1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Newton's Laws, Finding Correct Acceleration and Applied Force

  1. Mar 5, 2012 #1
    1. The problem statement, all variables and given/known data
    John (mass = 80 kg) rests on the top face of a light cube (mass = 0). A light rope passes horizontally from him over the front edge, and vertically down to Brian (mass = 60 kg) who hangs in contact with the front face as shown. The cube rests on a horizontal surface and there is no frictional force anywhere. Joe (mass = 70 kg) has his back against a wall and applies a horizontal push perpendicular to the rear face. Calculate the value of this force, which will prevent both John and Brian from sliding over the surface of the cube.

    2. Relevant equations
    FUN = ma
    Fg = mg

    3. The attempt at a solution
    John (80 kg) = m1
    Brian (60 kg) = m2

    Fg2 = m2g
    Fg2 = (60)(9.8)
    Fg2 = 588 N

    Since John and Brian should not slide off the cube, I'm guessing the force of tension in the rope should have the same magnitude as the force of gravity.

    FT = Fg2
    FT = 588 N

    FUN = m1a1
    FT = m1a1
    a1 = FT/ m1
    a1 = 588/ 80
    a1 = 7.35 m/s²

    The light box will need a force to accelerate it at a rate of 7.35m/s².

    The mass of the box will be equal to 60 kg because it is in contact with Brian.

    FUN = m3a1
    FA - FOn 3 due to 2 = m3a
    FA = m3a + FOn 3 due to 2
    FA = (0)(7.35) + (m2a2)
    FA = 0 + (60)(7.35)
    FA = 441 N

    I'm not sure where I went wrong, but it doesn't seem right. Thank you very much for any help.
  2. jcsd
  3. Mar 5, 2012 #2


    User Avatar

    Staff: Mentor

    Excellent up to here!
    Joe needs to accelerate the whole system, including both John and Brian. In fact, it's Brian's inertial force that counteracts the tension in the rope he's holding.
  4. Mar 6, 2012 #3
    Oh! I think I understand! Instead of putting a mass of 0, I should use 80 kg because that is John's mass?

    I would get:

    FUN = m3a1 - m2a1
    FA = (80)(7.35) - (60)(7.35)
    FA = 147 N

    Is 147 N correct?
  5. Mar 6, 2012 #4


    User Avatar

    Staff: Mentor

    No, Joe needs to produce this acceleration while pushing BOTH Brian and John. The masses add.
  6. Mar 6, 2012 #5
    If I add John and Brian's mass together, I think I should get:

    FUN = (m1 + m2)a1
    FA = (80 + 60)(7.35)
    FA = 1029 N
  7. Mar 6, 2012 #6


    User Avatar

    Staff: Mentor

    That looks good.
  8. Mar 6, 2012 #7
    Thank you very much!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook