# Why might the normal force on a box not be equal to its weight?

Very simple homework question which I managed to get wrong:

"The weight of a box sitting on the floor points directly down. The normal force of the floor on the box points directly up. Need these two forces have the same magnitude? (y/N)"

My answer was yes, the correct answer is no. I must be missing some important concept here, may someone help me think of a realistic scenario in which the magnitudes are not equal?

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The question is "as is," it was not mentioned that the floor is flat/horizontal plane, but I imagine this should be the case the way it is worded. – Leonardo Oct 25 '12 at 2:35

There are various possible reasons why the normal force of the floor on the box might not equal the weight of the box. For example:

• There could be another box on top of the box being asked about.

• The floor could be the floor of an elevator that is accelerating (so the net force on the box would not be zero).

• The "floor" could be a board, with a C-clamp pressing the box to the board.

• The "floor" could be the bottom of a tank of water (so there would be a buoyant force on the box).

And of course you could put these together in combinations.

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So this is a little bit of a trick question? Probably, due to the lack of information that has confused me. Otherwise, I believe you are correct. I assumed that "weight" would mean force due to gravity on the mass of the object in question. Then again, say there was another box then wouldn't its force on the floor vector increase, and therefore the floor on the box force vector increase equally? – Leonardo Oct 25 '12 at 2:59
@Leonardo - If you have a 2.00 kg block sitting directly on top of another 2.00 kg block, which in turn is sitting on the (level) floor, then the three forces on the bottom block are... 1) The weight of the bottom block (the gravitational force of the Earth on the bottom block): 19.6 N down. 2) The force of the top block on the bottom block (which is actually a normal force): 19.6 N down. 3) The normal force of the floor on the bottom block: 39.2 N up. – Psi Oct 25 '12 at 4:15
@Leonardo - I don't think of my first answer (a box on top of the box being asked about) as a trick answer, but rather as perhaps the simplest possibility to understand. However, inspired by Mark's answer (so I am giving his answer an up vote), I have added a few more possibilities. – Psi Oct 26 '12 at 6:33

Sometimes questions like this are asked to make the point that mechanical equilibrium (ME) should not always be assumed as a given. And also that the definition of ME states that both the net forces and net torques must total to zero. The question does not explicitly state that the system is in mechanical equilibrium, and it says nothing about the shape of the box.

If the system is not in ME, then there is no reason for the forces to have to be equal. The box could be in a state of translational or rotational acceleration.

Note there may be no ME even if the normal force on the floor and the weight are equal and opposite. The box could be tipping over because it has an asymmetric shape that causes the force normal to the floor to not be directly in line with the center of mass. The resulting net torque on the box could be tipping it over.

I voted up the previous answer because I liked its elegance (stack on a second box,) but decided to post this anyways because I suspect (hope) the question was suppose to be more about physics than a trick.

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