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!

Atwood Machine

  1. Jan 23, 2005 #1
    Need a check on this one please. In an Atwood Machine simulation, m1 is 1.0kg and m2 is 1.1kg. m2 rests on the floor that exerts a normal force, Fn, on m2. when the system is in equalibrium, the tension, T, in the rope at 2.54s is 9.789. If Fn +T- m2*g = 0 then Fn = .991. Is this correct?
     
  2. jcsd
  3. Jan 24, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    equilibrium

    If the question is "What normal force does the floor exert on m2 when m2 rests on the floor?" then you're probably OK. (Although I have no idea what 2.54s has to do with anything.)

    The forces on m2 are: [itex]F_n + T - m_2 g = 0[/itex]; similarly, the forces on m1 are: [itex]T - m_1 g = 0[/itex]. So [itex]F_n = (m_2 - m_1)g[/itex].
     
  4. Jan 24, 2005 #3
    equilibrium

    The question "What is the normal force, FN, and what is the tension, T, in the rope? T (at t = 2.54s) = ______. At equilibrium: FN + T – m2 • g = 0. Then FN = _______." This is a timed simulation and at 2.54s the rope tension,T, is 9.789
     
  5. Jan 24, 2005 #4

    Doc Al

    User Avatar

    Staff: Mentor

    OK, now I see what you're talking about. Since T is given and equilibrium is assumed, your answer is correct; rounded off, the normal force would be 1.0 N.
     
  6. Jan 25, 2005 #5
    equilibrium

    Thanks for the quick reply.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Atwood Machine
  1. Atwoods Machine (Replies: 4)

  2. Atwoods machine (Replies: 1)

  3. Atwood's machine (Replies: 12)

  4. The Atwood machine. (Replies: 1)

  5. Atwood Machine (Replies: 5)

Loading...