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(B) The force exerted by a fluid on a submerged surface is given by the product of the pressure at the centroid of the surface and the area of the sur

Vedclass · Accepted Answer

$\left( {{P_0} + h\rho g} \right)\frac{1}{2}ab + \frac{{{a^2}b}}{3}\rho g$

Vedclass · Answer

(B) The force exerted by a fluid on a submerged surface is given by the product of the pressure at the centroid of the surface and the area of the surface.$1$. The area of the triangular plate $ABC$ is $A = \frac{1}{2} 	imes b 	imes a$.$2$. The depth of the centroid of the triangle from the water surface is calculated as follows: The vertices are at depths $h$ (for $A$),$h+a$ (for $B$),and $h+a$ (for $C$). The depth of the centroid $h_c$ is the average of the depths of the vertices: $h_c = \frac{h + (h+a) + (h+a)}{3} = \frac{3h + 2a}{3} = h + \frac{2a}{3}$.$3$. The pressure at the centroid is $P_c = P_0 + h_c ho g = P_0 + (h + \frac{2a}{3}) ho g$.$4$. The total force $F$ is $P_c 	imes A = (P_0 + h ho g + \frac{2a}{3} ho g) 	imes \frac{1}{2} ab = (P_0 + h ho g) \frac{1}{2} ab + \frac{1}{2} ab 	imes \frac{2a}{3} ho g = (P_0 + h ho g) \frac{1}{2} ab + \frac{a^2 b}{3} ho g$.

$A$ vertical triangular plate $ABC$ is placed inside water with side $BC$ parallel to the water surface as shown. The force on one surface of the plate by water is (density of water is $\rho$ and atmospheric pressure is $P_0$).

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