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!

Homework Help: Acceleration of pulley system

  1. Jul 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider the systems of 2kg and 5kg blocks. The 2kg block is resting upon a smooth frictionless horizontal surface and friction in pulley bearing is negligible. The 5kg block is suspended by a cord connected to the 2kg block. Determine the acceleration of system and tension of cord.

    2. Relevant equations

    3. The attempt at a solution

    To find acceleration of system, I understand that I will have to get the external forces divided by the total mass of system.

    | a. system | = F. ext / m
    a = [5(9.81) - 2(a)] / 5+2
    a = (49.05 - 2a) / 7
    7a = 49.05 - 2a
    a = 5.45 m/s^2

    My answer above is incorrect although it appears to be logical to me. The correct answer for acceleration of system is 7m/s^2. I haven't attempted the second part of the qn. Anyone please enlighten me on what went wrong in my approach to this question
  2. jcsd
  3. Jul 30, 2016 #2


    User Avatar
    Science Advisor

    Can you explain where each of the terms in this equation came from and justify each one?

    If my guess is correct, the problem is that you have not carefully defined which pieces of your system count as part of the "m" and which count as external.
  4. Jul 30, 2016 #3
    Since the pulley will move in the direction of the 5kg block suspended, I let its force be positive, which is 5 x 9.81 Another force will be coming from the 2kg block moving horizontally in an unknown acceleration, which will be 2 x a Since this force opposes the direction of force the system is going in, it's negative. Therefore external forces acted on system is (5 x 9.81) - 2a

    Total mass in system will be 2+5 = 7kg

    Applying F = ma
    a = F/m
    That's how I came up with my workings.
  5. Jul 30, 2016 #4


    User Avatar
    Science Advisor

    You have still not defined what things are internal to your system and what things are external. You mention the force of the pulley. What force do you mean by this? I can think of three.
  6. Jul 30, 2016 #5
    Hmm to be honest I'm unsure of what does it mean by external forces or internal. It's just so that in a question earlier which gives that two weights are being suspended by a pulley, I used this approach and it somehow worked and I'm merely trying to reapply it to this question.
    * perhaps external forces means the weight of the blocks?
  7. Jul 30, 2016 #6


    User Avatar
    Science Advisor

    According to Newton's third law, all forces come in pairs (*). They involve two objects and the force of the one on the other is equal and opposite to the force on the one. An "external" force is one that acts between an object inside the system and an object outside. An "internal" force is one that acts between two objects both of which are inside the system. F=ma only concerns itself with external forces acting on internal bodies m. An internal force cancels out since the two sides of the force pair are equal and opposite. You get to pick which objects are inside the system and which are outside when you analyze the problem.

    Do you know how to draw a "free body diagram"? You pick one object and draw all of the forces that act on that object. This helps keep track of which forces matter and which do not.

    Are considering both the 2kg and the 5kg block as part of your system?
    Are you considering the force applied by the cord on the 5kg block?
    Are you considering the force applied by the cord on the 2kg block?

    (*) At some point in your education you may encounter inertial forces arising from the use of a non-inertial reference frame. Those do not have third law partners.
  8. Jul 30, 2016 #7
    kay... So I've realised that the 2kg block moves horizontally and that the only force that it exerts in the system is the tensional force of the cord.
    Correcting my previous workings will give me:
    a= [Ft - Ft + 5(9.81)] / 5+2
    = 49. 05/ 7
    ≈ 7m/s^2
    Which is the correct answer...
  9. Jul 30, 2016 #8
    But just as I thought I have managed to solve the problem.. Another popped up and I would like to clarify this...
    If T= ma
    T of cord horizontally is 2a while T of cord vertically would be 5a....
    But we know that the tensional force would have to cancel out each other somehow and what I've discovered as shown above shows I can't be.. Ahhhh would appreciate if you could help point out what I've wrongly understood... :H
  10. Jul 30, 2016 #9


    User Avatar
    Science Advisor

    You have to be clear on what body you are contemplating and what forces act on that body.

    Consider the 2kg mass on the table. The only force acting on that mass (in the x direction) is the tension T in the cord. The mass m is 2kg. The acceleration is a.

    ##F=ma## so ##T=ma## and T=14 kg m/sec2 as you have already calculated.

    Now consider the 5kg mass hanging from the cord. There are two forces acting on that mass.
  11. Jul 30, 2016 #10
    Ahhhhh, took me some time to figure it out.
    Tension 2a does not equate to 5a. It equates to 49.05 - 5a. That makes much more sense. Thank you for your time jbriggs444, appreciate it greatly :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted