# General Question on scales at grocery stores

1. Oct 31, 2013

### oneplusone

In grocery stores I often see the scales. It's basically a frame connected to a vertical spring, with the spring compressed at amount x. Next, I drop an apple on it from a certain distance above the bottom of the frame. I want to calculate how far the spring will extend.

So for this problem, does it make sense to use conservation of energy? (like you start off with the gravitational potentials, then the spring potential), or does it make more sense to use momentum to find the final velocity of the object striking the scale and then work from there?

Just was wondering :)

2. Oct 31, 2013

### SteamKing

Staff Emeritus
The compression or extension of a spring is proportional to the amount of force acting on the spring.

The equation is F = kx, where F is the force, x is the distance compressed (or extended), and k is called the spring constant.

Are you looking for something else?

3. Oct 31, 2013

### Staff: Mentor

Yes, I would use energies for this type of question. Before impact, the spring is just supporting the scale pan, and the apple has some kinetic energy from its motion. After impact and at full scale extension, the KE for the apple (and pan) are zero, the spring is storing some PE, and the gravitational PE of the pan and apple are different than they were at impact...

4. Oct 31, 2013

### oneplusone

@SteamKing, yes that is correct, however you are NOT given the spring constant :(

@berkeman that's what I was thinking also, except instead of having the first frame when the apple's about to hit the pan, you look at the gravitational potential energy instead (of the apple). that would work too, right?

5. Oct 31, 2013

### Staff: Mentor

Seems like it should work.

6. Oct 31, 2013

### oneplusone

Thanks, actually I plan to test this out tomorrow.
Just another question: why don't we have to factor in momentum?

7. Oct 31, 2013

### Staff: Mentor

This type of problem is usually easiest to solve using energy conservation. There are probably several other ways to solve it, but why make it harder?

BTW, if you are going to actually test it, be sure to factor in the non-negligible mass of the pan. Oh, and watch out for the produce guy -- they carry pruning knives...

8. Nov 2, 2013

### mrspeedybob

How is this solvable at all without knowing the spring constant? A very stiff spring may only deflect by a millimeter where a weaker spring my deflect by tens of centimeters given exactly the same apple drop.

9. Nov 2, 2013

### AlephZero

You can easily measure the spring constant. Just notice how far the scale moves for a given static load.

If your measure the frequency of the oscillations when you drop the apple, you can find the effective mass of the scale system as well. That will not be the same as the actual mass, unless you take into account all the internal mechanism.

This might seem a bit crude, but actually it is the same principle as is used for real-world measurements of the dynamic properties of structures, which are often done to check out computer models and tweak the model to give accurate results. Instead of dropping an apple, you would hit the scales with a hammer with a load cell build into it to measure the applied force, and use accelerometers or strain gauges (or maybe a seriously expensive gizmo like a scanning laser doppler vibrometer) to measure the movement.

Last edited: Nov 2, 2013
10. Nov 3, 2013

### CWatters

Unless the spring is damped there will be overshoot so the scales will briefly show a weight heavier than the actual weight. In the real world the scales will have mass (with which the apple has a collision, elastic or inelastic, the apple may bounce?) and possibly damping so it's never going to be easy to work out the actual momentary reading on the scale.

If you can ignore al these real world issues then yes conservation of energy is the way to go

11. Nov 3, 2013

### UltrafastPED

Spring scales react to the initial impulse and oscillate about the static equilibrium point - which we call the weight - so the max extension occurs on the first bounce.

PS: I grew up on a produce farm and learned early to lay things on the scale ... gives an accurate, repeatable measurement.

12. Nov 3, 2013

### oneplusone

Hi, actually this is based on a textbook question I was solving (instead of an Apple they used putty).
@above, when you say "initial impulse" would that be when the apple/putty strikes the scale? In that case, would that mean that conservation of energy could not be used?
I tried the experiment, however the scale moved too fast to measure anything, so i am redoing it in a few days :P

13. Nov 3, 2013

### Staff: Mentor

You can use the conservation of energy method if you compare the height of the panbwithout the apple to the height of the pan with the apple - after the scale has stopped bouncing back and forth and settled down into a steady state.

If instead you look at the initial deflection of the scale when you drop the apple, the problem is much more complicated because some of the kinetic energy of the falling apple is goes into accelerating the pan and other internal components of the scale.

14. Nov 3, 2013

### dauto

Using energy conservation wouldn't work because the collision of the putty with the plate of the scale is inelastic. The correct solution is a three step process.

1) Energy is conserved during the fall of the putty
2) Momentum is conserved during the collision
3) energy is conserved during the extension of the spring until the plate is at rest.

15. Nov 3, 2013

### oneplusone

Can someone else please verify this? I dont' see why it wouldn't work still.

16. Nov 3, 2013

### ZapperZ

Staff Emeritus
This is vague. Explain why YOU think it would still work, contrary to what you've been told in this thread!

Zz.

17. Nov 3, 2013

### oneplusone

Frame 1: silly putty elevated above scale, spring stretched out a little due to the mass of the scale.

Frame 2: max compression of the spring

Note: Set reference at full compression, initially silly putty is h above spring.

Equation 1: $$m_{\text{scale}}gh + m_{\text{silly}}g(h+h_0)+ \dfrac{1}{2}kx_0^2$$

Equation 2: $$\dfrac{1}{2}kx^2$$

Since h = x, we substitute and solve for x (or h)

EDIT: I have no idea to explain why it works, I'm asking why it doesn't work.

18. Nov 3, 2013

### Staff: Mentor

Your notation is unclear: Define h, h0, and x.

What about the collision of the silly putty with the scale? Mechanical energy will not be conserved.

19. Nov 4, 2013

### CWatters

That seems like the correct approach to me.

20. Nov 4, 2013

### Staff: Mentor

Sorry that I posted incorrectly. I was thinking rigid apple, not compressible putty.