Question

A cylindrical container, 1 meter long and 1 meter in radius, is placed on horizontally on its side and filled halfway with water. a) Calculate the horizontal component of the force acting on the cylindrical side-walls? b) Calculate the force acting on each round end? Then we are given the integral solutions of integral h *sqrt(1-h^2)dh = -1/3(1-h^2)^3/2 and integral sin(x)cos(x)dx = -1/2*cos^2 (x) to help out with the problem.

Answer 1

We have integrate the forces acting on the walls as it is not same on all levels.

So, we assume a plank with one side of length (d) parrallel to the diameter at circular end at a distance x from the centre and the other side running along the length of the cylinder with small thickness 'dx'

From properties of triangle we get

$d= 2 \sqrt{1-x^{2}}$

Here @ is angle subtended at the centre.

We can see at end point on the circular end P is acting

So, Force acting on the circular end = P * Area

Area = d * dx

P = $\rho$gx

Force

$\mathrm{d}F = \rho g x \mathbf{d}\: dx$

$\mathrm{d}F = 2\rho g x {}{\sqrt{1-x^{2}}} dx$

$\Rightarrow \oint \mathrm{d}F = 2\rho g\int_{0}^{1}{x}{\sqrt{1-x^{2}}} dx$

$\Rightarrow \oint \mathrm{d}F = 2\rho g \left |(-1/3)(1-x^{2})^{3/2} \right |_{0}^{1}$

$\Rightarrow F = \frac{2}{3}\rho g$

Force acting on curve end = P * Area

$\mathrm{d}F = \rho g x L\: dx$

dF )gRcos@ L dr

Rule of triangle. Here, x= Rcos@

$\Rightarrow \mathrm{d}F = \rho gL cos@ \: dx$

Since, x= Rcos@

dx = -R sin@ d@ = -sin@ d@

$\Rightarrow \mathrm{d}F =- \rho gL cos@ sin@ \: d@$

ρg Lcos@sin@ ddi

$\Rightarrow \oint \mathrm{d}F =- 2\rho gL\int_{0}^{\pi /2 } cos@ sin@ \: d@$

$\Rightarrow \oint \mathrm{d}F =- 2\rho gL\left |(-1/2) cos^{2}@ \right |_{0}^{\pi /2}$

$\Rightarrow \left |F \right | =\rho gL$

which are the required answer

If any doubt feel free to comment