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Finding the tension in a rope when acceleration is balanced

  1. Oct 8, 2014 #1
    1. The problem statement, all variables and given/known data
    A m1 = 7.3 kg block and a m2 = 10.2 kg block, connected by a rope that passes over a frictionless peg, slide on frictionless incline.

    4-p-077-alt.gif

    (a) Find the acceleration of the blocks and the tension in the rope.

    (b) The two blocks are replaced by two others of masses m1 and m2 such that there is no acceleration. Find whatever information you can about the masses of these two new blocks.

    m1/m2 = ?

    2. Relevant equations
    (a)
    http://www5a.wolframalpha.com/Calculate/MSP/MSP15211ddd67249f6c0h99000021236a56865d5b05?MSPStoreType=image/gif&s=30&w=226.&h=36. [Broken]

    T = (m1g)(1 + (((m2sin(50 deg)) - (m1sin(40 deg)))/(m1 + m2))

    3. The attempt at a solution
    (a) in order to find the acceleration of the blocks I used the acceleration equation above to get...
    http://www5a.wolframalpha.com/Calculate/MSP/MSP97820d5a6g74c9g0833000042fai256cg284eae?MSPStoreType=image/gif&s=54&w=353.&h=43. [Broken] = 1.75 m/s2

    But I am having trouble finding the tension. Using the equation above I do not get the right answer.
    http://www4c.wolframalpha.com/Calculate/MSP/MSP110351gc6963gg3d1i776000051bdi059eic3399d?MSPStoreType=image/gif&s=3&w=322.&h=37. [Broken] = 84.39 N
    Is this the right equation? What do I need to do differently?

    (b) I really have no idea what this part is asking for. If a = 0 wouldn't m1 and m2 both be equal to 0?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 8, 2014 #2
    You are not taking into account the gravitational force (F=mg).
     
  4. Oct 8, 2014 #3
    In the Tension equation? I thought I was because it is (m1 x g)(...)
     
  5. Oct 8, 2014 #4
    EDIT: Just noticed it was frictionless. My bad, sorry.

    For the (b) part if you take equation for the acceleration in case where a=0, you would get m2sin(50)-m1sin(40)=0, m1sin(40)=m2sin(50), m1/m2=sin(50)/sin(40).
     
  6. Oct 8, 2014 #5
    Ahhh, I see. Awesome. So do you think I did the tension part correctly then?
     
  7. Oct 8, 2014 #6
    Well, for the first object you would have m1a=T-m1gsin40, and for the second m2a=m2gsin50-T. If you solve for a in both cases, you would have a=(T-m1gsin40)/m1 and a=(m2gsin50-T)/m2. If you put these 2 in an equation (T-m1gsin40)m1=(m2gsin50-T)/m2, it should be fairly easy to solve for T in that case.

    By my calculations you should get ~58.77N
     
    Last edited: Oct 8, 2014
  8. Oct 8, 2014 #7
    Oh okay, that was similar to what I tried. I got ~58.80 N. Thank you very much!
     
  9. Oct 8, 2014 #8
    You're very welcome :D
     
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