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!

Dependent motion between 2 bodies to find velocity

  1. Mar 23, 2009 #1
    1. The problem statement, all variables and given/known data

    If the system is released from rest, determine the speeds of both masses after B has moved 1m. Neglect friction and the masses of the pulleys.

    http://img14.imageshack.us/img14/6321/fbd.jpg [Broken]

    3. The attempt at a solution
    I have drawn up the free body diagram for both blocks, as well as labelled the lengths Sa and Sb relative to their respective datum points/lines.

    I obtain the equation 3Sa + 2Sb = L* where L* is the length of rope that is constant
    this leads to the velocity and acceleration equations:

    3Va = -2Vb and 3Aa = -2Va

    My question is where do I go from here?
    Do I do newtons second law, sum of forces = ma?
    This only gives me acceleration in terms of tension T
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 23, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    First of all you have to find the tension in the rope.
    Since rope is single, acceleration of both the masses must proportional to the change in the lengths. .
    Calculate the accelerations of the both the masses and equate them. That gives you the tension. If the know the acceleration of either mass, using kinematic equation, you can find the the velocity.
     
    Last edited: Mar 23, 2009
  4. Mar 23, 2009 #3
    I have seen other examples with 2 hanging blocks suspended by a pulley system where there was only one rope and acceleration of the two blocks was different.

    Additionally, what is the point of the acceleration equation 3Aa = -2Va, if it is contrary to what you say?
     
  5. Mar 23, 2009 #4

    rl.bhat

    User Avatar
    Homework Helper

    OK. I have edited the previous post. When B moves 1 m downward how much A moves upward?
     
  6. Mar 23, 2009 #5
    Firstly, I would be inclinded (pardon the pun) to say block A moves downward and block B moves upwards.

    Secondly, I am not 100% sure mathematically how to determine how far A would move when B moves 1m. I am tempted to say 1.5m due to their being 3*Sa and 2*Sb
     
  7. Mar 23, 2009 #6

    rl.bhat

    User Avatar
    Homework Helper

    OK. Now find acceleration of A and B and equate them in the ratio 2Aa = 3Bb
    Take the acceleration positive in both cases i.e in the direction of increasing velocity.
     
    Last edited: Mar 23, 2009
  8. Mar 23, 2009 #7
    Ok, If B moves 1m up, Sb will shorten by 2m.
    Hence Sa will be 2/3m as sort of suggested by the displacement equation.

    Mathematically, how do we come to this conclusion from: 3Sa + 2Sb = L*

    => 3Sa = L* - 2Sb
    => Sa = (L* - 2Sb)/3

    What does this mean physically?
     
  9. Mar 23, 2009 #8

    rl.bhat

    User Avatar
    Homework Helper

    What does this mean physically?
    It means that if B moves x m, A moves 2/3 x m. Since they take same time for the displacement Bv =2/3Av and Ba = 2/3 Aa
     
    Last edited: Mar 23, 2009
  10. Mar 23, 2009 #9
    I could understand that if the displacement equation was Sa = 2Sb/3
    I am confused due to the constant length L*

    Sa = (L* - 2Sb)/3
    Sa = L*/3 - 2Sb/3

    Doesn't the L*/3 also affect the length of Sa?
     
  11. Mar 23, 2009 #10

    rl.bhat

    User Avatar
    Homework Helper

    Total length L = 2Sb + 3Sa.
    Take time derivative on both side. Note that L is constant. That gives you the velocity of A and B. Again take the time derivative of velocities. That gives you the accelerations of A and B.
     
  12. Mar 23, 2009 #11
    Ok, I found
    Aa = -1.099 m/s^2
    Ab = 1.649 m/s^2

    I know acceleration and initial velocity and I want final velocity when B has moved 1m.

    I want to use Vf^2 = Vi^2 + 2as, but acceleration isn't constant because it starts from rest?
     
  13. Mar 23, 2009 #12

    rl.bhat

    User Avatar
    Homework Helper

    When you drop any object from a height, it also starts from rest. Still it has a constant acceleration. Similarly here also the initial velocity is zero and acceleration is constant.
     
  14. Mar 24, 2009 #13
    how do I prove/show that acceleration is constant?
     
  15. Mar 24, 2009 #14

    rl.bhat

    User Avatar
    Homework Helper

    The movements of A and B are caused by the gravitational forces on them.
    When an objects slides on an inclined plane, its acceleration decreases but it remains constant.
     
  16. Mar 24, 2009 #15
    don't you mean it constantly increases since it starts from rest?
     
  17. Mar 24, 2009 #16

    rl.bhat

    User Avatar
    Homework Helper

    No. The velocity increases.
    When we write v = u + at, a remains constant.
    When one mass is pulling other, both must have the same acceleration.
     
  18. Mar 24, 2009 #17
    Ok, now that I have established acceleration is constant, I am able to use Vf^2 = Vi^2 + 2as.

    However, since my acceleration of block a is negative, I run into a problem with taking square root of a negative. Have I done somthing wrong?
     
  19. Mar 24, 2009 #18

    rl.bhat

    User Avatar
    Homework Helper

    Negative or positive sign indicates the direction of the acceleration. You have to find the magnitude of the acceleration. For that take the positive value.
     
  20. Mar 24, 2009 #19
    cool.

    I figured out V of block b = 1.82m/s (upwards) and V of block a = 1.21m/s (down the incline)
    These answers seem reasonable to me, I am confident they are correct.

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