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!

A tug-of-war on the ice

  1. Feb 16, 2008 #1
    James and Ramon are standing 20.0 m apart on the slippery surface of a frozen pond. Ramon has mass 60.0 kg and James has mass 90.0 kg. Ramon pulls on the rope to give himself a speed of 0.70 m/s. What is James's speed?

    From what I understand, since there is an external force due to Ramon's pull, momentum is not conserved, and I can't use the conservation of momentum equation. So, I'm thinking this is a center of mass problem, with V_cm = 0.70 m/s, and James's speed is 0.70 m/s. However, that's not the correct answer. Can anyone help me with this?
  2. jcsd
  3. Feb 16, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Well, there's certainly an "external" force on James, if you consider James and Ramon as separate systems. But what if you consider them as parts of the same system?

    Alternatively, consider Newton's 3rd law: Whatever force that Ramon exerts on James must equal the force that James exerts on Ramon.

    You can view this as a center of mass problem (note that this is equivalent to using conservation of momentum). But what makes you think that the V_cm = 0.70 m/s? That's James's speed after the pull, not the velocity of the cm.
  4. Feb 17, 2008 #3
    It turns out that I have to use center of mass for this problem.
    What I have so far:
    (m_J)(v_J)+(m_R)(v_R)=(m_J + m_R)V_cm
    The thing is, I have two unknowns here: v_J and V_cm. I suppose v_J is what I'm solving for. I'm stuck.
  5. Feb 17, 2008 #4
    first solve this prob to get a better understanding.....

    Two people of equal mass, 6 meters apart, attempt a tug of war on frictionless ice. If they pull on opposite ends of the rope with equal forces, each slides 3 meters to a point midway between them.Suppose instead that only one person pulls and the other fastens the rope around his or her waist. How for does each person slide? (Neglect any effects of the rope's mass.)
  6. Feb 17, 2008 #5

    Doc Al

    User Avatar

    Staff: Mentor

    What you're using is conservation of momentum. And you are given V_cm: They start from rest. :wink:
  7. Feb 17, 2008 #6
    Thanks, guys! I got it! =)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: A tug-of-war on the ice
  1. Tug Of War (Replies: 1)

  2. Tug-of-war paradox (Replies: 1)

  3. Tug-of-War Problem (Replies: 2)