# Pressure and normal force (1 Viewer)

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#### phrygian

1. The problem statement, all variables and given/known data

A rock of mass M with a density twice that of water is sitting on the bottom of an aquarium tank filled with water. The normal force exerted on the rock by the bottom of the tank is?

2. Relevant equations

3. The attempt at a solution

The answer is Mg/2, I understand how to arrive at that by subtracting the buoyant force which is equal to mg from the gravitational force on the object which is 2mg. but shouldn't the normal force on the object be different when there are different amounts of water pushing down on it? Since the amount of water is not given shouldn't the correct answer be "impossible to determine from the information given?

#### Q_Goest

Homework Helper
Gold Member
Hi phrygian. Are you familiar with the Archeimedes' principal? Let's not talk about "amount" of something. Let's give everything a specific, measurable quantity such as "volume" or "mass". Archimedes' principal simply states that any object, wholly or partly immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.

So the rock displaces some volume of water which has a given weight.

Does that help?

#### phrygian

I understand that but if I make a free body diagram for the rock wouldnt the forces be normal force + bouyant force - gravitational force - force due to pressure from the fluid above = 0 ? So doesnt the normal force change as the force due to pressure from the fluid above changes when there are different amounts of fluid above?

#### Q_Goest

Homework Helper
Gold Member
A free body diagram of the forces on the object in the water would consist of the forces down (weight of object) minus the forces up (integrate pressure over the area). This can be simplifed if we imagine a cubical volume. Forces on the sides cancel out. Force on top equals pressure times area. Force on bottom equals pressure times area. Use Bernoulli's equation to determine pressure. The force up (difference in pressure times area) minus the force down (weight) equals the weight of the fluid, which is exactly what Archimedes' principal tells us.

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