A triangular plate with height 4 ft and a base of 6 ft is submerged vertically in water so that the top is 3 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.)

The answer to the question is

 = 250 lb force

Explanation:

The given variables are

Height of triangular plate = 4 ft

Base of triangular plate = 6 ft

The hydro-static force on the triangle is the product of pressure at a particular depth and the area of the surface of the triangle at that depth

the width of the triangle is given by

slope =

Here we have y =  -4/3x +4  and the depth from the top at any y is 7 -y

Therefore pressure = 62.5×(7-y)

And the area of a strip is is given by 2 times  the x coordinate  and Δy

That is area = ×Δy

Force on a strip is given by Force = Pressure × Area

= 62.5×(7-y)× ×Δy Which gives 93.75 ×(7-y)(4-y)Δy

Hence hydrostatic force =  =

=  = 250 lb force

it may be wrong but i think its a

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more