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Homework Help: Newtons Laws - Boxes connected with a cord

  1. Feb 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A block A (mass 2.25k) rests on a tabletop. It is connected by a horizontal cord passomg over a light, frictionless pulley to a hanging block B (mass 1.3kg). The coeffisient of kinetic friction between block A and the tabletop is 0.45. After the block is released from the rest,
    fint the speed of each block after moving 3cm.

    2. Relevant equations

    3. The attempt at a solution
    I found the frictionforce by:

    Frictionforce = coeffisient * N (where N = G of block A).

    Frictionforce = 9,93 N.

    The force on the rope will be = the G (gravityforce) on block B.

    ForceOnRope = 12,75N

    Then I tried to use Newtons second law to find the acceleration, but this attempt failed when I tried to find the velocity of the boxes, which will be the same?
  2. jcsd
  3. Feb 17, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The tension in the rope does not equal the weight of block B. If it did, then the blocks would be in equilibrium.

    Call the tension T. Set up equations for both blocks and you'll be able to solve for the tension and the acceleration.
  4. Feb 17, 2008 #3

    Note that the Tension will be the same throughout the rope. So if you set up the forces on each block and solve them for the Tension, you can then set those 2 equations equal to solve for other variables, such as the acceleration. Then you can use kinematics to solve for the speed.
    Last edited: Feb 17, 2008
  5. Feb 17, 2008 #4
    Thanks, will try this tomorrow morning!
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