Homework Help: Torque of Pulley with Two Masses2

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1. Aug 4, 2017

BBA Biochemistry

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. Aug 4, 2017

haruspex

Not exactly. There is a crucial difference. Hint: acceleration is a vector.

3. Aug 4, 2017

Orodruin

Staff Emeritus
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.

4. Aug 4, 2017

BBA Biochemistry

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 by a moderator: Aug 4, 2017
5. Aug 4, 2017

haruspex

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.

6. Aug 4, 2017

BBA Biochemistry

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.

Last edited by a moderator: Aug 4, 2017
7. Aug 4, 2017

haruspex

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)