1. Not finding help here? Sign up for a free 30min 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!

Torque of Pulley with Two Masses2

  1. Aug 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Two weights are connected by a very light flexible cord that passes over a 40.0-N frictionless pulley of radius 0.300 m. Weight #1 is 125 N and Weight #2 is 75 N. The pulley is a solid uniform disk and is supported by a hook connected to the ceiling. What force does the ceiling exert on the hook?

    2. Relevant equations
    F=ma
    τ=FtanR=Iα

    3. The attempt at a solution

    The pulley applies a downward force to the hook. Since the pulley isn't accelerating, the net force on it must be zero. The pulley's downward force on the hook is due to its own weight, but also two different tension forces since the pulley isn't massless.

    First, I decide to try and find the two tension forces, T1 and T2. I set up equations for the motion of the hanging weights:

    F1=m1a=T1-m1g
    F2=m2a=T2-m2g

    Since they are connected by a rope, they should have the same acceleration.

    a= [T1-m1g] / m1

    I plugged this into the equation for F2, getting:

    m2[T1-m1g] / m1 = T2-m2g

    There are still two unknowns in this equation, so I set up a torque equation:

    τ=FtanR=Ipulleyα

    Using the moment of inertia of a solid cylinder (0.5McylR2) and a=αr, The torque force is (assuming no slipping):

    (T1-T2) x R = 0.5Mpulleya

    I solved this torque equation for T1, getting:

    T1=0.5(Mpa+T2

    I then plugged this into my F1 equation to solve for acceleration. I then used acceleration to solve for T2. After finding T2, I used its value to solve for T1. The algebra got a bit complex, and I ended up with 3.19 N for T2 and 63.2 N for T1, which are incorrect.

    Can anyone point out if my equation set up was incorrect or if I messed up in the algebra?
     
  2. jcsd
  3. Aug 4, 2017 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Not exactly. There is a crucial difference. Hint: acceleration is a vector.
     
  4. Aug 4, 2017 #3

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    On first glance, you have assumed that the accelerations are the same. This is not true. They have the same magnitude but in opposite directions.
     
  5. Aug 4, 2017 #4
    So the acceleration of weight 1 and weight 2 are opposite. That would make my F1 and F2 equations:

    F1=m1(-a)=T1-m1g
    F2=m2a=T2-m2g

    Would the linear acceleration of the rotating pulley also be negative?

    Edit: I did the math with the following equations:

    F1=m1(-a)=T1-m1g
    F2=m2a=T2-m2g
    τ=T1-T2=0.5Mp(-a)

    (With the pulley's linear acceleration in the negative x-direction)

    The algebra was quite involved (I got equations with 5-6 terms on each side and usually an indicator that I did something wrong), but I went through and got T2=-8.33N and T1=-33.7 N. I added these together along with the 40 N weight of the pulley, getting -82.03 N downwards force being applied by the pulley onto the hook, so the hook would apply 82.03 N back, but this was incorrect. Are my three equations incorrect or should I just double check my math?
     
    Last edited: Aug 4, 2017
  6. Aug 4, 2017 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That says you are taking a as positive down on the weight 1 side.
    If weight 1 goes down then T1>T2, but this equation would make a negative.
    Negative tension? That should have been a hint.
     
  7. Aug 4, 2017 #6
    I'm still confused on the signs. Weight 1 (125 N) would go down, hence it's acceleration is downward. Hence the negative sign in front of the a term. Weight 2 (75 N) goes up, so its acceleration is positive.

    I'm shooting in the dark on the sign of the acceleration for the pulley. For angular acceleration, counterclockwise is positive, but for I'm really not sure on linear acceleration. Can I just use the magnitude of acceleration, since the pulley isn't really translating?

    Edit:

    Instead of setting up my equations like above, worrying about the sign of acceleration, can I set up my equations like this and have my signs be coordinating along the pulley?

    F1=W1-T1=m1a (it will be moving downward W1>T1)

    F2= T2-W2=m2a (will move upwards since T2>W2)

    I can set up my torque equation to where it's rotating with a positive torque:

    T1-T2R=0.5mpR2(a/R)

    Which cancels out to:

    T1-T2=0.5(Wp/g)a

    The directions make more sense to me here. Is this fine?

    Edit #2:

    These equations turned out to be correct. It took me so long to solve it as I kept messing up the algebra. I ended up having to plug all my numbers in to get a much more simple-looking system of equations initially, and then solve them instead of carrying symbols around like we normally do in physics.

    Thanks for your help everyone!
     
    Last edited: Aug 4, 2017
  8. Aug 4, 2017 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes.
    To be clear, you do not need to guess correctly in advance which way the system will move; you just have to make sure the equations are internally consistent.
    If we take a as the downward acceleration for m1, then
    m1a = m1g-T1
    (T1-T2)r= (moment of inertia of pulley)(a/r)
    If we switch to taking a as up for m1 and down for m2 then it is also reversed for the pulley:
    m1a=T1-m1g
    (T2-T1)r=(moment of inertia of pulley)(a/r)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Torque of Pulley with Two Masses2
  1. Torque on a Pulley (Replies: 17)

  2. Torque on a Pulley (Replies: 6)

  3. Torque on a pulley (Replies: 1)

Loading...