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Simple Question About Newton's Laws and Forces

  1. Jun 1, 2014 #1
    1. The problem statement, all variables and given/known data

    A 90 kg box is pushed by a horizontal force F at constant speed up a ramp inclined at 28°, as shown. Determine the magnitude of the applied force:
    a. when the ramp is frictionless.
    b. when the coefficient of kinetic friction is 0.18.

    2. Relevant equations

    F= ma, Ffr = μk * Fn

    3. The attempt at a solution

    Everywhere I look online I keep getting told that for part (a) I need to do sin28 * 882N (i.e. sin28 * mg) but I have NO idea why. Could somebody draw me a picture of the vector decomposition because I really really do not understand why you'd multiply.

    For part (b) my course hasn't even taught us about kinetic friction yet, but through looking around I've found out that Ffr = μk * Fn. So once I've found the force of friction, what do I do with it?

    Thanks for any help...
     
  2. jcsd
  3. Jun 1, 2014 #2
    Have you tried drawing a diagram?

    If you upload a picture of your attempt I can see where you're having the issue.

    The key to these questions are good force diagrams, so having one drawn for you won't solve anything in the long run.
     
  4. Jun 1, 2014 #3
    This is the closest guess that I have. I don't have a clue really. Sorry for the bad drawing.
     

    Attached Files:

  5. Jun 1, 2014 #4

    Nathanael

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    Homework Helper

    Just focus on the force of gravity for a moment. Draw it so that the force of gravity is the hypotenuse and the two legs are parallel and perpendicular to the ramp. (This way you've broken up the force of gravity into the components perpendicular to the ramp and parallel to the ramp.)

    Then just think about the frictionless case. If the box is kept at a constant velocity, then what must be true? How can you make this true?
     
  6. Jun 1, 2014 #5
    Okay, I've done that. What are the components of the gravity vector though? Isn't it just pointing down?
     
  7. Jun 1, 2014 #6
    The weight of the box does point straight down, but think about the direction of the normal force, and you should see that you need to decompose mg into two components.
     
  8. Jun 1, 2014 #7

    CAF123

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    Gold Member

    Yes, but the coordinate system imposed in the solution is one in which the axes are parallel and perpendicular to the plane of the incline. In this case, you can decompose the gravity vector along these mutually orthogonal directions.
     
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