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Homework Help: Kinetic Friction Question

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data

    A crate on a floor has an applied force (12 degrees below horizontal) of 524N. The crate moves with constant velocity. The coefficient of kinetic friction is 0.15. What is the mass of the crate?

    2. Relevant equations

    Fk=uk*Fn, Fn= m * g

    3. The attempt at a solution

    OKay so this is what I did. Apparently it is not right, my teacher said it is 80% right. Can someone please tell me what I am doing wrong. As soon as possible would be really helpful. Thanks

    Fn= 524cos12
    = 512.5493428

    m= ukFn/ g
    = (0.15)(512.5493428N)/(9.81m/s^2)
    = 7.8kg
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 16, 2009 #2
    Can someone please help me
  4. Nov 16, 2009 #3


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    You are forgetting to include the weight of the crate in your calculations. Draw a free body diagram.
  5. Nov 16, 2009 #4
    but the question is asking for the weight...is it not?
  6. Nov 16, 2009 #5


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    It is asking for the mass. In your work you have the normal force based on the applied force only, you must also include the weight of the crate, that is acting downwards.
  7. Nov 16, 2009 #6
    So that would be the Fy value? Which is 524Sin12= 108.945726 then I add this value to the Fx value? So,

    F(normal force) = 108.945726+512.5493428 =621.5 N ?
  8. Nov 16, 2009 #7


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    No, you are just using the same force for everything, that makes no sense.

    1. Do you know what the normal force represents?

    2. You need to identify ALL of the forces acting on the crate. You have the normal force, the applied force (which you must resolve into components, like you've done), you have friction, and you have the weight. You need to sum up the forces in both directions (horizontal and vertical) to get two equations. The "vertical" equation will give you an expression for the normal force. You can then use this expression in your "horizontal" equation so you can solve for the mass of the crate.

    3. Draw a diagram with all the forces. There should be examples of this in your textbook.
  9. Nov 16, 2009 #8
    I do have a free body diagram that my teacher said is correct. I know normal force =mg. But if I don't have the mass how can I find the normal force? I know I have to substract the frictional force but again I need the mass to figure this out. Right?
  10. Nov 16, 2009 #9
    I don't know. I don't understand. Thanks anyway.
  11. Nov 16, 2009 #10


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    The normal force does not equal just mg, though. This is because you have the applied force there as well. You need to add up ALL of the forces in the vertical direction.
    So N - mg - Fy = 0
    where Fy is the vertical component of the 524 N force.

    Does that help?
  12. Nov 16, 2009 #11
    I think so...524-109=mg?
  13. Nov 16, 2009 #12
    oh! F(normal force) = mg = 524-109
  14. Nov 16, 2009 #13


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    You don't know N directly. Maybe my notation is confusing you. When I wrote "N" in that equation, I meant Fn [F(normal force)]. Is that what you are used to? Fn is not 524 Newtons. 524 Newtons is the applied force. They are not the same thing.

    Looking at your free body diagram, you should have Fn upwards, Fy downwards, and mg downwards. You are correct that Fy = 109 Newtons.

    You want to express this as Fn = Fy + mg for now.

    The second thing you need to do is look at the forces in the horizontal direction to get another equation. So what forces do you have there, based on your free body diagram? Add them up, like we did for the vertical direction.
  15. Nov 17, 2009 #14
    Ok this is what I have come up with:::

    F(friction) = 524cos12= 513N

    so now I have,

    F(friction) =(uk)(Fn)

    Fn= F(friction)/ (uk)
    = 513/0.15

    Weight= 3420- 524sin12


    Weight = mg


    ?????? I hope...
  16. Nov 18, 2009 #15
    Can anyone tell me if this is right?
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