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Acceleration due to Tension

  1. Jul 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider the following diagram:

    5970334255_f6653f74cb_m.jpg

    Assume the strings and pulleys have negligible masses and the coefficient of kinetic friction between the 2.0 kg block and the table ([itex]\mu[/itex][itex]_{k}[/itex]) is .12. What is the acceleration of the 2.0 kg block?


    2. Relevant equations

    Net force = T[itex]_{2}[/itex] - (T[itex]_{1}[/itex] + [itex]\mu[/itex][itex]_{k}[/itex]F[itex]_{g}[/itex]), where T[itex]_{2}[/itex] is the tension from the 3 kg block, T[itex]_{1}[/itex] is the tension from the 1 kg block, and F[itex]_{g}[/itex] is the force of gravity on the 2 kg block.

    a= Net force/mass

    3. The attempt at a solution

    This seems like a really straightforward problem, but none of the multiple choice answers match my result so I'm confused as to what exactly I'm missing.

    I drew a free-body diagram for the 2 kg block with the normal force pointing up, the force of gravity pointing down, T[itex]_{1}[/itex] + [itex]\mu[/itex][itex]_{k}[/itex]*F[itex]_{g}[/itex] pointing left (presumably the block will accelerate to the right, so the friction force points in the opposite direction), and T[itex]_{2}[/itex] pointing right.

    T[itex]_{1}[/itex] = 1 kg * 9.8 m/s^2 = 9.8 N.

    [itex]\mu[/itex][itex]_{k}[/itex]*F[itex]_{g}[/itex] = .12 * 2 * 9.8 m/s^2 = 2.4 N

    T[itex]_{2}[/itex] = 3 kg * 9.8m/s^2 = 29.4 N.

    Net force = 29.4 N - (9.8+2.4) = 17.2 N

    a = 17.2 N/ 2 kg = 8.6 m/s^2.

    All of the options for answers are below 4.6 m/s^2.

    I'm not looking for the answer; I'd like to just find out conceptually what I'm missing so I can solve it for myself.

    What am I missing?

    Thanks in advance,
    m3rc
     
    Last edited: Jul 24, 2011
  2. jcsd
  3. Jul 24, 2011 #2
    Hi, m3rc! Good question, and thanks for explaining your work so clearly.

    Where you're going wrong is that you assume that the two hanging blocks exert a force equal to their weight on the middle block. That would be true if the system was stationary, but it is not true if the system is accelerating, as in this case. The clearest way to see this is to imagine all three blocks were falling freely -- in that case there would be no tension in either string.

    What you need to do is draw a free-body diagram for each block, not just the middle one. This will give you three equations of motion, with three unknowns: the tension in each string, and the acceleration. (We know that the accelerations of the three blocks are all equal because they are tied together by the string). Solve the system of equations to obtain the acceleration and the two tension forces.
     
  4. Jul 24, 2011 #3
    That did it, thanks so much for your help, Mike!

    I drew free-body diagrams for each of the blocks, which led to the equations:

    1 kg: a = (T1-9.8 N)/1 kg, where 9.8 N is the force of gravity on the block
    2 kg: a = (T2-T1+[itex]\mu[/itex]k(19.8 N))/2 kg, where 19.8 N is the force of gravity on the block.
    3 kg: a = (T2 - 29.4 N)/3 kg, where 29.4 N is the force of gravity on the block.

    I then solved for T2 in terms of T1 by setting the second equation equal to the third because the accelerations are equal. I plugged my answer back into the second equation and solved for T1 by setting it equal to the first equation. I then plugged both answers back into the second equation to come up with the correct answer, 2.8 m/s^2.

    Thanks again, Mike!
     
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