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Find the mass of m1 on an inclined pulley system

  1. Feb 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Two masses ##m_1## and ##m_2## are connected by a massless cord and a pulley, on a rough ramp, tilted at angle ∅ The pulley is massless and frictionless. The coefficient of kinetic friction between the ramp and m1 is ##μ_k.## Derive an expression for ##m_1,## given that ##m_2## accelerates downward with acceleration of magnitude ##a.##

    Diagram:
    mEv7VFo.png


    2. Relevant equations
    ##F=ma##
    ##g=9.8 m/s^2##


    3. The attempt at a solution
    I drew my free body diagrams for each mass.
    For ##m_1,## the axes are aligned with the ramp, not the ground. ##m_1g## is broken up into its ##x## and ##y## components: ##m_1gsin∅## and ##m_1gcos∅,## respectively. Tension ##T## is larger than frictional force ##F_f## since the object is accelerating up the ramp despite the frictional force in the opposite direction.

    For ##m_2,## the axes are aligned with the ground with the up direction being the positive ##y## direction. The weight ##mg## is larger than the tension ##T## due to its downward acceleration.

    I think that the tension will be the same everywhere in the rope, so ##T_{m_1}=T_{m_2}.##

    With regard to ##m_1:##
    ##T=m_2a##
    ##F_{net}=T-F_f,##
    or
    ##F_{net}=m_2a-μ_km_1gcos∅.##

    With regard to ##m_2:##
    ##F_{net}=m_2g-T,##
    or
    ##F_{net}=m_2g-m_2a.##

    That's about as far as I got. It's not much of a start. I don't know what steps I need to take towards isolating ##m_1.##
     
    Last edited: Feb 21, 2013
  2. jcsd
  3. Feb 21, 2013 #2
    Where did you get T = m2a from?
    For the net force on m1, you forgot the component of gravity acting down the ramp.
     
  4. Feb 21, 2013 #3
    I derived tension ##T## by applying ##F=ma## and getting ##T=ma## since ##T## is a force. I saw that since the mass of ##m_2## is given and its downward acceleration ##a## is given and on the same axis as ##T## (the ##y## axis), I could plug those into the force equation and get the tension.

    And yeah, I can see what you're saying about the gravity. Since ##m_1##'s motion is completely horizontal, it should actually be ##F_{net_{m_1}}=m_2a−μ_km_1gsin∅.## Not sure how I managed to mix up sin and cos, sorry.
     
  5. Feb 21, 2013 #4
    Have another look at your free-body diagrams.
    The one for m2 should have the tension pointing up and the weight down.
    The sum of both of these forces will be m2a.
    So, T = m2a is not correct.

    The motion of m1 may be horizontal with respect to the ramp, but it's not horizontal with respect to the ground, so there is a component of the weight acting to pull the mass back down the incline.
     
  6. Feb 21, 2013 #5
    Ok, so ##T+m_2g=m_2a,##
    or
    ##T=m_2a-m_2g.##

    So, correct me if I'm misunderstanding what you're saying, does that mean that ##mg_x## should actually be subtracted, making ##F_{net_{m_1}}=T-F_f-mg_x##?
     
  7. Feb 21, 2013 #6
    Remember that T and the weight are in opposite directions, so
    m2g - T = m2a
    ( my y-axis points down here )

    Your expression for Fnet for m1 is correct.
    When you equate this with m1a, be careful with the signs.
     
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