Friction of a sliding box on a slope

  • Thread starter J-dizzal
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  • #1
J-dizzal
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Homework Statement


A loaded penguin sled weighing 75 N rests on a plane inclined at angle θ = 21° to the horizontal (see the figure). Between the sled and the plane, the coefficient of static friction is 0.30, and the coefficient of kinetic friction is 0.18. (a) What is the minimum magnitude of the force F , parallel to the plane, that will prevent the sled from slipping down the plane? (b) What is the minimum magnitude F that will start the sled moving up the plane? (c) What value of F is required to move the sled up the plane at constant velocity?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c06/fig06_29.gif

Homework Equations


fss Fn is the magnitude of the maximum static force.
fkk Fn is the kinetic friction force.

The Attempt at a Solution


http://s1294.photobucket.com/user/jhoversten/media/20150626_193750_zpsvqv6akeu.jpg.html [Broken]

i can see why my attempt at solving a is not working. thanks[/B]
 
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Answers and Replies

  • #2
Nathanael
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You did it right, except you miscalculated FT and FN

The angle is 21 degrees not 20 degrees.
 
  • #3
J-dizzal
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opps thanks
 
  • #4
J-dizzal
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You did it right, except you miscalculated FT and FN

The angle is 21 degrees not 20 degrees.
im getting 4.646 N but this is still wrong.
 
  • #5
Nathanael
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im getting 4.646 N but this is still wrong.
That's not what I get. Did you correct both FT and FN?
 
  • #6
J-dizzal
394
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That's not what I get. Did you correct both FT and FN?
oh yea i forgot to correct the right side of my equation. thanks
 
  • #7
J-dizzal
394
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for part b; i set up and equation identical to the equation in part a except is used μk and got an answer 14.276 N. And this would be the force required to start the sled moving up the ramp after the static force was applied. Im not sure how to set it up

i added the force from part a to part b and that was not good.
 
  • #8
J-dizzal
394
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That's not what I get. Did you correct both FT and FN?
i initiall thought it would be the same answer as part a, because once the block is broken free from gravitational forces and static frictional force it should be ready to move with any increase in force applied right?
 
  • #9
Nathanael
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for part b; i set up and equation identical to the equation in part a except is used μk
μk does not come into play yet. You want to find the force F which puts it on the verge of slipping upwards, so you will still want to use μs

i initiall thought it would be the same answer as part a, because once the block is broken free from gravitational forces and static frictional force it should be ready to move with any increase in force applied right?
It's similar to part a but it's not the same:
Friction opposes the motion of the block; in part a, the block was about slip downwards. In part b, the block is about to slip upwards.
 
  • #10
J-dizzal
394
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μk does not come into play yet. You want to find the force F which puts it on the verge of slipping upwards, so you will still want to use μs


It's similar to part a but it's not the same:
Friction opposes the motion of the block; in part a, the block was about slip downwards. In part b, the block is about to slip upwards.
oh right friction is a non conservative force so it works in both directions.
 
  • #11
J-dizzal
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oh right friction is a non conservative force so it works in both directions.
Thanks Nathanael
 
  • #12
Nathanael
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No problem
 

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