1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

System of moving bodies

  1. Jun 5, 2004 #1
    i'm working without my textbook and going by the some solutions that i have so i need some help

    1) Two masses, m1 and m2, joined with a cord are dragged along the horizontal plane surface. the applied force is F=20N, mass m1=m2=3kg, and the acceleration is 0.50 m/s. what is the tension, T, in the connecting cord if the frictional forces on the two blocks blocks are equal? How large is the frictional force, F, on each block?
    diagram:
    | m1 |-----| m2|------>F
     
  2. jcsd
  3. Jun 6, 2004 #2
    can you explain what you have done to try and solve this?
     
  4. Jun 6, 2004 #3
    i've solved simpled problems--problems with only one body moving. so i tried to apply those principles here:

    for tension:
    T=3kg*.50
    i know know that is wrong since there is a frictional force. i just dont know how to compute it since the coefficient of friction is not given
    so, there it is. i didnt get very far into it.
     
  5. Jun 6, 2004 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Attack the problem step by step. Start, as always, by identifying all the forces acting on each of the masses. Draw a picture with those forces labeled. (All we care about are the horizontal forces.)

    Once you've done that, there are several ways to proceed. Here's one way. Treat each mass separately and apply Newton's 2nd law. You'll get an equation for each mass. Combine them and you can solve for the tension in the cord and the fricitional force. Give it a shot.
     
  6. Jun 6, 2004 #5
    i did give it a shot, please i'm totally lost here, my Final that i must Ace is tomorrow. please work this problem out for me, somebody
     
  7. Jun 6, 2004 #6

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    What forces are the front block subject to?
    Clearly, the applied force F, a tensile force T, and a frictional force f.
    Obviously, both the tensile force and the frictional force works in opposite direction than F, so we have for particle 2:
    [tex]F-T-f=ma[/tex]
    For the rear block, the tensile force has the same magnitude T, but works in the opposite direction.
    Hence, we have:
    [tex]T-f=ma[/tex]
    You have only 2 unknowns here, T and f, and 2 equations..
     
  8. Jun 6, 2004 #7
    from the equations i know how get get T and f. i have another question, how do you know T is opposite the force. i acutally thought T was going the same direction as the force for the box in the front. or is the tension always opposite the force. also, for the box in the rear, why did you leave out F out of the equation and not write F-T-f=m*a. i dont understand because the force is also acting on this box.
     
  9. Jun 6, 2004 #8

    Doc Al

    User Avatar

    Staff: Mentor

    The tension in the cord exerts a pull (T) at both ends of the cord. According to your diagram, the cord pulls m1 to the right and m2 to the left.
    The force F is only applied to mass m2; it does not touch m1. It will, of course, affect the tension in the connecting cord--but that's a separate force.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: System of moving bodies
Loading...