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Why is this wrong?

  1. Nov 22, 2005 #1
    I just recently got my midterm back and I felt this was graded wrong...

    I dont have a picture..but just picture a ramp inclined to [tex] \phi=30[/tex] One mass is sitting on the incline...And the other is dangling from a rope which runs over a pulley

    Two masses [tex]m_1[/tex] & [tex]m_2[/tex] both have a mass of 2Kg. If [tex]m_1[/tex] slide up the ramp at a constant velocity, what is the kinetic frictional force on [tex]m_1[/tex]? Assume that the masses of the rope and pulley are negligible, and that the pulley is frictionless.
    How I did it...
    [tex]m_1 x/ T-f_k = m_1a[/tex] (Eq1)
    [tex] y/ N-mgsin \phi=0[/tex]
    [tex]m_2 x/ T-mgsin \phi=m_2a[/tex]
    [tex] y/ T=mgsin \phi[/tex] (Eq2)
    -------------------------------------------------
    isolate T in Eq1
    [tex]T-f_k=m_1a[/tex] Since a=0
    [tex]T=f_k[/tex]
    Sub Eq2 into Eq1
    [tex]mgsin \phi=f_k[/tex]
    [tex]2Kg(9.8m/s)sin \phi=f_k[/tex]
    [tex]f_k=9.8N[/tex]
    Although i got the right answer he says that it needs to be expressed like this
    [tex]T-f_k-mgsin \phi=0[/tex]
    [tex]f_k=T-mgsin \phi[/tex] T=mg
    [tex]f_k=mg-mgsin \phi = 9.8N[/tex]
    Why is my way wrong?
     
    Last edited: Nov 22, 2005
  2. jcsd
  3. Nov 22, 2005 #2

    mezarashi

    User Avatar
    Homework Helper

    Could you explain what your variables x, y, and T mean?

    You should analyze it through the use of force diagrams. The hanging mass is there only to tell you that the tension in the string, T is equal to mg. From there you free body diagram should focus entirely on the mass on the incline.
     
  4. Nov 22, 2005 #3
    What is this x ? Is that x=gsin(30°) ?

    Besides, in the x-direction (ie along the incline) you have 3 forces
    1) T [tex]m_2g[/tex]
    2) friction [tex] \mu_kN[/tex]
    3) the x-component of gravity : gsin(30°)
    I did not put in the right signs of these forces...
    I only see two forces in your equation

    What is x/T ?

    The acceleration of object 1 is indeed 0

    You should have [tex]m_1a_1=0=m_2g - \mu_kN -mgsin(30°)[/tex]

    What is m_2 x/T ???
    This is incorrect.

    In the y direction (perpendicular to the incline) you have
    [tex]m_1a_1 = 0 =-mgcos(30°) + N[/tex]

    Where N is the normal force.


    But you only need the equation in the x direction to get your answer

    marlon
     
    Last edited: Nov 22, 2005
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