A round conical flask is filled with water of a depth $h$. The radius of the upper water surface is $R_1$ and that of the lower surface is $R_2$.
- What is the net force that the water exerts on the sides of the flask?
- What is the force on the bottom of the flask?
- What is the sum of these forces?
Ignore atmospheric pressure.
I know the force on the bottom is:
$$F = pgh\pi R_2^2$$ since $F$ is the pressure times cross section area and pressure is $pgh$.
However, I don’t know how to find the force on the sides of the jar. I think some integration is involved.
The radius $r$ of the flask at the water depth $l$ is $$ r= r_1+\frac{l}h(r_2-r_1)$$ Let $\theta $ be the side slant angle of the flask. Over the depth $dl$, the area $dA$ of the flask is $$dA= \frac{2\pi r}{\cos \theta}dl$$ Per the symmetry, the horizontal force exerted on the side of the flask neutralizes, while vertical force exerted over the area $dA$ is \begin{align} dF &=p g l \sin\theta dA=2\pi pg\tan\theta \ lrdl\\ &= 2\pi pg\frac{r_2-r_1}{h}l\left[r_1+\frac{l}h(r_2-r_1)\right]dl \end{align} Then, the total net force is integrated over the water depth $h$ \begin{align} F& =\int_0^h dF= 2\pi pg\frac{r_2-r_1}{h} \int_0^h\left[lr_1+\frac{l^2}h(r_2-r_1)\right]dl\\ &= 2\pi pg\frac{r_2-r_1}{h} \left[\frac12r_1h^2+\frac{h^2}3( r_2-r_1)\right]=\frac\pi3 pgh(2r_2^2-r_1^2-r_1r_2) \end{align}