Apparent weight in an elevator intuition

[Normalization]

Homework Statement

The numbers aren't important because I'm after getting a more intuitive sense of the problem.

When an elevator is accelerating upwards with a mass in it on a scale. Why is the apparent weight the normal force?

Homework Equations

F=ma W=mg

The Attempt at a Solution

I get that the resultant force upwards is equal to the pushing force (normal force) of the elevator - the weight of the mass so that: +ma-mg = ƩF. But why is the term ma equal to the apparent weight of the mass?

 

[technician]

If the mass is on a scale inside the elevator and it is accelerating upwards there must be a resultant upward force.
Can you identify the forces acting on the mass (there are 2) and write down an expression for the resultant...
(the apparent weight is what the scale indicates)

 

[Normalization]

ƩF = N - W Where N is the normal force.

So N - W = Ma

So N (apparent weight) = Ma + Mg

so N = M(a+g)

But why is N the apparent weight? Is it because of Newtons third law?

 

[technician]

'Weight' is a strange topic in physics ! I suppose the sensible definition of weight is the downwards pull due to gravity, in which case weight does not change (unless height changes).
BUT what you EXPERIENCE as weight is the reading on a balance. You FEEL heavier in an elevator accelerating upwards and you feel lighter in an elevator accelerating downwards.
IT is possible to feel 'weightless' in a falling elevator.
In your analysis you have found that the reading on the balance is greater than the downwards force of gravity (mg) so weight seems greater (ma + mg)

 

[Normalization]

Hmm... Still not quite satisfied.

So suppose instead of a mass I have a mass on a spring. The spring will extend more as there is an upwards acceleration right? Because the apparent weight then is larger than the weight at equilibrium. So is it the increased push on the retort stand and spring and mass (N = M(a+g) compared to N = Mg) which causes the retort stand, spring and mass to push at M(a+g) on the elevator due to Newtons third law which makes the spring extend?

 

[Chestermiller]

ƩF = N - W Where N is the normal force.

So N - W = Ma

So N (apparent weight) = Ma + Mg

so N = M(a+g)

But why is N the apparent weight? Is it because of Newtons third law?

This is just a matter of terminology. The scale reads N = M(a+g), and whenever a scale shows some kind of a reading, we call it a weight. But it really isn't weight. It is just the reading that the scale shows. Maybe that's where the word "apparent" comes in.

Chet

 

[technician]

agree

 

[Normalization]

Thank you chet and technition, but is this right though?

So suppose instead of a mass I have a mass on a spring. The spring will extend more as there is an upwards acceleration right? Because the apparent weight then is larger than the weight at equilibrium. So is it the increased push on the retort stand and spring and mass (N = M(a+g) compared to N = Mg) which causes the retort stand, spring and mass to push at M(a+g) on the elevator due to Newtons third law which makes the spring extend?

 

[technician]

yes means the same as a greater reading on a balance. A balance contains a spring !

 

[Normalization]

Ok thank you, just wanted to make sure that an upwards resultant force results in an increase apparent weight because of N3.