest, that is, the condition for which static friction ceases to be valid. I used kinetic friction though, so I don't know what to do now.
 
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Use the static coefficient of friction in your formula.
Once it has started moving, the lower dynamic coefficient comes into play and a smaller force will suffice to keep it moving.
 

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