Simple Fluids Problem: Cylinder on Oil with Added Weight

[BlueDevil14]

Homework Statement

A cylindrical disk of wood weighing 47.0 N and having a diameter of 30 cm floats on a cylinder of oil of density 0.850 g/cm^3 (the figure). The cylinder of oil is 75.0 cm deep and has a diameter the same as that of the wood.

a) What is the gauge pressure at the top of the oil column?

b) Suppose now that someone puts a weight of 88.0 N on top of the wood, but no oil seeps around the edge of the wood. What is the change in pressure at the bottom of the oil?

c) What is the change in pressure at the halfway down in the oil?

The Attempt at a Solution

a) $$P=\frac{F}{A} \Rightarrow P=\frac{47.0 \mathrm{N}}{.071 \mathrm{ m}^3}=665 \mathrm{ Pa}$$

this answer is correct

b) why would the pressure change at the bottom not be equal to the pressure change at top. i.e 1245 Pa - 665 Pa?

 

[Doc Al]

b) why would the pressure change at the bottom not be equal to the pressure change at top. i.e 1245 Pa - 665 Pa?

Why are you subtracting 665 Pa?

 

[BlueDevil14]

I was thinking the change in pressure a the bottom would be equal to the change in pressure at the top, and the final-initial at top would be 1245-665 Pa.

 

[Doc Al]

I was thinking the change in pressure a the bottom would be equal to the change in pressure at the top,

That makes sense.

and the final-initial at top would be 1245-665 Pa.

When you add the weight onto the wood, how much does the pressure increase?

 

[BlueDevil14]

oh... I forgot to add the weights!

then in should be $$P_{2}=\frac{(47.0+88) \mathrm{N}}{.071 \mathrm{ m}^3}=1901 \mathrm{ Pa}$$

It is therefore 1901-665 not 1245-665

 

[BlueDevil14]

Thanks, it sometimes just requires someone to point out the obvious for you