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Tension between two blocks!

  1. Feb 8, 2010 #1
    Hey, I'm just a tad confused about tension and how it works...

    1. The problem statement, all variables and given/known data

    Two blocks m1 (11 kg) and m2 (18 kg) are connected by strings, as shown in the below. String 2 breaks when T2 = 4.9033 N. Find T1 when string 2 breaks.
    Answer in units of N. (Blocks are on a flat surface, not hanging)


    2. Relevant equations
    F = ma
    Not sure if any others are necessary, I don't believe so. Friction not included in this problem.

    3. The attempt at a solution
    Before the string snaps, if T2 = 4.9033 N, at that point, the acceleration of m1 = T2/m1 = .44575454545 m/s/s. I assume acceleration across the whole system would be the same, making the T1 = (18 kg + 11 kg)(.445754545 m/s/s), but I am completely stuck on what happens after the string breaks...does the acceleration remain the same and make T1 = .445754545 m/s/s * 18 kg = 8.02358 N?

    Also, its my first time here, did I include enough relevant info in my post? Any feedback greatly appreciated, thanks!
    Last edited: Feb 8, 2010
  2. jcsd
  3. Feb 8, 2010 #2


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    Homework Helper

    I think the question is asking what T1 is the moment before string 2 breaks, not the moment after. I think it's reasonable to assume that T1 stays the same the moment after the other string breaks; after all, whatever is applying the force can't instantly change the force it applies.
  4. Feb 8, 2010 #3

    Doc Al

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    Staff: Mentor

    Why are you worried about what happens after the string breaks? I think all they want is the value of T1 at the moment that the string breaks (or just before), which you already figured out.
  5. Feb 8, 2010 #4
    You haven't told us what the nature is of the force that's pulling on both masses. If it's a constant force then your first andwer would be right. If it's a mass hanging off the end
    of the table, you would have to know the mass.
  6. Feb 8, 2010 #5
    Sure enough...that's right. Thank you very much. I think I over-thunk it...

    Theoretically, what would you have to do to calculate T1 after it broke?
  7. Feb 8, 2010 #6
    @willem2, it was a constant horizontal force.
  8. Feb 8, 2010 #7

    Doc Al

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    Staff: Mentor

    As willem2 notes, it depends on how the force T1 is applied. If you can assume that T1 is fixed, then you're done. Otherwise, you'll have to see what changes.
  9. Feb 8, 2010 #8
    Thank you all :)
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