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Homework Help: Two objects, and where they meet

  1. Mar 6, 2014 #1
    1. The problem statement, all variables and given/known data
    A worker with a mass of Mw is pulling on a mass-less rope that is attached to a box with a mass of mb on a friction-less surface. The worker pulls with a constant force starting at rest. The Worker is at x = 0, and the box is at xb Find the position at which they meet in terms given.

    2. Relevant equations

    x = xi + vi * t + 1/2 a * t2
    F = m * a

    3. The attempt at a solution

    Using newton's laws I know that the force on each object is equal, so

    Fw = Fb

    xw f = xw i + vw i* t + 1/2 a * t2

    thus for the worker's side of the equation
    xw f = 1/2 a * t2

    and the box moves

    xb f = xb i + vb i* t - 1/2 a * t2


    xb f = xb i - 1/2 a * t2

    since they meet xb f is equal to xw f


    1/2 a * t2 = xb i - 1/2 a * t2

    xb i = 1/2 aw * t2 + 1/2 ab * t2

    2 xb i = (aw + ab) * t2

    t = √( ( 2 * xb i) / aw + ab)

    then taking what I have just solved for time, and plugging that back into the basic kinematic equation to find the distance I get jibberish. So I'm not sure where to go from here.

    xmeet = xinitial + 1/2 aworker * ( ( 2 * xb i) / aw + ab)
  2. jcsd
  3. Mar 6, 2014 #2
    They don't have the same acceleration, do they?
    But do you really need to go through this?
    Have you learned about center of mass?
  4. Mar 6, 2014 #3
    Okay, your approach could work eventually, but the problem is that you don't know what your acceleration is, since you aren't given the force with which the worker pulls. There's a simpler way to go about this though. Think about the worker-block system. Is the tension in the rope external or internal to this system?
  5. Mar 6, 2014 #4


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    Staff: Mentor

    Use F=m•a to determine the acceleration of each of the bodies.
  6. Mar 7, 2014 #5
    fun mass calculations

    I know the force on the person, and block is the same but I do not know the mass, or ratio of mass. So the closest I can get to knowing the constant accelerations is a = F/mworker/block depending on which mass is used.

    While I know about the center of mass, it isn't something I am supposed to use. Nor would it be okay to use as it is not a given, or something simple as a kinematic equation.
    Last edited: Mar 7, 2014
  7. Mar 7, 2014 #6


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    Staff: Mentor

    That's right. The two bodies experience different accelerations.
  8. Mar 7, 2014 #7
    If you don't use center of mass (by the way, you don't have to be "given" it to be able to use it) then your approach can work. However, with the answer you have you have it in terms of acceleration, but you want to have it in terms of only things given in the problem, so replace the accelerations by force and mass.
  9. Mar 10, 2014 #8
    relative displacement and velocity concept would be nice here.
    What would be the relative displacement of the man wrt the box, taking origin to be at the initial position of the man ?
  10. Mar 10, 2014 #9
    @jack per the question I actually have to be given that to use it, or the answer is not correct. The whole if your method does not match it is not correct issue.

    @phoenix The worker starts at x, which happens to be zero, and the box starts at x b. They meet at xf, and x box final(the same spot). Now this occurs at the same instant so the displacement of time delta T is therefore equal. My "solutions" just don't work.
  11. Mar 10, 2014 #10
    Your solutions are OK now.
    Just replace the accelerations in terms of the known masses.and the tension in the rope.
    The tension will simplify.
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