Question

Name: 1. A cattle trough has a semi-circular cross section with a height of 1 foot. The length of the trough is 10 feet. The trough is half-full of water. Assume the density of water is 62.4 lb/ft3 (a) (8 points) How much work is required to pump the water out of the trough? (b) (7 points) Determine the total force of the water on a semi-circular face of the trough.

Answer 1

(a)

Consider an elementary disc of water. Volume of this elementary disc = where dy = thickness of the elementary disc.

(a) Consider an elementary disc of water. Volume of this elementary disc = pi (5^2) dy where dy = thickness of the elementary disc. Mass = Density times Volume = 62.4 times pi (5^2) dy Potential Energy of the elementary disc of water = (pi (5^2) dy) (62.4)g (1 - y) where y = height of the elementary disc of water from the ground Potential Energy of the of all the water = y = 0.5 integral y = 0(pi (5^2) dy) (62.4)g (1 - y) = 18029.1 lbf ft We integrate till y = 0.5 as the trough is half full, i.e., full till y = 0.5. Potential energy of all water = Work done to pump out all the water to the top = 18029.1 lbf ft (b) Force of the water on the semicircular face = y = 0.5 integral y = 0(pi (5^2) dy) (62.4)g = 24038.8 lbf

Mass = Density x Volume = 62.4 x

(a) Consider an elementary disc of water. Volume of this elementary disc = pi (5^2) dy where dy = thickness of the elementary disc. Mass = Density times Volume = 62.4 times pi (5^2) dy Potential Energy of the elementary disc of water = (pi (5^2) dy) (62.4)g (1 - y) where y = height of the elementary disc of water from the ground Potential Energy of the of all the water = y = 0.5 integral y = 0(pi (5^2) dy) (62.4)g (1 - y) = 18029.1 lbf ft We integrate till y = 0.5 as the trough is half full, i.e., full till y = 0.5. Potential energy of all water = Work done to pump out all the water to the top = 18029.1 lbf ft (b) Force of the water on the semicircular face = y = 0.5 integral y = 0(pi (5^2) dy) (62.4)g = 24038.8 lbf

Potential Energy of the elementary disc of water =  

where y =height of the elementary disc of wter from the ground

Potential Energy of the of all the water =   $\int_{y=0}^{y=0.5}(\pi(5^2)dy)(62.4)g(1-y)$ = 18029.1 lbf ft

We integrate till y=0.5 as the trough is hallf full,i.e, full till y=0.5.

Potential energy of all water = Work done to pump out all the water to the top = 18029.1 lbf ft

(b)

Force of the water on the semicircular face = $\int_{y=0}^{y=0.5}(\pi(5^2)dy)(62.4)g$

= 24038.8 lbf