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Easy newtons law problem

  1. May 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Check out my attachment to get an idea what is happening. What we have is an inclined plane with 2 masses (m1 and m2) which are touching one another and the angle of the incline is 30 degrees.
    The coefficient of kinetic friction between m1 and the plane is 0.23 = mu_k1
    The coefficient of kinetic friction between m2 and the plane is 0.51 = mu_k2
    m1 = 2.5kg
    m2 = 3.1kg
    theta = 30 degrees
    The question is to find the acceleration of the pair of masses, and the second question is find the force of m1 on m2

    2. Relevant equations
    F=ma


    3. The attempt at a solution
    I applied Mr Newton on each mass individually to get
    for mass m1:
    x direction (parallel to the inclined plane):
    [tex]-F_{21}-\mu_{k1}N_{1}+m_{1}g\sin(30)=m_{1}a[/tex]
    y direction (perpendicular to the inclined plane):
    [tex]-m_{1}g\cos(30)+N_{1}=0[/tex]

    for mass m2:
    x direction:
    [tex]F_{12}-\mu_{k2}N_{2}+m_{2}g\sin(30)=m_{2}a[/tex]
    y direction:
    [tex]-m_{2}g\cos(30)+N_{2}=0[/tex]

    I used [tex]F_{12}=-F_{21}[/tex].

    I put all that into the mix and out popped a = 9.33m/s^2, which is wrong.
    The answer says a = 1.6m/s^2. I am pretty sure my maths is ok. I'd say the problem is in my application of Newton but can't see anything wrong.

    Thanks in advance
     

    Attached Files:

  2. jcsd
  3. May 24, 2008 #2

    Chi Meson

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    Science Advisor
    Homework Helper

    Well, I don't know what you did to get that answer.

    On all these "sliding down incline" problems, you usually can assume that all forces perpendicular to the surface will balance, so all you need to look at the parallel forces.

    I can't tell if you allowed the normal forces between blocks to cancel each other. You should have.
    Did you substitute [tex]m_g{1}\cos(30)[/tex] for the N to find the frictional force? (for each block?

    Did you divide the net force by the total mass?

    I can't tell what you did.
     
    Last edited: May 24, 2008
  4. May 24, 2008 #3

    alphysicist

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    Homework Helper

    Hi vladmir69,

    I noticed that you said that [itex]F_{12}=-F_{21}[/itex]; I think that is where your error is, because with the notation you have in your problem that is not quite true.

    What is true is that [tex]\vec F_{12}= - \vec F_{21}[/tex]; the vectors are equal in magnitude and opposite in direction. But in your equations, [itex]F_{12}[/itex] and [itex]F_{21}[/itex] just represent the magnitudes of those forces; you have already taken care of the direction by making one term negative and one term positive. Now you just have to force the magnitudes to be equal, so you need to use [itex]F_{12}= F_{21}[/itex].
     
  5. May 27, 2008 #4
    thanks alphysicist you have solved the problem
     
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