What is the coefficient of friction?

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SUMMARY

The coefficient of friction between the floor and a 500kg box being pushed with an applied force of 8400N at a 35-degree angle below the horizontal is calculated to be 0.71. The applied force is balanced by the frictional force due to the box moving at a constant velocity. It is crucial to understand that the friction force is not equal to the total applied force but rather a component of it, influenced by the vertical component affecting the normal force.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of force components
  • Knowledge of friction and its coefficients
  • Basic algebra for solving equations
NEXT STEPS
  • Study the calculation of force components using trigonometry
  • Learn about the relationship between normal force and friction
  • Explore the differences between static and kinetic friction
  • Investigate real-world applications of friction in mechanical systems
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of friction calculations in real-world scenarios.

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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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