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Tension On An Inclined Slope

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Figure P8.36 shows a block of mass m resting on a 20° slope. The block has coefficients of friction µs = 0.83 and µk = 0.47 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass m2 = 2.0 kg.

    http://img6.imageshack.us/img6/9688/p836.gif [Broken]


    (a) What is the minimum mass m that will stick and not slip?
    m_1=1.783kg
    (b) If this mimimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?
    a_x= ???




    2. Relevant equations

    F=ma
    T=ma
    f_k=μ_k*n
    n=m*g*cos(θ)

    m_1 = 1.783kg
    m_2 = 2.0kg
    θ = 20 degrees
    T = 19.6N
    μ_k = 0.47
    μ_s = 0.83



    3. The attempt at a solution


    I found the mass m_1 using.. (part A)

    m_1=(T/(μ_s*g*cos(θ)+gsin(θ))

    ..and I tried to find the a_x using forces.. (Part B begins here)

    T - f_k - m*g*sin(θ) = m*a_x

    ..which gave me the equation...

    a_x=(T/m_1)-(μ*g*cos(θ))-(g*sin(θ))

    ..and I am not sure if I am doing the math wrong but my answers are rejected.

    Could anyone perhaps check my force diagram and algebra?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 17, 2009 #2

    LowlyPion

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    I get a similar answer for a) of 1.78 kg. Maybe it's a significant digit issue?

    for b) I just figured what the net force difference was on the basis of the change in μ. Then I applied that to the total mass to figure the acceleration.
     
  4. Mar 17, 2009 #3
    Well yea, sorry I didn't say it clearly but answer A was accepted.

    For B yea, I think I'm doing it right..
     
  5. Mar 17, 2009 #4

    LowlyPion

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    Your formula for acceleration can't be correct because you have to consider the total mass of the 2 blocks.

    Maybe try my previous suggestion?
     
  6. Mar 18, 2009 #5
    Hmmm okay. I'll have to go back and re-read the section on interaction between two bodies. I'm having difficulty understanding that concept and that is currently what all my problems consist of, hehe~
     
  7. Mar 18, 2009 #6

    LowlyPion

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    Since both blocks are tied together, if they are to accelerate together, then their mass must be accelerated together and whatever the motive force then acceleration will be a consequence of the combined masses and the force.
     
  8. Mar 18, 2009 #7
    If I was creating an equation I would simply add the idea of F=m2g to the acceleration then, Would I then subtract the T force suspending the block?

    Well I guess that in turn would cancel the Tension force I was using to accelerate it up the ramp... wait no that doesn't make sense..

    Oh.. in my equation maybe I just need to change..

    ax=(T/m1)-(μ*g*cos(θ))-(g*sin(θ))

    to

    ax=(T/mboth)-(μ*g*cos(θ))-(g*sin(θ))

    Would that take into account both blocks? I don't think so... I think I need a new equation don't I?
     
  9. Mar 18, 2009 #8

    LowlyPion

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    In the problem you have solved for the weight at which it is just stable using the static coefficient of friction.

    If you take that equation and instead use the kinetic coefficient of friction then you have an instability right? You have a net force that you can calculate directly as merely the Δ of the μ's times the normal force of the mass on the incline.

    F = (μs - μk)*cosθ*m1*g

    That yields a net force. Fnet

    That force is free to accelerate the system now. But the system has a mass of m1 + m2

    So then isn't your a = Fnet/(m1+m2)
     
  10. Mar 18, 2009 #9
    μ θ

    Ok so I ran through the problem with the idea that both of the masses must be taken into consideration when accelerating the mass on the slope. So.

    I came up with..

    ax=((T-μkmgCos(θ)-mgSin(θ))/MBoth)

    Would that take it into account the way you meant for me to?
     
  11. Mar 18, 2009 #10
    I see what you are saying here, if the net force worked to hold the system in place, well then when μs changes to μk then we can use the same formula. We just need to consider both masses.

    My main question at the moment is.. when taking into account both masses and dividing by them, it is t hen necessary to add a special subscript (ie: mboth) to differentiate them from the others. Now, in dividing by that mass instead of the original mass the equation becomes a bit more complicated since the masses are not going to cancel out the way they would in a normal "F=ma" to "a=F/m" step, right?

    So instead of going from something like..

    axm=F-mgCos(θ)

    to

    ax=(F/m)-gCos(θ)

    instead it works like..

    axmboth=F-mgCos(θ)

    to

    ax=((F-mgCos(θ))/mboth)

    All I'm trying to say is the masses don't cancel out now, which they used to, which makes sense, right?
     
  12. Mar 18, 2009 #11
    Thanks =)

    I know I talk to myself a lot but it helps me the most I guess

    The answer to part 2 was..


    ax=((T-μkmgCos(θ)-mgSin(θ))/MBoth)
     
  13. Mar 18, 2009 #12

    LowlyPion

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    Honestly I haven't worked it out with the formalism of the variables. The use of T in the equation seems a little counter intuitive to me because it seems itself affected by the acceleration.
     
  14. Mar 18, 2009 #13
    In my head I saw that the T was the Force behind the acceleration which was made by the second mass and its reaction to gravity.. so the second mass is part of T itself, I think?

    Anyway, I mean to elaborate as to why I agree with you, I'm just not good at explaining myself.

    But thank you, I doubt I would've figured it out without your help.
     
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