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Easy question has me confused (Inclined plane)

  1. Dec 15, 2009 #1
    It seems to be an easy question but for some reason I can't figure it out.

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


    If the 10 lb block A slides down the plane with a constant velocity when θ = 30°, determine the acceleration of the block when θ = 45°.

    2. Relevant equations

    F = ma

    3. The attempt at a solution

    I actually am very puzzled by this question. I know that acceleration = 0 at 30° but I don't know how to make use of it. I tried setting the parallel force at 45° to equal ma, then solve for a, but that didn't do it either. Any insight? Thanks.
  2. jcsd
  3. Dec 15, 2009 #2
    first resolve the forces around the block
  4. Dec 15, 2009 #3
    FNormal = 10cosθ
    Fparallel = 10sinθ

    Just sub in 30 and 45 for θ.

    What else is there to use?
  5. Dec 15, 2009 #4
    where is gravity in your equation remeber its the force of the block acting down the slope not the mass. Just want to check do you assume that the plane is smooth?
  6. Dec 15, 2009 #5
    Well the 10sinθ is the force of gravity down the ramp. Also, yes I'm assuming the plane is smooth. It doesn't say anything to lead me to believe there's friction.
  7. Dec 15, 2009 #6
    it should be the force of the block acting down the ramp so 10gsin45=ma

    but since the mass is the same
    should get gsin45=a
  8. Dec 15, 2009 #7
    Hmm I though 10lb is referring to a force not a mass in this case.

    Regardless, gsin45 = a is not working out.

    32.2sin45 = 22.77 ft/s^2

    The real answer is 9.62 ft/s^2.
  9. Dec 15, 2009 #8
    hmmm not sure then. ill have another look
    we both must be missing something obvious :/
  10. Dec 15, 2009 #9
    Yeah lol this is really confusing for me too. It doesn't seem to be very hard but something is obviously not clicking.
  11. Dec 15, 2009 #10
    Any new ideas?
  12. Dec 15, 2009 #11


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    Homework Helper

    The only reason I think they would give you the 30 degree condition is that you need to account for friction.

    try using the 30 degree criteria to get the coefficient of friction.
  13. Dec 16, 2009 #12
    You got it rock.freak667!

    Kind of a weird question since it didn't say anything about friction. It's not in the friction chapter either. It's a few chapters later. I guess they expect us to use ideas from previous chapters even without making mention of them.

  14. Oct 6, 2010 #13
    You would use the 30 degrees to solve for the kinetic friction and then use the value for kinetic friction to solve for the acceleration. I believe its supposed to be a trick question since nowhere is friction mentioned. Cannot be solved without first finding the friction which happens to be 0.5774. Find the Normal force, and with the friction you should come out with the answer. Hope this helps.
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