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

`F=4673.95lbs`

Step-by-step explanation:

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

The width of the triangle

`(W(x))/(7)=((8-x))/(2)`

The hydrostatic force is the pressure times area of the submerged

`W(x)=(7*(8-x))/(4)`

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

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

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

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

`F=109.375[4x^2-(1)/(3)x^3]|4,8`

`F=109.375*42.73=4673.95lbs`