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Two forces held by pulley's on an inclined plane.

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose the angles on an inclined plane are 60*(left-hand side) and 20* (right-hand side). If the left-hand mass is 2.1 kg, what should the right-hand mass be so that it accelerates downslope at 0.64 m/s/s?

    2. Relevant equations

    F=ma

    3. The attempt at a solution

    a=.64
    Ml (left-side) = 2.1kg
    Mr (right-side) = ?
    θl = 60*
    θr = 20*

    Ml -> F = Mla -> T - Mlgsinθl = Mla
    Mr -> F = Mra -> Mrgsinθr - T = Mra

    I understand that since acceleration is the same then Mla + Mra = (Ml + Mr)a

    I also know that T = Mla + Mlgsinθl

    So I tried to sub in for T -> Mrgsinθr - Mla + Mlgsinθl = Mra but this doesn't make any sense to me. No matter how I look at it I can't seem to isolate Mr by itself. This is causing me considerable frustration.
     
  2. jcsd
  3. Sep 17, 2011 #2
    I'm assuming the pulley is negligible? I would suggest drawing a picture and displaying the forces acting on the system.
     
  4. Sep 18, 2011 #3
    Yes the pulley is negligible, as is the rope. I drew a picture and free-body diagram and labeled all the forces accordingly.

    I think my issue here is figuring out how to do the algebra correctly to end up with the Mass of the right-hand side by itself so I can solve the problem. Anywhere in my steps that you see an error?
     
  5. Sep 18, 2011 #4
    Have you tried:
    [itex] \large M_r = \frac{M_l(a+g\sin{\theta_l})}{g\sin{\theta_r}+a} [/itex]?
     
  6. Sep 18, 2011 #5
    You think I could get you to run through how you managed to get that? I still just can't see how you managed to get Mr by itself. I can tell there's some factoring in there somewhere but I guess I'm just not making the right calculations.
     
  7. Sep 18, 2011 #6
    It's very simple, and there's no secret to it.
    It always helps to arrange the needed elements on one side, and the rest, the known ones on the other.
    So we would get,
    [itex]
    \LARGE
    M_rg\sin{\theta_r} + M_ra = M_lgsin{\theta_l}+M_la
    [/itex]
    All that's left is to extract M_r by removing it as a common factor/coefficient, and dividng by the remainder, namely, g sin(theta_r)+a.
    Hope that's clear enough,
    Daniel
    P.S
    I might help if you pointed exactly which step you don't follow...
     
  8. Sep 18, 2011 #7
    I seem to be having trouble figuring out how you get a positives on both sides of the equation above. I managed to get there but I had a negative the left side of it, (right-hand side of the problem).

    I'll write down all my steps...

    [itex]
    \LARGE
    T - M_lgsin{\theta_l} = M_la
    [/itex]
    [itex]
    \LARGE
    M_rgsin{\theta_r} - T = M_ra
    [/itex]
    Adding both equations, the tensions cancel leaving...

    [itex]
    \LARGE
    M_rgsin{\theta_r} - M_lgsin{\theta_l} = M_la + M_ra
    [/itex]

    I subtracted the M_ra from the right-side of the equation, and added M_lgsin(theta_l) to get

    [itex]
    \LARGE
    M_rgsin{\theta_r} - M_ra = M_lgsin{\theta_l} + M_la
    [/itex]

    I just don't see how you ended up with the equation you posted, with positives on both sides.
     
  9. Sep 19, 2011 #8
    Hi,
    You're positively right! there's a minus there, but that shouldn't matter. Anyway, can you extract m_r now?
     
  10. Sep 19, 2011 #9
    Yes I can. I ended up getting...

    [itex]
    \LARGE
    M_r(gsin{\theta_r} - a) = M_lgsin{\theta_l} + M_la
    [/itex]

    If divide by the
    [itex]
    gsin{\theta_r} - a
    [/itex]

    the answer still comes out negative for me. Am I right in thinking that if I reversed the order of...
    [itex]
    gsin{\theta_r} - a
    [/itex]

    it would make it positive like so...

    [itex]
    a + gsin{\theta_r}
    [/itex]
     
  11. Sep 19, 2011 #10
    It can't possibly be negative!
    With your given data, g*sin(theta_r) is larger(much!) than a.
    Check your arithmatics...
     
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