# Hydrostatics - pressure increases with depth

• DieCommie
In summary, Derive the fact that hydrostatic pressure increases linearly with depth under the surface of a fluid (assumed to be of constant density). If density is also allowed to vary with depth, then your answer would change. The force per unit volume in a fluid is simply -\nabla p. A block of mass 'm' and density ' \rho ' is suspended from a spring (with spring constant k) in a fluid of density \rho_f < \rho . Drive an expression for the extension of the spring from its equilibrium length.f

## Homework Statement

i)Derive the fact that hydrostatic pressure increases linearly with depth under the surface of a fluid (assumed to be of constant density). Comment on how your answer would change if density is also allowed to vary with depth.

ii) Show that the force per unit volume in a fluid is simply $$-\nabla p$$

iii) A block of mass 'm' and density '$$\rho$$' is suspended from a spring (with spring constant k) in a fluid of density $$\rho_f < \rho$$. Drive an expression for the extension of the spring from its equilibrium length.

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## The Attempt at a Solution

Honestly I have been staring at this for a while not knowing how to start. I've only had one lecture on this in the class so far. It was pretty basic, archemide's principle, force is normal to object, pressure is a function of depth, yada yada yada...

Also I remember the del notation from Calculus III, but haven't used it yet in Physics. I remember its $$\nabla f = < \frac{\partial f}{\partial x} , \frac {\partial f}{\partial y} , \frac {\partial f}{\partial z} >$$ and I think this is the direction and slope of greatest rate of change? I can't think of how this applies to part ii.

Any tips on any of the parts would be GREATLY appreciated!

## The Attempt at a Solution

Lets try part one first. Just a hint but consider the forces acting on either ends of a cylinder of liquid that is in equilibrium and has a uniform density. The force acts over the area of the cylinder ends hence introducing pressure.

Yeah, for the third part, $$kx_0=mg$$ (initially). When its immersed it the liquid, by the archemedes principle, it experiences a loss in weight. Use that for the weight, subtract the two and you'll have your $$dx$$

Lets try part one first. Just a hint but consider the forces acting on either ends of a cylinder of liquid that is in equilibrium and has a uniform density. The force acts over the area of the cylinder ends hence introducing pressure.

I tried using a small box, and for force I get
$$F = dy dz ((p (x,y,z) - p (x+dx,y,z)) \check{i} + dz dx ((p(x,y,z) - p(x, y+ dy, z)) \check{j} + dy dz ((p(x.y.z) - p(x,y,z+dz)) \check{k}$$

Is this at all helpful/along the lines of where I need to go?

Yeah, for the third part, $$kx_0=mg$$ (initially). When its immersed it the liquid, by the archemedes principle, it experiences a loss in weight. Use that for the weight, subtract the two and you'll have your $$dx$$
I found the force on the object to be

$$F = vg(\rho - \rho_f)$$

set that equal to kx, and get $$x = \frac{vg(\rho - \rho_f)}{k}$$

I think that's it...

I think you're over-complicating things here.

Say the force on the top of the cylinder (or cuboid) at the surface is P0A and the force on the bottom of the cylinder supporting it is PA. Then the bottom face of the cylinder must be in equilibrium therefore PA = P0A + Mg.

Solve for P.

Actually, I think you've made a mistake somewhere. $$mg=kx_0$$...(1)
$$m=\rho V$$
$$V=\frac{m}{\rho}$$
$$F_b=\rho _v V$$
$$kx=mg-F_b$$...(2)
(2)-(1)
$$k(x-x_0)=\rho _v \frac{m}{\rho}$$
$$x-x_0=\rho_v \frac{m}{k\rho}$$