# Lift, reading on the scale

1. May 14, 2008

### Hevonen

[SOLVED] Lift, reading on the scale

1. The problem statement, all variables and given/known data
Mandy stands on a weighing scale inside a lift (elevator) that accelerates vertically upwards. The forces on Mandy are her weight W and the reaction force from the scale R.

The reading of the scale is
A. R + W.
B. W.
C. R.
D. R – W.
3. The attempt at a solution
The right answer is C, but I consider that the scale records Mandy's weight that must either weight at rest or weight at acceleration (there is two types of definition for weight; here I assume that it is weight >= weight(rest) ). So, the answer which follows this way should be A. However, it is wrong.

2. May 14, 2008

### Hootenanny

Staff Emeritus
There is only one defintion of weight and that is the product of mass and acceleration due to gravity. It is easy to confuse the term weight with apparent weight (which I think you refer to as weight with acceleration), however, when we say weight we always mean the former case rather than the latter.

Let's modify the question is a little, assume that the lift is stationary. What is the magnitude of the normal reaction force?

3. May 14, 2008

### Hevonen

It must be R.
But how can the reading be the same when stationary and when moving?

4. May 14, 2008

### Hootenanny

Staff Emeritus
Let's take a step back here. When you stand on your bathroom scales, how do they measure your weight? Which force do they actually measure?

Last edited: May 14, 2008
5. May 14, 2008

### Hevonen

They measure my weight that is the gravitational force.
I considered that my weight = reaction force from the scale R.

6. May 14, 2008

### Hootenanny

Staff Emeritus
when the elevator is stationary the reading on the scales would be,

$$R + W = W+W = 2W$$

?

7. May 14, 2008

### Hevonen

You have the point, but why the outcome is the same although Mandy is accelerating (R > W)?
It seems that the scale always records the reaction force.

8. May 14, 2008

### Hootenanny

Staff Emeritus
And you've got the point: A set of scales determines your weight by measuring the reaction force. That's what I was getting at before, you cannot directly measure an object's weight. Instead, the weight must be determined by measuring another force, the reaction force .

9. May 14, 2008

### chemisttree

The scale has a counterweight that experiences the same reaction force as Mandy. Therefore, the reading will only indicate the weight of Mandy.

This assumes that the scale doesn't utilize a massless spring to measure force. In that case the scale would read R+W.

Last edited: May 14, 2008
10. May 14, 2008

### Hootenanny

Staff Emeritus
Mandy stands on the scales, so I think that it's safe to assume that the type of scales she is using doesn't have a counter-weight.

11. May 14, 2008

### chemisttree

You can stand on a scale (like in the doctor's office) that does have a counterweight.

12. May 14, 2008

### Hootenanny

Staff Emeritus
No! The scale would only ever read the normal reaction force.

13. May 14, 2008

### chemisttree

I see your point. Weight must be defined as the reaction force in this problem which is a function of the total acceleration applied to that which is measured. Mass is not measured... only weight. The acceleration due to gravity plus the acceleration due to the upward movement of the lift multiplied by Mandy's mass will give her weight (force). A trick(y) question indeed.

BTW, the scale (actually a balance as I described it) measures mass independent of acceleration.

Last edited: May 14, 2008
14. May 14, 2008

### Hootenanny

Staff Emeritus
Yes, it can often be confusing when this type of problem is first encountered. The solution can often seem counter-intuitive initially.

Explicitly, for the benefit of Hevonen: there are only two forces acting on Mandy, her weight (mg) and the normal reaction force (R). It is this normal reaction force that is measured by the (spring) scale to determine the weight of Mandy.

Now consider Newton's second law,

$$\Sum F = ma$$

And in our case (defining up as positive),

$$R - mg = ma$$

$$\Rightarrow R = mg + ma$$

So the normal reaction force provides both the force required to accelerate Mandy and the force to stop her falling through the floor of the elevator. The key points to remember are that the scales measure the reaction force R and that the reaction force must account for both Mandy's weight and her acceleration.

I hope that helps clear things up.
I agree that a counter-weighted balance measures the 'true weight', independent of acceleration.

15. May 14, 2008

### Hevonen

Your reasonings clarify the question thoroughly.
So the key to this question is that the reading in the scale = normal reaction force.

Thanks for both of you!

Last edited: May 14, 2008