# Differing weight on scales depending on approach

1. Dec 14, 2013

### Fineline00

1. The problem statement, all variables and given/known data

Draw graphs of weight against time for the following

1) man stands on scale very gently and then gets off again very gently
2) man jumps onto scales and then jumps off again
3) man stands on scale and lets his knees unlock so that he drops, then stops.

For each of these describe the reasoning behind the graph.

2. Relevant equations

mg = R
Resultant force causes an acceleration F=ma

3. The attempt at a solution
1)A slow acceleration possibly constant velocity of approach means the change in weight over time is small and the maximum weight is equal to the weight of the man
2)The man is originally accelerating downwards therefore the reaction force provided by the scales must increase therefore a simulated increase in weight. Vice-versa for jumping upwards
3)Weight must decrease before returning to original weight

My attempts at the graphs are attached going 1 -> 3, left to right. I'm not sure about 2 and 3 as I think I contradict myself

Thanks for the help

#### Attached Files:

• ###### w over time.png
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2. Dec 14, 2013

### CWatters

Have another think about 2 and 3.

2) If you jump in the air and land on the scales you will hit the scales with some velocity. Some time later that velocity has gone and you are just left standing on the scales. What will the scales read during the impact phase? Same/higher/lower than your weight? What will the scales read when you have come to rest and are just standing on them (lets say you stand on them for 5 seconds before jumping off).

3. Dec 14, 2013

### CWatters

3) You have this partly right. The weight will reduce while you are dropping but what happens after that? See Q2.

4. Dec 14, 2013

### Fineline00

For number 2, should the weight actually decrease then because the man is accelerating downwards (during the impact stage). When you are at rest, then your weight should be correct as you are stationary.

For number 3, I can't see why the weight of the man should not return to normal after he becomes stationary again. The only change I can think of is that the man's centre of gravity has now changed but I cannot see how this affects the reading

5. Dec 14, 2013

### haruspex

Just after landing on the scales, he's moving downwards, but which way is he accelerating?
Remember that what the scales actually read is how far the spring is depressed. There is a certain depression which corresponds to his rest weight. When the scales are lower than that he will show as weighing more, when above that, less.
I would assume that between landing on the scales and jumping off again the man will be 'rigid' for a while. That means the system will behave as a damped oscillator, but everything stops in a finite time because static friction takes over.
You need to think carefully about what happens when he jumps off. What's the first thing he will do?
It depends what you mean by the man becoming stationary. When he first becomes rigid again, things are still moving.

6. Dec 14, 2013

### CWatters

No it won't be less during the impact phase. Think about the difference between placing a weight on your foot and dropping a weight on your foot :-) Which is going to hurt more?

Correct.

It does, but as you said that's "after he becomes stationary again". What happens between him being in free fall/weightless and him becoming stationary again?

7. Dec 15, 2013

### Fineline00

For Number 2

The weight of the man would increase as he hits the scales but then as he becomes stationary the spring becomes less depressed and therefore shows a smaller weight (F=-kx). I'm unclear as to whether the weight shown should actually decrease for a while because there needs to be a resultant force upwards in order to decelerate the man.

When he jumps off, the reading should first increase(?) because he applies more force downwards in order to jump off. When he starts to accelerate upwards, the reading should decrease because there is less force acting on the scale

Number 3

Would the reading slightly increase because there needs to be a resultant force upwards in order to decelerate the man so he becomes stationary

8. Dec 15, 2013

### haruspex

Clearly the registered weight will decrease until it shows his actual weight. The spring will then be in its equilibrium position for the man's actual weight. Is that it? Will it stop right there?
Stand upright, then jump. What is the first thing you do?
What about before that, when he is accelerating downwards?

9. Dec 15, 2013

### Fineline00

I'm not sure what you mean by that statement, do you mean the spring or the man?

I think the first thing you would do is then decelerate downwards, meaning a reduction in the reading on the weight before you then start the motion of jumping upwards

When he is accelerating downwards, the reading should be lower than his normal weight

10. Dec 15, 2013

### CWatters

Do you have any scales at home? Might be easier to just try it :-) I think it would be easier to see what's going on if you have an old fashioned type with a moving pointer rather than a modern digital display. Failing that try kitchen scales.

Normally there isn't much damping so if you drop something on scales the reading oscillates before settling at the correct value.

Last edited: Dec 15, 2013
11. Dec 15, 2013

### Fineline00

Unfortunately both of my scales are digital and tend to blur/crash when the change is sudden

From what I can tell for number three is that when I am stationary my weight is constant. When I unlock my knees the reading decreases from equilibrum and then increases when I start to stop and slow down before returning the original weight which makes sense having now discussed it. It looks a bit like image wt2.

For number 2, I can only try what happens when I am stationary and I am preparing myself to jump as when I jump on my scales it just crashes. The graph seems to look like wt3 which is essentially wt2 but a bit extra when I start to jump as I increase the force because I push downwards and then as the contact reduces the weight reduces. I just can't imagine what would happen before I am stationary and I am jumping onto the scales

In the pictures I have draw the changes linearly when they should be curved

#### Attached Files:

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• ###### wt3.png
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12. Dec 15, 2013

### haruspex

I man the system, spring+man. Springs imply oscillation.
accelerate downwards
Yes.