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!

Forces On A Ramp Finding Acceleration Question?

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Jane, mass of 55kg holds one end of a vertical rope that passes up over smooth rock and down along a 30 degrees sloping hill as shown. But Josh, mass of 95kg tightly holds the other end of the rope. Assume no friction. Calculate the acceleration. The picture is included





    Chatham-Kent-20120327-01002.jpg




    3. The attempt at a solution

    Fjane = m x g = 55 x 9.81 = 539.5 N
    Fjosh = m x g = 95 x 9.81 x sin 30 degrees = 466 N
    I know what to do from here but my main question is, will jane go up or will she go down??
    According to the forces of gravity jane will go down, right? But the heavier mass belongs to Josh.
    If Jane was to get pulled up than the equations would be:

    1.T(tension) - mjane(9.81) = mjane x acceleration

    2.mjosh x sin(30) - T(tension) = mjosh x acceleration

    Is that correct?

    if it is i end up with -0.5m/s^2
    However, im not sure if its right please correct me
     
  2. jcsd
  3. Mar 27, 2012 #2

    PeterO

    User Avatar
    Homework Helper



    Clearly if Josh was on a level surface, jane would go down, and Josh would go towards the rock.

    If the rock was a magic free floating rock, so the the rope to both Josh and Jane was vertical (like over a pulley) Josh would go down, Jane would go up.

    Somewhere between the (for Josh) vertical and horizontal extremes they will go no-where - the system will be in balance.

    The problem to comtemplate, which side of the balance point is a 30 degree slope.
    Your calculated figures are good - and should lead you to a correct decision.
     
  4. Mar 27, 2012 #3
    But in this situation doesnt Jane go down and Josh go up because of the forces

    Janes force = 539.5 Newtons
    Josh's Force = 466 Newtons

    Since Janes force is greater than Joshs force, wont Jane go down and Josh go up??
     
  5. Mar 27, 2012 #4

    PeterO

    User Avatar
    Homework Helper

    That is a correct analysis of the situation you presented.
    There will be a tension in the rope, the same throughout. It will be an appropriate value between the two forces you have - lets say 500N

    At that value the force on Jane are 539.5N down (weight) 500N up (tension) so net force 39.5 Down.
    Force on Josh 466N down the slope (weight component) 500N up the slope (tension) so net force 34 N up the slope.

    I think you will find that those net forces will give different accelerations for Josh and Jane, which is not possible; they are tied together. BUT if you had the correct tension (I just made up the 500N value), the net force on each would give the appropriate, common acceleration.
     
  6. Mar 27, 2012 #5
    But when you go and solve it wont tension cancel, and initially since tension is the same for both objects wouldnt it be the same answer anyways? Oh and also i would have to divide by the total mass of both Jane and josh right? Or would i just use Jane because it only asks for her VERTICAL acceleration upwards?
     
  7. Mar 27, 2012 #6

    PeterO

    User Avatar
    Homework Helper

    Won't tension cancel with what?

    If the tension in the rope was 500N, that means the rope pulls up on Jane with a force of 500N, and pulls along the slope with a force of 500N on Josh. Now a force on Jane is certainly not going to cancel out a force on Josh ???

    The tension also means a very complicated force effect on the Rock - but the rock is part of the Earth so the force won't achieve very much due to the ENORMOUS mass of the Earth, so in generally ignored.
     
  8. Mar 27, 2012 #7

    PeterO

    User Avatar
    Homework Helper

    You will need to consider the mass of both when calculating the Tension, if you wish to use Tension during your calculations.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook