A door in a tank in the form of quadrant of a cylinder of 1.5 m radius and 8 m wide. Calculate the resultant force on the door and its location.

Data:-

Radius R=1.5 m

Width = 8 m

To find:-

Resultant Force $(F_R)=?$

location $\theta=?$

HINT:-

$F_R=\sqrt{F_x^2+F_y^2}$

$\theta=\tan ^{-1}(\frac{F_y}{F_x})$

Solution:

a) Horizontal component $(F_x)$

$F_x=W_w\cdot A\cdot F$

Here, $F=\frac{1.5}{2}=0.75m$

$A=1.5\times 8=12 m^2$

$W_w=9810 N/m^2$

$\therefore F_x=9810\times 12]times 0.75$

$=88.29\times 10^3N$

b) Vertical component $(F_y)$

$F_y=W_w\times \text{ cls area of ABC}\times \text{ width of door}$

$=W_w\times \frac{\pi}{4}\times 1.5^2\times 8$

$138.685\times 10^3N$

Now,

Resultant force R$=\sqrt{F_x^2+F_y^2}=\sqrt{(88.29\times 10^3)^2+(138.685\times 10^3)^2}$

$=164.68\times 10^3N$

$\theta=\tan ^{-1}\frac{F_y}{F_x}=57.51^\circ$