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Question regarding basic dynamics and forces involved?

  1. Sep 13, 2013 #1
    1. The problem statement, all variables and given/known data

    "A 60kg student lifts herself from rest to a speed of 1.5m/s in 2.1s. The chair has a mass of 35kg."

    All information is included in the accompanying image. In short, it is a basic kinematic and dynamic equation. Currently, we are only working on problems that involve constant acceleration or zero acceleration. Anyway, the book only gives one answer to one part of the problem, and it states that the force of the rope is 500N, while I'm getting ~998N as can be seen in the image below. I am at a complete loss as to where I'm going wrong. Hopefully, my writing is legible enough for you to understand what I did.

    Also, though I know I'd need to indicate position on my force diagrams, this isn't homework, just practice for an exam this Monday. I was wondering though if the chair needs a force for both gravity and the student since both are acting on it? If so, did I label it right? Our instructor never showed us how to label multiple forces acting on an object in one direction yet.

    2. Relevant equations

    The only equations I am concerned about are the ones involving the dynamics. I feel I have the kinematics down sound. However, I am not sure of how to go about setting up the dynamics for this problem to solve for the force of the rope?

    3. The attempt at a solution

    Well, here is what I was doing:
    1) Setting the rope as a positive force in the equation
    2) Calculated the gravitational force of the student to be 588N (60kg*9.8m/s^2)
    3) Calculated the gravitational force of the chair to be 343N (35kg*9.8m/s^2)
    4) Set up my dynamic equation as:
    F(rope) - F(student) - F(chair) = [m(chair) + m(student)] * .71 (rate of acceleration)
    F(rope) - 588N - 343N = 95 * .71
    F(rope) = 998.45N

    Since the answer the solution page had for the force of the rope was 500N, I know something is wrong here, but I have no clue what?

    CDFLiV1.jpg
     
  2. jcsd
  3. Sep 13, 2013 #2
    It looks like one of the forces you forgot on your student diagram was the force of the rope pulling upward on the student. I believe if you include this, you should get something like ## 2 F_{rope} = 998.45 ##, which works out to about the 500 N you expect. I havn't checked that out yet, but try it and see if that solves your problem.

    Just a free body issue, the rest of it (the kinematics/weights, at first glance) looks great.
     
    Last edited: Sep 13, 2013
  4. Sep 13, 2013 #3
    For starters, thanks for the feedback.

    As far as your suggestion, I am still at a loss. The total gravitational force of both the student and the chair (95kg) comes out to be 931N. So, in order for there to be an acceleration upwards (positive), the force of the rope would have to be greater than 931N, wouldn't it? When you multiply the acceleration by the total mass of the student and chair, it comes out to 67.45N. So, from what I was thinking, the 67.45N would be the additional N necessary to attain the acceleration above and beyond the 931N necessary to equal the gravitational force. Please tell me where my reasoning is wrong here, as I am still confused?
     
  5. Sep 14, 2013 #4
    Hmm, well let's try thinking about the free body diagrams again--I may have misunderstood your labeling which might be causing your confusion if you are missing some things too.

    For the student: ## F_{rope}## and the normal force exerted by the chair point upward, while the weight of the student points downward.

    For the chair: again, ## F_{rope} ## points upward, while the weight of the chair as well as the normal force of the student point downward.

    Try doing a dymanic analysis on the student first, with their mass and acceleration you discovered, and then the chair with its mass and the same acceleration.

    Keep note that the two normal forces are equal from Newton's 3rd law, so you should be able to substitute one set of equations into the other to solve for the rope force alone, since you know the weight forces for each too.

    See if that helps your understanding-->hope I didn't make it worse!
     
  6. Sep 14, 2013 #5
    Ok, I now see where I went wrong. Thank you so much for the clarification. My problem was that I was looking at the force of the rope as acting collectively on both the student and the chair due to not taking into consideration the normal force of the chair exerted on the student.

    I see now why they say it sometimes takes a while to wrap your mind around the concepts. Thanks again for the help.
     
  7. Sep 14, 2013 #6
    Sure thing, glad we could work it out!
     
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