What is the coefficient of friction question?

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SUMMARY

The discussion focuses on calculating the coefficient of friction for a box weighing 300 Newtons being pulled with a tension of 100 Newtons at a 30-degree angle. The applied horizontal force is determined to be 86.6 Newtons, which equals the force of friction since the box moves at a constant velocity. The relationship between the force of friction and the coefficient of friction is established as Ff = μmg, leading to the equation 86.6N = μ(300N). Solving this yields the coefficient of friction (μ).

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  • Understanding of free body diagrams
  • Knowledge of Newton's laws of motion
  • Familiarity with trigonometric functions (specifically cosine)
  • Basic principles of friction and its calculation
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Homework Statement




a boy uses a rope to pull a box weighing 300 Newtons across a surface with a constant velocity . The rope makes an angle of 30 degrees with the horizontal, and the tension in the rope is 100 Newtons. What is the coefficient of friction ??


The Attempt at a Solution



300 Newtons - 100 Newtons = 200netowns

200 Newtons x Cos30 = 173 ?




is this the correct way to go about it ?
 
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Well the easiest way to solve these kind of problems is to draw out a free body diagram. The thing to know about friction is that it only opposes the force parallel to the motion. Since the rope is at a 30 degree angle, the horizontal applied F= (100N)cos(30) = 86.6 N.
Now since the box is moving at constant speed the Force of friction must be equal to the applied force. So Ff = 86.6 N in opposite direction. Then you must know Ff = umg where mg is the weight. 86.6N = u(300N) and solve for the coefficient of friction
 

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