Question

A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 4 ft below the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)

Answer 1

Answer:

$F=4673.95lbs$

Explanation:

Recall weight density of water 62.5 lb/ft^3 and the height 6ft base 8ft

The width of the triangle

$\frac{W(x)}{7}=\frac{(8-x)}{2}$

The hydrostatic force is the pressure times area of the submerged

$W(x)=\frac{7*(8-x)}{4}$

$F=\int\limits {p} \, dA$

$F=\int\limits^8_2 {pgx*\frac{7}{4}*(8-x)} \, dx$

$F=\int\limits^8_2 {62.5*\frac{7}{4}*x*(8-x)} \, dx$

$F=109.375\int\limits^8_4{(8x-x^2)} \, dx$

$F=109.375[4x^2-\frac{1}{3}x^3]|4,8$

$F=109.375*42.73=4673.95lbs$