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Find the acceleration of the box.

  1. Jan 17, 2009 #1
    I really need some help with this problem. PLEASE HELP!!

    Here it is:

    You are pushing a box uphill the hill has a pitch of 15 degrees above the horizontal, and the static and kinetic friction coefficients between the box and the hill are 0.45 and 0.35. The box has a mass of 40 kg. You push horizontally with a push of 180 N, and the box is initially moving uphill. Find the acceleration of the box.


    ANY HELP WOULD BE GREAT!!!
     
  2. jcsd
  3. Jan 17, 2009 #2
    I really need some help with this problem. PLEASE HELP!!

    Here it is:

    You are pushing a box uphill the hill has a pitch of 15 degrees above the horizontal, and the static and kinetic friction coefficients between the box and the hill are 0.45 and 0.35. The box has a mass of 40 kg. You push horizontally with a push of 180 N, and the box is initially moving uphill. Find the acceleration of the box.


    ANY HELP WOULD BE GREAT!!!
     
  4. Jan 17, 2009 #3

    tiny-tim

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    Welcome to PF!

    Hi rickylin89! Welcome to PF! :wink:

    Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:

    (but please don't double-post :frown:)
     
  5. Jan 17, 2009 #4
    Honestly, I don't even know how to start the problem. I really don't know what to do lol.
     
  6. Jan 17, 2009 #5
    Perhaps you could begin by drawing a free body diagram and summing forces parallel and perpendicular to the inclined surface.
     
  7. Jan 17, 2009 #6
    I'm not really sure what the free body diagram would look like.
     
  8. Jan 17, 2009 #7
    The free body diagram will show the box and all of the forces that act on the box. These include the force of gravity (the weight), the normal reaction force from the surface, the friction force parallel to the surface, and the horizontal applied force of 180 N. Once you have this figure, it will help you see what forces act in each direction so that you can properly write the force sums for Newton's second law.
     
  9. Jan 18, 2009 #8
    Alright I drew the free body diagram. I believe that the net force is the up force, the weight is the down force, the force of friction is to the leff of the box, and the force of the push is to the right of the box. The force of friction and force of the push are both at an angle though. Does this seem right?
     
  10. Jan 18, 2009 #9
    Who (what?) is this fellow "net" force? Where does he (it) come from? (Do you know what the term "net force" means? It really does not sound like you do, and that is something you need to understand. Please look up the definition.)

    The problem statement said, "You push horizontally," but now you tell me, "The ...force of the push are both at an angle..." Which way is it?

    You are correct in saying that the weight acts down. I think you still have some thinking to do about the direction of the other forces, and just what they come from.

    A term that you ought to know is "normal reaction." It describes the reaction force normal (perpendicular) to a surface. It is a useful term.
     
  11. Jan 18, 2009 #10
    I know that the force of the push is equal to 180sin(15) but I don't understand how to incorporate the force of friction.
     
  12. Jan 18, 2009 #11
    Unless you have changed the problem statement, you don't know this. The original problem statement said that the push was 180 N.

    You still need to sort out your FBD, and until you do that, you are not going to get very far. Please don't try to cut corners. It is costing you time.
     
  13. Jan 18, 2009 #12
    I don't understand what you are asking me to do.
     
  14. Jan 18, 2009 #13
    What I meant to say was.... I know that the NORMAL force is mgcos(15). I believe that is right because I drew out the components. I am just sure if it is cos(15) or sin(15)?
     
  15. Jan 19, 2009 #14

    tiny-tim

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    Hi rickylin89! :smile:

    :rolleyes: hmm … there are two ways to work this out …

    i] suppose the angle was 0º instead of 15º … would the normal force be mgcos(0) or mgsin(0)? :wink:

    ii] components are always cos of the angle between the force and the direction … sin only comes in when for example you've only defined one angle, θ say, and you need to use it for two perpendicular directions … then it'll be cos(θ) and cos(90º - θ), and we write the latter as sin(θ)! :smile:
     
  16. Jan 19, 2009 #15
    So then it's mgcos(15)?
     
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