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2.7 kg wood box

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A m = 2.7 kg wood box slides down a vertical wood wall while you push on it at a θ = 35° angle. What magnitude of force should you apply to cause the box to slide down at a constant speed?


    2. Relevant equations



    3. The attempt at a solution

    I got 64.5766 N for the answer. Don't know how to do it.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 9, 2009 #2
    Do you know what the net force on the box is equal to if it's travelling at a constant speed?

    What force is the box moving downwards with? (if your hands aren't there).

    just some things to consider

    remember that for it to not be accelerating up or down the force you push it up with (vertically) must be equal to the force it's coming down with. think about splitting the 35 degree force into vertical and horizontal forcess
     
  4. Oct 9, 2009 #3
    don't really understand that.... could you tell me what formulas to use? thanks
     
  5. Oct 9, 2009 #4
    okay, do you understand that in order for something to be static, or travelling with a constant velocity, there must be no net force acting on it?

    i think it's far more important to understand what to do here than simply use formulae.

    also do you know how to resolve forces into vertical and horizontal?
     
  6. Oct 9, 2009 #5
    yeah I understand the no net force thing...and resolving vertical and horizontal forces...using sin and cos. yes...
     
  7. Oct 9, 2009 #6
    Okay, so you must therefore realise that if it's moving at a constant speed (downwards) it's net force will be 0.

    therefore the force of gravity down, and the vertical component of your pushing (upward) will be equal

    your equation will look something like so;

    mg - Fcos(theta) = 0

    is friction mentioned at all?
     
    Last edited: Oct 9, 2009
  8. Oct 9, 2009 #7
    I got 29.2799 for the final answer. it is wrong
     
  9. Oct 10, 2009 #8
    that's not terribley helpful...

    what working out did you use to get that? if friction mentioned at all?
     
  10. Oct 10, 2009 #9
    I have this exact same question. Friction is not mentioned at all. He stated the exact wording already. I need help on this question as well. Is it ok to hijack this thread, or do I have to start a new one?
     
  11. Oct 11, 2009 #10
    No that's fine, it's the same question.

    Hmm when i tried it I got 32.3N, is that the correct answer? If it is, see my post above, which says this

    Okay, so you must therefore realise that if it's moving at a constant speed (downwards) it's net force will be 0.

    therefore the force of gravity down, and the vertical component of your pushing (upward) will be equal

    your equation will look something like so;

    mg - Fsin(theta) = 0


    if not tell me, the answer may be wrong
     
    Last edited: Oct 11, 2009
  12. Oct 11, 2009 #11
    Actually its the sine component that's balancing the weight of the body. So your answer might be around 47 or so, I reckon.
     
  13. Oct 11, 2009 #12
    Hm, would presume they mean 35 degrees from the vertical, not horizontal.

    edit: oh right just saw the picture attached, my bad :P
     
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