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Tension in String between two blocks

  1. Oct 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Block 1, of mass [itex]m_1 =2.6[/itex]kg and block 2 of mass [itex]m_2=1.4[/itex]kg are connected by a massless string. Block 2 then experiences a force [itex]F=18[/itex]N acting in the postivive x direction and at an angle of [itex]\theta = 38°[/itex] . The coefficiant of kinetic friction between the floor and both of the blocks is 0.24.

    What is the tension in the string?

    2. Relevant equations
    [itex]F_{net}=ma \\
    F_f = \mu_k F_N

    3. The attempt at a solution
    Before I did anything else, as standard, drew a diagram.

    Then found the normal force acting on block 2...
    F_{N_2} =m_2 g + F \sin{\theta} \\
    F_{N_2} = (1.4)(9.81) + (18) \sin{38} = 24.8N

    Then I tried to find the tension by doing...
    T=F \cos{\theta} - \mu_k (F_{N_1} + F_{N_2}) \\
    T=F \cos{\theta} - \mu_k (m_1 g +24.8) \\
    T=(18) \cos{38} - 0.24(2.6(9.81)+24.8) = 2.11 N

    I do not think I have done this correctly so would appreciate any feedback/advice, thanks :)
  2. jcsd
  3. Oct 13, 2014 #2


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    Did the diagram give you reason to add the ##m_2 g## and the ##F \sin{\theta}## in that way ?

    And are both friction forces contributing to T with the same sign ?

    Perhaps showing the diagram you made can help out here. Otherwise: make a separate diagram for each block.
  4. Oct 13, 2014 #3
    I cant post the diagram, at least not until way after the coursework is due, because I have not got a camera phone at the moment (waiting for a replacement from my insurance) and do not have access to a scanner.

    With regards to the frictional forces, I had them both acting in the negative x-direction, because both blocks are moving in the positive x-direction.

    And with regards to the [itex]m_2 g[/itex] and the [itex] F \sin{\theta} [/itex], I did one diagram and equated the forces and the maths told me it was [itex]m_2 g - F \sin{\theta}[/itex] But then for some reason I thought I made a mistake and scribbled it out and when I started again I had them added together. Was I correct first time around?

    EDIT: I am trying to draw a crude version of my diagram using ms-paint and will post it as soon as I can.
    Last edited: Oct 13, 2014
  5. Oct 13, 2014 #4


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    Which way does m2 g work ? And F sin(theta) ? Well then. The block doesn't accelerate up or down, so the net force in the y-direction is zero, which helps you to calculate the normal force. That, in turn helps you to calculate the maximum friction force.

    Write out (and post) the components of all forces on m1 and the components of all forces on m2 in your drawings as a check. Make sure you get the signs right !
  6. Oct 13, 2014 #5
    Heres my first diagram which is just a diagram of the setup, plus the friction forces which I added on. (Link http://s28.postimg.org/61l56znt9/diagram.jpg )

    I appreciate your help a lot, thanks. But its midnight (at least in the UK) and need to get some sleep! So will have another crack at it tomorrow when I can.
  7. Oct 13, 2014 #6
    Are you sure of the location and direction of the force F? I imagined from the problem statement that the force was on the right hand side of m2, and was up and to the right.

    There are big problems with your horizontal force balances. You left out the ma's. Also, if the force balance is being applied to the combination of the two masses, then the tension should not be in the equation. Do yourself a favor and write a force balance on each mass individually. Then, if you really want to do the combined force balance, add the individual force balances together.

  8. Oct 14, 2014 #7
    Yes. The problem came with a diagram and the diagram I posted up is exactly the same, with F in the exact same place (i.e. the force is "pushing" rather than pulling.); the only difference between the two diagrams is that the one in the problem only had F, whereas I added in the weight, normal and frictional forces.

    I left of the ma as in my head when attempting it, I thought I could calculate the tension without it, however it was the main reason why I did not think I was correct.

    I am off to uni now for the day (and have other coursework) so will have another go when I get back later in the day :)
  9. Oct 14, 2014 #8
    Ok now this is what I have done.

    First Equated forces for block 1 (the one on the right):
    Vertically for block 2:
    F_{N_2}=m_2 g + F \sin{\theta}

    Horizontally for block 2
    F \cos{\theta} - F_{f_2} - T = m_2 a

    Then for block 1 (the one of the left)
    F_{N_1}=m_1 g

    T-F_{f_1}=m_1 a

    Then what I did was rearranged the second equation above (the one for horizontal forces on block 2) for T, and substituted that into the equation for the horizontal forces on the other block, with the aim of calculating the acceleration; so then I could go back and pop a into either one to get the Tension.

    T=F \cos{\theta} - F_{f_2} - m_2 a

    And then subbing it into the other one gives...
    F \cos{\theta}- F_{f_2} - F_{f_1} - m_2 a=m_1 a \\
    F \cos{\theta}- F_{f_2} - F_{f_1} = (m_1+m_2)a \\
    a = \frac{F \cos{\theta}- F_{f_2} - F_{f_1}}{m_1+m_2} \\

    I wont do any more as its time consuming with all the latex markup, but is the thought process correct, and also is what I have already done correct?

    It has to be submitted tomorrow and this problem seems to be driving me nuts! lol

    Side Note: I kept with [itex]m_2 g + F \sin{\theta} [/itex] for the normal force of block 2 just from looking at the diagram and the way that F is placed (acting down into the ground), that the vertical component of it would be pushing the block into the ground and therefore making it "heavier"

    Thanks! :)
    Last edited: Oct 14, 2014
  10. Oct 14, 2014 #9
    Yes. This is correct. Now, you also need to substitute the coefficient of friction times the normal forces to get the frictional forces. Once you know the acceleration, you can get the tension T.

  11. Oct 14, 2014 #10
    Yup. Thank you so much! :)
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