Find the hydrastatic force on one end of the trough?

Question

A watering trough measures 1.2 m wide, 0.8 m deep, and has ends that are triangles. If completely filled with water, what is the hydrostatic force on one end of the trough?

HELP!!!!!!

Answer 1

The pressure acting on the on "liquid" side of the end is the hydrostatic pressure, which increases by the depth z:

p = p₀ + ρ·g·z

(where p₀ is the atmospheric at the liquid surface).

On the "air" side atmospheric pressure acts in opposite direction. So the net pressure acting on the ends of the through is:

Δp = ρ·g·z

Consider a differential segment of the triangle, which has height dz and is cut parallel parallel to the liquid surface at depth z. The width of such a segment decreases linearly from a=1.2m at z=0 to 0 at z=b= 0.8m. So its width at z in between is:

w = a·(1 - z/b)

Hence its area this differential segment is:

dA = w dz = a·(1 - z/b) dz

The force acting on that differential part of the end is:

dF = Δp dA

= ρ·g·z ·a·(1 - z/b) dz

= ρ·g·a·(z - z²/b) dz

Integrate this forces over the whole depth and you get the total force acting on the end:

F = ∫ dF

= ∫0→b [ ρ·g·a·(z - z²/b) ] dz

= ρ·g·a· ∫0→b [ (z - z²/b) ] dz

= ρ·g·a· { (1/2)·b² - (1/3)·b³/b - (1/2)·0² + (1/3)·0³/b) }

= (1/6)·ρ·g·a·b²

=>

F = (1/6) · 1000kg/m³ · 9.81m/s² · 1.2m · (0.8m)²

= 1255.68 kgm/s²

= 1255.68 N