Balti Physics!!

figure Downward force acting on an elemental strip as shown is equal to $\rho*g*y*2*\pi*x*\dfrac{dy}{\cos\theta}*\sin\theta.\Rightarrow$ total downward force on the sides of the bucket is equal to $\int_{0}^{h}\rho*g*y*2*\pi(b-y*\tan\theta)*\tan\theta*dy=\rho*g*2*\pi*\tan\theta*\int_{0}^{h}(b-y\tan\theta)*y*dy=$ $\rho*g**2*\pi*\tan\theta[\dfrac{b*h^{2}}{2}-\tan\theta*\dfrac{h^{3}}{3}].$ Now, $\tan\theta=\dfrac{b-a}{h}=\dfrac{\pi}{6},$ $x=(b-y*\tan\theta.)$ Substituting values,we getnet downward force on the sides of the bucket $=10,000N.$ Solution by my Dad not by me!

The volume of a conical frustum is given by:

$V_f = 1/3 \times h \times \pi \times (r_1^2 + r_1 \times r_2 + r_2^2 )$

For our case substituting $a$ and $b$ for $r_1$ and $r_2$ we get a total volume of:

$1.75 m^3$

The lower horizontal surface holds the weight of the water placed exacly above it. The rest of the water is hold by the sides. We need to determine the volume of the water which is $NOT$ above the lower horizontal surface(the circular disk with radius $a$ )

The volume of the water above the lower horizontal surface is the volume of a cylinder with radius $a$ and height $h$ .

So we have $V_{cyl} = \pi \times a^2 \times h = 0.75 m^3$

Taking the difference we get:

$V_{side} = V_f - V_{cyl} = 1 m^3$

but the water density is $10^3 kg/m^3$

So $Mass_{side} = 10^3 \times 1 = 10^3 kg$

$Weight = Mass \times g = 10^4 N = \boxed{10,000 N}$