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Two Hanging Masses (TENSION)

  1. Apr 2, 2008 #1
    Two Hanging Masses (TENSION) :)

    1. The problem statement, all variables and given/known data
    Two blocks with masses M1 and M2 hag one under the other. For this problem take the positive direction to be upward and use g for the magnitude of the acceleration due to gravity. The blocks are now accelerating upwads (due to the tension in the strings) with acceleration of magnitude a. find the tension in the lower and upper rope.
    [​IMG]

    2. Relevant equations
    F=ma
    W=mg


    3. The attempt at a solution

    I found the tension in the lower and upper rope when it was stationary and therefore in equilibrium.
    T2=M2g
    T1=M1g
    Then to find the tension when accelerating I just used
    T2=Ma+Mg
    T1=Ma+M1g+M2g

    For some reason it says that the tension in rope 2 does not depend on the variable M1.... So I'm just really lost now on which direction to take..
    Thanks in advance
     
  2. jcsd
  3. Apr 2, 2008 #2
    Welcome to PF>

    While acceleration T2 is the normal force arising only due to the weight of M2.
    Hence T2 does not depend on variable M1.
     
  4. Apr 2, 2008 #3
    Oh thank you! That clears up the first one but I'm still stuck on the T1 when accelerating. Would I just go T1=(M1+M2)a + (M1+M2)g?
     
  5. Apr 2, 2008 #4
    Yes becoz the total downward force acting on T1 while acceleration is the sum of the weights of M1 and M2.
     
  6. Apr 2, 2008 #5

    Doc Al

    User Avatar

    Staff: Mentor

    That would be correct for M2 but not for T1 (three forces act on M1). In any case, the equilibrium situation is not relevant to this problem.
    Carefully apply Newton's 2nd law to each mass. Start by identifying all the forces that act on each mass. (Hint: three forces act on M1.)

    (The diagram is misleading because it does not show all the tension forces acting on the masses.)
     
  7. Apr 23, 2009 #6
    Re: Two Hanging Masses (TENSION) :)

    first, solve for the T2.
    T2=M1(g)
    then, for T1,
    since T1 is carrying the M2,
    ΣF=ma
    substitute ΣF to T2 + M1(g) -T1
    so,
    T2 + M1(g) -T1= ma (at rest)
    T2 + M1(g) -T1= 0
    T1=T2 + M1(g)
     
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