Question involving coefficient of friction and force of friction

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Homework Statement


"A 1250 kg slipper hippo slides down a mud covered hill inclined at an angle of 18 degrees to the horizontal. A) If the coefficient of friction is 0.0900, what force of friction impedes the hippo's motion down the hill? B) If the hill were steeper, how would this affect the coefficient of sliding friction?"

Homework Equations



mg x sin(θ)
mg x cos(θ)
ƩF = ma
μmg x sin(θ)
μmg x cos(θ)


The Attempt at a Solution



Our professor/teacher taught us by just using mg x sin(θ) and plugging in the values. Someone in the class used the formula μmg x sin(θ) from a physics book and got a completely different answer than the teacher. So my main question is: Does coefficient of friction affect the force of friction and if so, would I use μmg x sin(θ)? I am confused because if the coefficient of friction doesn't affect the force of friction why bother list it in the problem.
 
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Answers and Replies

  • #2
Doc Al
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So my main question is: Does coefficient of friction affect the force of friction
Of course.
and if so, would I use μmg x sin(θ)?
No. The kinetic friction force is μ*N, where N is the normal force. What's the normal force equal? (You need the normal component of the weight.)
 
  • #3
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No. The kinetic friction force is μ*N, where N is the normal force. What's the normal force equal? (You need the normal component of the weight.)
If I'm correct, the normal force is equal to the weight which equals Mass x Gravity. Therefore wouldn't μ(mg) be correct? The sin portion is in there due to an angle in the problem statement.

And a classmate just told me that the force of friction is only affected by the coefficient of friction when there is an acceleration value involved. Due to this problem not issuing a value for acceleration and our teacher notifying us of it being constant velocity earlier, does that coefficient even matter in this problem?
 
  • #4
Doc Al
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If I'm correct, the normal force is equal to the weight which equals Mass x Gravity. Therefore wouldn't μ(mg) be correct? The sin portion is in there due to an angle in the problem statement.
The normal force is equal to the component of the weight perpendicular to the surface, which is mgcosθ. Thus the friction will equal μmgcosθ. (mgsinθ is the component of the weight parallel to the surface.)
And a classmate just told me that the force of friction is only affected by the coefficient of friction when there is an acceleration value involved.
That's incorrect. The kinetic friction is proportional to the coefficient of friction.
 
  • #5
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The normal force is equal to the component of the weight perpendicular to the surface, which is mgcosθ. Thus the friction will equal μmgcosθ. (mgsinθ is the component of the weight parallel to the surface.)

That's incorrect. The kinetic friction is proportional to the coefficient of friction.
Okie dokie. Well thank you for solving my confusion. :wink:
 

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