# Waterloo SIN contest question (springs and energy)

1. Nov 10, 2012

### fchen720

1. The problem statement, all variables and given/known data
Both the question and solution are on this link (Question 7):
http://sin.uwaterloo.ca/solutions_2012.pdf

I understand the solution they provided, but I think the question is misleading and should not have been worded that way. I just want to clarify if i'm right or not.

2. Relevant equations

3. The attempt at a solution
I initially thought that no matter what the spring was doing in the middle, the force applied must be the sum of the frictions of both boxes, or 2umg. If the boxes were connected I know the force would definitely be 2umg, and it seemed illogical that adding a little spring would give me free energy.

In case the handwriting is hard to read:
Wtotal = Wspring + Wbox1
Ftotal * x = 0.5kx^2 + Fbox1 * x
Ftotal = 0.5kx + Fbox1

Then they said that since that the force applied by the spring is equal to the force of friction on the box2:

kx = umg

Ftotal = 0.5umg + umg
= 1.5 umg

I don't think it's fair to say that the force I need to apply is 1.5umg because I can rewrite: Wspring = Faverage * x
therefore
Ftotal = Faverage + Fbox1

So shouldn't Ftotal just be the average force that I need to apply, and not the full force once the spring is fully stretched?
If I can only apply 1.5umg of force and no more, shouldn't I not be able to move the box since I can't apply the full 2umg for the spring to stretch fully?

2. Nov 10, 2012

### lewando

I'm looking for a spring in the PDF. Cannonballs instead.

3. Nov 10, 2012

### lewando

NVM. My problem. Now I see it.

4. Nov 10, 2012

### lewando

What part of the question did you think was misleading or not worded properly? How should it have been stated to be more clear?

5. Nov 10, 2012

### fchen720

My problem with the question is that I shouldn't be able to move both boxes by applying
1.5umg. Maybe the average force is 1.5umg as I'm stretching the spring, but shouldn't I eventually need to apply 2umg for the spring to fully stretch? If i'm correct, the answer should be 2umg.

6. Nov 10, 2012

### lewando

7. Nov 10, 2012

### lewando

I think I see where the logic breaks down:
When you decompose Wtotal into Ftotal * x, Ftotal does not represent the maximum applied force. The force applied varies from umg when x = 0+ to (umg + kx) at x.

What you end up with is the average force, not the peak force required to dislodge m2.

[the term "Ftotal" is what is misleading-- it should be "Faverage" to be precise, and that is not what you are looking for]

Last edited: Nov 10, 2012
8. Nov 10, 2012

### fchen720

Thank you for replying Lewando. Much appreciated!

9. Nov 10, 2012

### aralbrec

This is not correct. The force being applied is constant.

I think what is confusing is that this is not a statics problem. The mass on the right is accelerating and the force the right side of the spring is applying is not simply kx at any instant in time.

You can see this by drawing a free body diagram for the right mass, positive x coordinates to the right.

You have the constant (minimum) force F pulling to the right and you have some force Fs pulling to the left. Then:

(F - Fs) = ma
Fs = F - ma

where acceleration is positive to the right. Presumably a minimum applied force would have the right mass slowing down (ie decelerating) just as the first mass was about to move. This means a < 0 and Fs applied by the spring is greater than F.

10. Nov 10, 2012

### lewando

Now that I re-read the problem statement, I see-- "minimum constant force". Clear as day. This was missed. Apologies to fchen720 for not picking that up. Good catch, aralbrec.

11. Nov 10, 2012

### lewando

So the minimum steady-state force required to continuously move m2 is indeed 2umg, but that is not what the question is asking. If I understand, a constant force just incrementally larger than 1.5umg will move m2 incrementally, satisfying the question, but then m2 will stop, permanently.

12. Nov 10, 2012

### aralbrec

Yes, I'd agree with that. What will happen if (1.5umg < F < 2umg) is that the left mass will stop and slip repeatedly (slipping both left and right) as the right mass shuffles back and forth in a damped sinusoidal motion. The friction will eventually cause all motion to stop and the right mass will be located at a fixed distance to the right of the left mass such that kx = F. If F > 2umg, you will see the left and right masses initially have a sinusoidal element superimposed on their right motion but eventually the friction damps out the sinusoidal motion and the masses settle into a constant separation set by F=kx, but both masses continue to accelerate to the right and never stop moving once they start to move.

The surprising part of the question is the realization that a moving mass has inertia so in the beginning, the right mass accelerates according to F=ma and as the spring stretches, the acceleration decreases but because of the motion, the spring stretches much more than F=kx before the right mass comes to a rest. This means the force experience at the left end of the spring is greater than the applied constant force F at peak separation.

We can quickly solve for this case, take a free body diagram for the right mass, applied constant force F pulling to the right and both friction and force of the spring pulling to the left. Then we have:

(F - fr) - Fs = ma

Set Feq = F - fr (subtract constant friction from constant applied force)

Feq - Fs = ma = m d2x/dt2
m d2x/dt2 + kx - Feq = 0 ;; Fs = kx

solving this:

x(t) = (Feq/k) * (1 - cos(t√(k/m)))

x(t) is maximum when cos is -1.

xmax = 2 Feq / k

The maximum force applied by the spring is:

Fsmax = k xmax = 2 Feq = 2*(F - fr)

At the edge of motion for mass 1, the Fsmax = umg

umg = 2*(F - umg) ;; friction experienced by mass 2 is the same
F = 1.5 umg

So, same answer for the minimum force. The equation for x(t) derived above is only accurate up to the right mass stopping to move. This is because if the right mass begins to move left (and it will since at this point F < Fspring) the direction of friction force on the right mass will change from the direction assumed in the equation we started with.

Last edited: Nov 11, 2012