Pushing books - force, kinetic and static friction

  • Thread starter mybrohshi5
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  • #1
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Homework Statement



A woman attempts to push a box of books that has mass 36.5 kg up a ramp inclined at an angle 27.5 degrees above the horizontal. The coefficients of friction between the ramp and the box are mu_s and mu_k. The force F applied by the woman is horizontal.

Assume that mu_s is 0.860 and that mu_k is 0.320. What magnitude of force must the woman apply to keep the box moving up the ramp at constant speed?

Homework Equations



acceleration = 0

w = 36.5(9.8) = 357.7

μs = fs/n

μk = fk/n

n - 357.7cos(27.5) - Fsin(27.5) = 0

n=Fsin(27.5)+357.7cos(27.5)

Fcos(27.5) - 357.7sin(27.5) - fs = 0

(fs = μs*n)

The Attempt at a Solution



I know μs= [Fcos(27.5) - 357.7sin(27.5)] / [357.7cos(27.5) + Fsin(27.5)]

I tried solving for F and i got 219.8 but that is wrong and i think it may have to do with something about not factoring in the μk.

Any suggestions?

Thank you
 
Last edited:

Answers and Replies

  • #2
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Constant speed up the ramp. Fnet= zero

So she applies a horizontal force to push the book up at constant velocity. Some of this horizontal force is used to push the books in the direction of the ramp.

Since the object is moving, you use kinetic friction coefficient. There is frictional force and weight force opposing her hand.
 
  • #3
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so this is what i did but im not sure how to use the kinetic friction coefficient....

Fcos(27.5) - 357.7sin(27.5) - 0.86(357.7cos(27.5)) = 0

i solved for F = 493.8

now how do i use this with the kinetic friction force to get the force i need?

or do i not even need this and am on the wrong track?

thank you
 
Last edited:
  • #4
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Fcos(27.5) - 357.7sin(27.5) - 0.86(357.7cos(27.5)) = 0

That looks good, except that you should use the coefficient for kinetic. Don't use static at all in that equation. Because the block is already moving, and it is only feeling kinetic friction.

If the question asked.....what is the minimum force needed by the woman to hold the block at rest. We can imagine that the block is at rest and we would static friction.

It always depends on the scenario, so picture the motion (if there is any) of the block and go from there.
 
  • #5
365
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My online homework is telling me its wrong.

This is what i got.

F= [ 0.32(357.7cos(27.5)) + 357.7sin(27.5) ] / cos(27.5)

F = 266.7/cos(27.5)

F = 300.7

That is wrong though. Can you see anything wrong with it?

Thank you
 
  • #6
14
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Opps, forgot that she applies a horizontal force that creates an additional force between the block and the ramp surface. The normal force is larger, and so the frictional force is larger.
 
  • #7
365
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I think i know what is wrong. can anyone please check this?

The normal force isnt just the cos component of Fg it is also the sin component of the force being pushed on the box and that is what i didnt include....

Fcos(27.5) - 357.7sin(27.5) - 0.32(Fsin(27.7) + 357.7cos(27.5)) = 0

Fcos(27.5) - 165.2 - 0.32Fsin(27.5) - 101.5 = 0

solve for F

Fcos(27.5) - 0.32Fsin(27.5) = 266.7

F(cos(27.5) - 0.32sin(27.5)) = 266.7

F = 266.7 / (cos(27.5) - 0.32sin(27.5))

F = 266.7 / 0.73925

F = 360.8 N

Does that look right?

Thanks for any help :)
 
  • #8
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I think that is what i just did in my calculations. can you look it over for me and see if it looks ok?
 
  • #9
14
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Fcos(27.5) - 357.7sin(27.5) - 0.32(357.7cos(27.5)) -0.32Fsin27.5 =0

Something like that, check the sin/cos stuff.
 
  • #10
14
0
Yeah, i got the same answer. 360.76N, hopefully they accept that.
 
  • #11
365
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i will try it. Thanks for all the help. i appreciate it :)
 
  • #12
365
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That was correct. Thank you
 

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