# Help with a lab practical

1. Jan 3, 2010

### Centurion1

1. The problem statement, all variables and given/known data
Well i have a lab practical due for physics and im having trouble with it. we have to assemble a tri level mobile. that wasn't that hard. the difficult part is doing the math to support my answers. i need to use translational and rotational equilibrium. i looked through my books but i couldnt find anything past a single help problem that doesnt apply to me and a brief explanation

2. Relevant equations

i dont have any equations because i have no idea how to set it up. can anyone walk me through a practice problem? i would really appreciate it.

3. The attempt at a solution

sorry. i understand that it needs to equal out so the net force equals zero so that it is in balance (equilibrium) im just not sure how to back it up with my numbers.

2. Jan 3, 2010

### Oddbio

Well for a mobile you don't need to worry about translational equilibrium..

But for rotational equilibrium, you need to make sure that the torques are balanced.

Torque = F*d
The component of force must be perpendicular to the surface (which is always true for a mobile so just use the force due to gravity).

So basically the only way that none of the arms would be moving, is if the net torque on that arm is zero. But on each side of the arm you will have two downward torques, so you just need to make sure that the magnitude of the torque on one side is equal to the magnitude of the torque on the other side (for each arm).

3. Jan 3, 2010

### Centurion1

how does gravity factor in?

so would there be a total of three forces?

and it would be like this?

Fnet= Fa + Fb + Fg = 0

also my mobile is tri-leveled so should i start at the top and work down when calculating? And would i factor in the pivot (whether it is counter or clockwise)

sorry, im just having trouble with this for some reason.

4. Jan 3, 2010

### Oddbio

You can really start at either the top or the bottom.
First make up your mind of exactly how it should look before you even start the calculations. What do you want to hang where. Then you should know their masses and all that is left to do is to find where the pivot should be on each arm.

I don't know what you mean by:
Fnet = Fa + Fb + Fg
So you need to be more descriptive. However, I'll make an assumption and go from there, if my assumption is wrong then disregard what I say next and I will correct it when you provide me more details.
Assuming that you are saying that Fa is the force on side A of the arm and Fb is the force on side B of the arm, then that is all you need. But the fact that you wrote it Fa+Fb+Fg tells me that you don't know how to find Fa and Fb. Because Fa and Fb ARE Fg, but on each side. The force pushing down on side A is the force of gravity due to the mass hanging on side A, and similarly for side B.
So you should have:
Fa * A = Fb * B (where A and B are the lengths of the arms on each side of the pivot)
NOTE: Fnet = 0 is the condition for translational equilibrium, which should be true as long as your mobile is not moving around.
The condition from which "Fa*A = Fb*B" is derived is rotational equilibrium, in which the NetTorque = 0.
Since Torque = Force*distance (again the force must be perpendicular) then we know that the net torque (torque from side A and side B) is Fa*A+Fb*B = 0
So really, Fa * A = -Fb * B the negative implies the torque spins the arm in the opposite direction, but it is now on the other side, so physically both forces are pointing down, and you can just simply use the magnitudes (not vectors) and say:
Fa * A = Fb * B
Notice, that this applies to all the mass hanging on the arm at a point, so if you are calculating it for the top arm of the mobile then the force on one side will have to include the mass for all of the arms hanging below it.

Also, I don't know how picky your teacher is, but technically having the pivot not directly in the middle will throw off the rotational equilibrium so you would have to calculate the mass of the portion of the lever arm on each side of the pivots, but that is a bit more difficult, your teacher probably won't mind you treating the lever arms as massless only where torque is concerned, am I right?

5. Jan 4, 2010

### Centurion1

i understand what your saying i think, i realized it after posting what i said though. for instance to demonstrate equlibrium would it go like this?

i have a single level mobile. on one end is a .040 kg weight. on the other end is a .060 weight. the .040 kg weight is .2 m away from the center and the .060 kg weight is .135 m

so then

Fa = (.04kg)*(9.81m/s^2)
= .39N

torque = .39N * .2m
= .08

Fb = (.06kg) * (9.81m/s^2)
= .58

torque = (.58N) * (.135m)
= .08

I'm throwing these numbers around they are sort of close to one level of my mobile but would this demonstrate rotational equilibrium because they are the same?

am i missing anything else? like how do the pivots fit in?

6. Jan 4, 2010

### Oddbio

That is exactly what I was saying, you got it.

Just to be sure though, your weights should include the mass of all the other arms of the mobile hanging below it.

One weight is trying to rotate the arm clockwise, the other weight is trying to rotate it counter clockwise, they are both doing it with a torque of 0.08 in opposite directions so they cancel out and the net torque is zero, so there is no rotation.