How Do You Calculate the Coefficient of Kinetic Friction for a Sled?

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Homework Help Overview

The discussion revolves around calculating the coefficient of kinetic friction for a sled being pulled across a horizontal surface at a constant velocity. The sled has a mass of 20.0 kg, and the pulling force is 80.0 N at an angle of 30 degrees above the horizontal.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Newton's 1st Law and the use of a Free Body Diagram to analyze the forces acting on the sled. There are questions about understanding these concepts and how they relate to the problem. Some participants attempt to clarify the relationship between the applied force, normal force, and frictional force.

Discussion Status

There is an ongoing exploration of the correct approach to calculate the coefficient of kinetic friction. Some participants have provided guidance on considering the components of the applied force and the implications for the normal force. Multiple interpretations of the problem are being discussed, with some participants expressing confusion about the calculations.

Contextual Notes

Participants note discrepancies in the expected answer for the coefficient of kinetic friction, with one participant indicating that their calculations yield a different result than anticipated. There is acknowledgment of the need to adjust for the vertical component of the applied force affecting the normal force.

PhunWithPhysics
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Can some one help explain?
A 20.0kg sled is being pulled across a horizontal surface at a CONSTANT VELOCITY. The pulling force has a magnitude of 80.0 Newtons(N) and is directed at an angle of 30 degreesabove the horizontal. What is the coefficient of kinetic friction?
 
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Do a FreeBody Diagram and apply Newton's 1st Law.
 
what? can you explain?
 
Do you know Newton's 1st Law?

Do you know what is a Freebody diagram?
 
Since the object is dragged along the horizontal, the normal reaction will be equal and opposite to the object's weight (W = mg).

The force is applied at an angle to the horizintal, but the friction will only act across the horizontal, so you need to find the horizontal component of the force (Fx = F cos theta).

As the object moves at a constant velocity, there are no unbalanced forces acting on the object, so the friction force must be equal in magnitude to the pulling force.

The friction force of a sliding object is proportional to the coefficient of kinetic friction and the normal reaction, so:

coefficient = -(F cos theta) / -(mg),

F being the applied pulling force.
 
still not working

It says that I should come up with an answer of 0.444 but I keep getting 0.357, What is going wrong?
 
El Hombre Invisible said:
The friction force of a sliding object is proportional to the coefficient of kinetic friction and the normal reaction, so:

coefficient = -(F cos theta) / -(mg),

F being the applied pulling force.
Since the 80N force is applied at an angle, the normal force is not mg. The normal force is reduced by the vertical component of the applied force.
 
just write it out for him

Just write the equation out i don't think he understands!
 
Doc Al said:
Since the 80N force is applied at an angle, the normal force is not mg. The normal force is reduced by the vertical component of the applied force.
*^$*! Very true. Sorry, Phun, I forgot that part. You'll need to subtracted F sin theta from -mg. Thanks, Doc Al, for spotting my clumsiness.
 
  • #10
bloccubed said:
Just write the equation out i don't think he understands!
He does, I think, I just missed off part of the solution and confused him. Entirely my fault, not his.
 
  • #11
got it

Thats much better! Thanks guys!
 

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