What is the coefficient of friction?

AI Thread Summary
The discussion revolves around calculating the coefficient of friction for a 500kg box being pushed with an applied force of 8400N at a 35-degree angle below the horizontal. The key equation used is Ff = (coefficient of kinetic friction) * mass * gravity, noting that the applied force equals the friction force at constant velocity. It is emphasized that the applied force's magnitude does not directly equal the friction force, as only its horizontal component does. Additionally, the vertical component of the applied force affects the normal force, which is crucial for determining the coefficient of friction. The final answer for the coefficient of friction is identified as 0.71.
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Homework Statement



A 500kg box is being pushed across a floor, with an applied force of 8400N, at a constant velocity, at an angle of 35 degrees below the horizontal. What is the coefficient friction between the floor and the box.

Homework Equations



I really do not know where to start.

The Attempt at a Solution



I know that Ff= (coefficient of kinetic friction)*mass*gravity
and because it's at constant velocity Fp=Ff right?
the answer is 0.71
 
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You understand the most important part already.

It's also important to realize though, that the applied force magnitude does not equal friction force, but it is a component of applied force that equals friction force.

Also, we have a vertical component of the applied force that will effect your normal force.

Can you find the i and j components of this 8400N force applied at your 35* angle?
 
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