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Pushing at a downward angle with friction

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A factory worker pushes a 32.0 kg crate a distance of 5.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.25. The factory worker pushes downward at an angle of 40° below the horizontal.
    What magnitude of force must the worker apply to move the crate at a constant velocity?



    2. Relevant equations
    Fn= mg+Fy
    fk=.25Fn
    Fy=Fsin(40)
    3. The attempt at a solution
    I spent a long time trying to figure this out, I know mg=313.6N and i need to add the force of the y component to that to find the normal force, but I don't know how to find Fy, or Fx. If there was no vertical push it would just be 78.4N in the x direction, which equals the force of friction. But with the vertical push it seems like the harder you push, the more friction increases, and thus you need to push harder. I saw other people have had the same question on this forum before, but with different numbers, but they don't help me because people only told them that the vertical push makes the normal force bigger, but I don't know how to calculate the vertical push.
     
    Last edited: Feb 7, 2009
  2. jcsd
  3. Feb 7, 2009 #2

    AEM

    User Avatar

    Would you believe that you have already written the equation that gives it to you???
    Before I tell you where it is, do you know how to break vectors into components? You use trig. --you know sines and cosines hypotenuses and opposite and adjacent sides and all that good stuff :smile:

    Up above you wrote [tex] F_y = F sin(40) [/tex] It seems you knew how to do it all along...
     
  4. Feb 7, 2009 #3
    But I don't know what Fy is or F, what I'm trying to find is F, if I could find Fy, Fx, fk or Fn then I would be able to find F
     
  5. Feb 7, 2009 #4

    AEM

    User Avatar

    Well, first since the velocity is constant, the acceleration is zero and so the sum of the forces in the x direction is zero. That implies that the frictional force is equal in magnitude to the applied horizontal force or,

    [tex] f = F cos(40) [/tex]

    But,as you know, the frictional force is also given by

    [tex] f = \mu_k(mg + Fsin(40) )[/tex]

    You know [tex] \mu_k [/tex] and m and g and you can set those two equations equal and get F.
     
  6. Feb 7, 2009 #5
    F=129.5N
    Thank You very much!

    I had been trying to find two equations like that which equaled each other so I would only have one variable.
     
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