physics/calculus problem?

ilovecoach

a flat triangular plate whose dimensions are 5, 5, and 6 ft. is submerged vertically in water (density 62.4 lbs/ft³) so that its longer side is at the bottom and parallel to the surface and its vertex is 2 ft. below the surface. find the force against the surface of the plate.

alam-ko-iyan

F = ρ ∫ h x dx .. . .

where x is the length of the partition

and h is the distance of the horizontal partition from the surface

ρ = 62.4

h = ax + b

h(2)=0

h(6) = 6 ... if the triangular plate is 5,5,6 in dimension

then there is a right triangle whose dimension is 5,3,4

thus 4 ft is the height of the vertical triangle

then

2a + b = 0

6a + b = 6

4a = 6

a = 3/2

b = -3

thus

F = ρ ∫ (from 2 to 6) (3/2 x - 3) x dx

.. . .. .