Solving the Elevator Problem: Determining the Normal Force on a 100kg Bag

[devanlevin]

in a problem where there is a 100kg bag placed on the floor of a 1000kg elevator, going up with an acceleration of 0.8m/s^2, what is the normal force applied on the bag by the elevator.

i thought that the normal force would be mg(using the mass 1100kg)-F(the force lifting the elevator) but its not, its the mg(using only the 100kg) + ma(again using 100kg)

therefore
N=980+80=1060

can anyone explain the logic in this, if it is correct.

 

[alphysicist]

Hi devanlevin,

It is correct. From the equation in your post, I think the answer you're looking for is this:

When we draw a force diagram for an object (such as the 100kg bag here), we only include those forces that are actually acting on the object itself. The force lifting the elevator (from the cable) is acting on the elevator, and so is not included in the force diagram for the bag. (It obviously has an effect on the bag, but it does not directly act on it.)

Does this answer your question?

 

[Moetasim]

Yes, your calculation is correct. The normal force on the bag is equal to the weight of the bag, mg, plus the force needed to accelerate the bag, ma. This is because the elevator is accelerating upward with an acceleration of 0.8m/s^2, so the bag will also experience this acceleration. In order for the bag to remain in contact with the elevator floor, the normal force must be equal to the weight of the bag plus the force needed to accelerate it. This is due to Newton's second law, which states that the net force on an object is equal to its mass times its acceleration. In this case, the net force on the bag is the normal force, which is equal to its weight plus the force needed to accelerate it. Therefore, your calculation is correct and your understanding of the concept is accurate.