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Homework Help: Inclined slope, b degrees

  1. Nov 26, 2008 #1
    as in the diagram below, mass m1 is placed above mass m2 on a sloped incline with an angle of b degrees. the frictional coefficient between the two masses is C1 and the coeficient between mass 2 and the slope is C2.
    find the acceleration of the masses.

    http://picasaweb.google.com/devanlevin/DropBox?authkey=sbH95pBl_D8#5273088118963403586 [Broken]

    after dividing the vectors into components i said
    N1=(m1)gcosb
    f1=(C1)(m1)gcosb

    N2=N1+(M2)gcosb=(M1+M2)gcosb
    f2=(C2)(M1+M2)gcosb

    for mass 1
    F=m1gsinb-f1=m1a
    =(m1)gsinb-(C1)(m1)gcosb=m1a

    a1=g(sinb-C1cosb)=gcosb(tgb-C1)
    meanin that when C1 is equal to tanb, body 1 will not move relative to body 2 or will not move relative to the slope?

    now for body2

    F=m2gsinb+f1-f2=m2a------> is this correct??
    after opening i get

    a2=m2gcosb(tgb-C2)+m1gcosb(C1-C2) is this correct?
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 26, 2008 #2

    Redbelly98

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    The expression for a1 is correct. However, when C1=tan(b), a1 will be zero. a1 is not the velocity, it is the ______ .

    Should be "m2 a2" on the left side. Otherwise it looks good.
     
  4. Nov 27, 2008 #3
    a1 is the acceleration, sorry, m1 will not accelerate, but is it relative to the slope or rellative to m2, ie will a1 relative to the a person standing on the slope be
    1) a1=g(sinb-C1cosb)=gcosb(tgb-C1) or
    2) a1=g(sinb-C1cosb)=gcosb(tgb-C1) + a2

    and a2=[m2gcosb(tgb-C2)+m1gcosb(C1-C2)]m2

    so a2=gcosb[(tgb-C2)+(m1/m2)(c1-C2)] relative to the slope
     
  5. Nov 27, 2008 #4

    Redbelly98

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    a1: Since you have accounted for all the forces when you got a=Fnet/m, the expression you had is the acceleration. (1) is correct.

    a2: looks good.
     
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