Potential Energy function for this system

[FruitNinja]

Homework Statement

Find the potential energy function for the three mass, earth, and pulley system as shown. The potential energy will be as a function of the vertical position downward as shown in the diagram. Also, find the equilibrium position of this system. The two outside masses are equal.

Homework Equations

ΔW=ΔE
W=∫F• dr[/B]

The Attempt at a Solution

Actually I do not conceptually understand this question. Why would the gravitational potential energy ΔGPE change based on what y is?
I guess we could do ΔWnet=ΔKE+ΔGPE and we could make the system everything so that it is equal to 0 so 0=ΔKE+ΔGPE, but I don't see what is CHANGING in the problem. Everything looks static to me

 

[Bystander]

Think about the total including the third mass; give you any ideas?

 

[FruitNinja]

Think about the total including the third mass; give you any ideas?

Hmm so I guess if we think of them as 3 separate PEs then 2 of them are decreasing when the 3rd is increasing. so that would change pe. i see that now. But still, the whole system looks static, like it is not moving or changing. Is there some kind of force that would push the middle one down along y? The problem says nothing about an external force applied. I need something to change

 

[Bystander]

How is the center mass supported?

 

[FruitNinja]

How is the center mass supported?

Supported? By the 2 tension forces of the 2 outer blocks upward. But that would be internal

 

[Bystander]

upward

...?

 

[FruitNinja]

...?

I mean the tension force is upward (direction)
Edit: I mean for the outer blocks it is upward. for the middle one it is at an angle

 

[Bystander]

Yeessss ... and to left and right ... a "sling" ... need any help resolving the forces?

 

[FruitNinja]

Yeessss ... and to left and right ... a "sling" ... need any help resolving the forces?

But I don not understand how that would help with finding the change in potential energy of the system since this isn't an f=ma problem

 

[forScience]

I'm trying to figure out this problem too.

I think what you're supposed to realize here is that the tension forces are doing work on the three masses, but since your system is everything, work = 0, and so you have to express that as a potential energy... so in a sense, you're trying to find TPE (tension potential energy).

 

[Bystander]

It is; there is only one angle for which the vertical components of the forces balance/are balanced.

 

[forScience]

It is; there is only one angle for which the vertical components of the forces balance/are balanced.

Ok, so I took your hint, and found that angle that would yield the equilibrium position, aka forces balance, and accel = 0.
using that angle, I found the y position that would yielf equilibrium.

I labeled the mass in the middle as m1 and the masses on the outside as m2.

F = ma System m2
There is a tension force up, and a gravity force down on m2. accel is 0 because it's in equilibrium. Take up to be positive.
Ft - Fg = 0
Ft = m2 * g

F = ma System m1
There are 2 equal tension forces angle (#) (<- i'll just write my theta as #) above the horizontal, and a gravity force down. accel = 0; in equilibrium. Up positive.
2Ft sin(#) - Fg = 0
2 (m2 * g) sin(#) = m1 * g
sin(#) = (1/2) m1/m2
(#) = sin^(-1) (m1/(2*m2))

So, in the diagram, it says that the 2 outer masses are a distance 2d apart, so the horizontal distance between m1 and either m2 is d.
tan(#) = y/d
so y = d * tan( sin^(-1) (m1/(2*m2)))

is this correct?

Edit: for clarity, this is what I'm seeing:

So if all my work is correct, I've solved the 2nd part of the problem.

But I'm still not sure how to solve the first part of the problem, which is finding the potential energy as a function of y...

 

[LeLop]

Well.. I found the angle of the tension forces to be arcTan(mg / 2D). So therefore, the equilibrium position would be... D tan(theta). Would this be correct?

 

[ehild]

You need the potential energy of the three-mass system, in the gravitational field of the Earth, in terms of y, position of the middle mass, assuming symmetric arrangement, when the position of the other masses are determined by y.
How would the potential energy change if you move the middle block downward by Δy?

 

[FruitNinja]

So is it asking for the change of PE from the y position to the equilibrium? Because there needs to be some kind of change

 

[forScience]

You need the potential energy of the three-mass system, in the gravitational field of the Earth, in terms of y, position of the middle mass, assuming symmetric arrangement, when the position of the other masses are determined by y.
How would the potential energy change if you move the middle block downward by Δy?

The middle block's GPE decreases when moved down by Δy.
ΔGPE = -m₁gΔy

 

[ehild]

The middle block's GPE decreases when moved down by Δy.
ΔGPE = -m₁gΔy

I meant the total potential energy of the whole system. How do the potential energies the other two blocks change if the middle block goes downward by Δy?