I Mass of Backpack in Moving Elevator

[annamal]

A backpack is attached to a spring scale which is attached to the ceiling of an elevator. The elevator is moving downwards with an acceleration of 3.8 m/s^2. The scale reads 60 N (Fscale). What is the mass of the backpack?
The solution to this problem says that Fscale - w (weight) = Fnet.
This confuses me because the elevator is causing a downward acceleration and so the spring scale reading should be the Fnet since it is the result of all the forces in play.
I think the answer should be Fscale (also known as Fnet now) = Fup (an upwards force on the spring produced by the moving elevator and the spring itself) - w (weight). Can someone explain how I am wrong?

 

[nasu]

The spring sclae reding depends on the extension of the spring which is related to the elastic force. It has nothing to do with gravity or any other force. If the elevator is in free fall the scale will read zero even though the net force on the body is mg.

 

[annamal]

I am thinking
Forig (force of scale with non moving elevator) + m*a (accceleration of elevator) = Fs ( resulting force of scale)

How is that wrong?

 

[Herman Trivilino]

Think of the backpack hanging from a rope attached to the ceiling. Two forces are exerted on the backpack. There's the upward force exerted by the rope and the downward force exerted by gravity. The net force is the vector sum of these two forces.

In your scenario the scale reads the force exerted on the backpack by the rope. It does not read the net force.

 

[annamal]

Think of the backpack hanging from a rope attached to the ceiling. Two forces are exerted on the backpack. There's the upward force exerted by the rope and the downward force exerted by gravity. The net force is the vector sum of these two forces.

In your scenario the scale reads the force exerted on the backpack by the rope. It does not read the net force.

Yes, but doesn't Forig (force of scale with non moving elevator) + m*a (accceleration of elevator) = Fs ( resulting force of scale) --> that's what I mean by net force for Fs

 

[Lnewqban]

... Can someone explain how I am wrong?

With the elevator fully stopped or moving at constant velocity, what would happen if you removed the scale (or Fs, or Fup or Fnet (the way you have re-define it)?
The only force acting on the backpack would be the weight (mg).
Then, the backpack would free fall at 9.81 m/s each second until hitting the elevator floor.
The backpack is falling down in an accelerated way, the elevator is not.

With the elevator moving down at constant acceleration of 9.81 m/s^2, what would happen if you removed the scale (or Fs, or Fup or Fnet)?
Then, both, the elevator and the backpack would free fall at 9.81 m/s each second, none of those two things moving respect to each other until the elevator reaches the ground.
Both, the elevator and the backpack are falling down in an accelerated way, and at the same rate.
Fs would be zero in the event that the scale remains in place rather than being removed.

"The elevator is moving downwards with an acceleration of 3.8 m/s^2" means that the elevator is not free falling, but that its fall is restricted to some degree by its motor or brake.

The scale is the solid link between elevator and backpack that is preventing the backpack from free falling onto the elevator floor.
That solid link happens to be elastic in this case, and its deformation is proportional to its internal tension (Fs).

 

[annamal]

I am thinking
Forig (force of scale with non moving elevator) + m*a (accceleration of elevator) = Fs ( resulting force of scale)

How is that wrong?

I process this better like this:
Forig + m*a - m*g = m*a --> Forig = m*g

 

[annamal]

With the elevator fully stopped or moving at constant velocity, what would happen if you removed the scale (or Fs, or Fup or Fnet (the way you have re-define it)?
The only force acting on the backpack would be the weight (mg).
Then, the backpack would free fall at 9.81 m/s each second until hitting the elevator floor.
The backpack is falling down in an accelerated way, the elevator is not.

With the elevator moving down at constant acceleration of 9.81 m/s^2, what would happen if you removed the scale (or Fs, or Fup or Fnet)?
Then, both, the elevator and the backpack would free fall at 9.81 m/s each second, none of those two things moving respect to each other until the elevator reaches the ground.
Both, the elevator and the backpack are falling down in an accelerated way, and at the same rate.
Fs would be zero in the event that the scale remains in place rather than being removed.

"The elevator is moving downwards with an acceleration of 3.8 m/s^2" means that the elevator is not free falling, but that its fall is restricted to some degree by its motor or brake.

The scale is the solid link between elevator and backpack that is preventing the backpack from free falling onto the elevator floor.
That solid link happens to be elastic in this case, and its deformation is proportional to its internal tension (Fs).

Ok, for an elevator dropping downwards with acceleration 3.8 m/s^2, in the elevator's frame of reference would Fs-mg = 0? Because that would mean Fs = m*g which doesn't make sense esp if the elevator is accelerating downwards.

 

[Herman Trivilino]

that's what I mean by net force for Fs

That's not the standard meaning of net force.

 

[Lnewqban]

Ok, for an elevator dropping downwards with acceleration 3.8 m/s^2, in the elevator's frame of reference would Fs-mg = 0? Because that would mean Fs = m*g which doesn't make sense esp if the elevator is accelerating downwards.

Again, the backpack is linked to the ceiling of the elevator via a spring (scale).
What the acceleration of the backpack would be respect to the elevator (its frame of reference)?
9.81-3.80 ≠ 0
If you have acceleration and a mass, you have a force.

 

[Lnewqban]

Please, read this:
https://www.physicsforums.com/insights/frequently-made-errors-pseudo-resultant-forces/