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Blocks on Floor - Tension between blocks

  1. Oct 21, 2008 #1
    1. The problem statement, all variables and given/known data
    A block of mass m1 = 2 kg and a block of mass m2 = 3 kg are tied together and are pulled from rest across the floor by a force of Fp = 25 N. The coefficient of friction of the blocks with the floor is µ = 0.3.

    a) What is the acceleration of the two blocks? Answer: 2.06m/s^2


    b) What is the tension in the string between the blocks?

    2. Relevant equations

    For a) F=ma

    For b) F=ma

    3. The attempt at a solution

    a) m=5 and a=9.8 the answer was 49. I then multiplied 49 by 0.3 to get 14.7N. 14.7N was then subtracted from 25N to get 10.3N. 10.3N was then divided by 5 to get 2.06m/s^2.

    b) Each block sould be calculated seperatley. Block m1 is 2kg and block m2 is 5kg. They are being pulled to the right with T (tension) between m1 and m2. Each block has different normal forces, m1=2 and m2=5. I used F=ma: 2(9.8)=19.6N and 5(9.8)=49N.

    The front block will have friction, the pulling force, and the tension of the rope acting on it. The back block will only have the tension of the rope and friction. Both will have an acceleration, a.

    Any help is much appreciated!
     
  2. jcsd
  3. Oct 21, 2008 #2

    tiny-tim

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    Hi PhysicslyDSBL! :smile:
    That's right! :smile:

    So, using the acceleration from part a), do F = ma (in the horizontal direction) for the back block. :wink:
     
  4. Oct 21, 2008 #3
    Thanks Tiny Tim!

    Maybe I have been staring at physics too long today 'cause I'm sure this is more obvious than it seems to me right now.

    So a=2.06, m=2 (the back block) which equals 4.12N. a does not equal 9.8 because this is the horizontal.

    It makes sense to me that the above should be the answer but its not. If I am still missing something can someone please fill me in?

    Thanks much!
     
  5. Oct 21, 2008 #4

    Doc Al

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    This is the net force on the back block. Note that the rope tension is only one of the horizontal forces acting on the block. What's the other?
     
  6. Oct 21, 2008 #5
    The other force would be the pulling force on both blocks. I am unsure about what to do with this.

    My book gives: F-T=m1a and T=m2a

    m1 and m2 seem to be backwards in this problem as the rope is connected between m1 and m2 [m2]------[m1]-----

    F=force (net force? 4.12)
    T=tension
    m1 = 2kg
    m2 = 3kg

    10.3-T=2(2.06)
    T=3(2.06)

    Substituting, this would equal 0.

    Sorry I'm getting so confused
     
  7. Oct 22, 2008 #6

    tiny-tim

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    Hi PhysicslyDSBL! :smile:

    No … Doc Al :smile: is asking you for the other (horizontal) force on the back block.

    In F = ma, F is the net force, so you must find all the horizontal forces, and add them.

    Look at the question again, and draw a diagram, with all the forces on it! :wink:
     
  8. Oct 23, 2008 #7
    Thanks for all the help everyone!

    After a few more calculations and substitutions I finally got the answer of 10N.
     
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