Dynamics - sled up a hill with several angles?

  • Thread starter optoracko
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  • #1
optoracko
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Homework Statement



A boy drags his sled of 90.0N at a constant speed up a 18 degree hill. He pulls with a 40.0N force on a rope which is 35 degrees above the horizontal.

a) What is coefficient of kinetic friction between the sled and the snow?

Homework Equations



Coefficieint = Force of Kinetic Friction / Normal Force

The Attempt at a Solution



I can find the normal force, no problem. It's just the kinetic friction I'm having trouble finding. I would think that the kinetic friction would be whatever the x component of the force applied is, since that is what is opposing the sled. However, if that was true, the Force net of the x prime would be 0. How would this be the kinetic friction then? Wouldn't it be for static since there is no movement?

I thought of doing this and checked yahoo answers, a response had a similar method. I just don't understand why to do this though.
 

Answers and Replies

  • #2
JaWiB
285
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I'm not sure I understand. What is your coordinate system? Is the x-direction parallel to the slope, or parallel to the horizontal? What is x prime?
 
  • #3
optoracko
18
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I'm not sure I understand. What is your coordinate system? Is the x-direction parallel to the slope, or parallel to the horizontal? What is x prime?
The x direction is parallel to the slope. X basically the x direction (the slope, back and forth). Don't know why I said prime now, lol.
 
  • #4
JaWiB
285
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Ok, well first of all the x-component of the pulling force isn't equal to the friction force because gravity also has a component in the x-direction. Second, zero net force means zero acceleration, not necessarily no movement.
 

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