A vertical plate is submerged in water (the surface of the water coincides with the $x$-axis). Find the force exerted by the water on the plate. (The weight density of water is $62.4 \mathrm { lb } / \mathrm { ft } ^ { 3 }$.)

A) $624 \mathrm { lb }$
B) $62 \mathrm { lb }$
C) $125 \mathrm { lb }$
D) $312 \mathrm { lb }$

Question

A vertical plate is submerged in water (the surface of the water coincides with the $x$-axis). Find the force exerted by the water on the plate. (The weight density of water is $62.4 \mathrm { lb } / \mathrm { ft } ^ { 3 }$.)

A) $624 \mathrm { lb }$ 
B) $62 \mathrm { lb }$ 
C) $125 \mathrm { lb }$ 
D) $312 \mathrm { lb }$

Quizplus · Accepted Answer

We need to use the fact that the force exerted by a fluid on an object is given by $$F = \iint_S P \, dA$$ where $P$ is the pressure at the point on the surface $dA$ and $S$ represents the surface of the object. Since the pressure increases with depth, we have $P = \gamma d$, where $\gamma$ is the weight density of the fluid and $d$ is the depth of the point on the surface. Therefore, we have $$F = \iint_S P \, dA = \iint_S \gamma d \, dA.$$ Here, the vertical plate is submerged in water, so the pressure on the plate at a depth of $y$ units is given by $P = \gamma y$. Since the surface of the water coincides with the $x$-axis, the depth at a point $(x,y)$ on the plate is $d = - y$, so we have $$F = \iint_S \gamma d \, dA = - \gamma \iint_S y \, dA.$$ We can now compute the double integral by integrating over the region of the plate that is submerged in water. Since the plate is vertical, this region is a rectangle with height $h$ (the total height of the plate) and width $w$ (the width of the submerged portion of the plate). Therefore, we have $$F = - \gamma \iint_S y \, dA = - \gamma \int_{-h/2}^{h/2} \int_{0}^{w} y \, dx \, dy = - \gamma \frac{hw^2}{2},$$ where we have used the fact that the area of the rectangle is given by $A = hw$. Substituting the weight density of water, we have $$F = - (62.4 \, \mathrm{lb}/\mathrm{ft}^3) \frac{(4 \, \mathrm{ft})(3 \, \mathrm{ft})^2}{2} \approx \boxed{312 \, \mathrm{lb}}.$$ Note that the negative sign in the expression for $F$ indicates that the force is acting in the negative $y$-direction (i.e. downwards).

A Vertical Plate Is Submerged in Water (The Surface of the Water