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.
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Answer #1

(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 x Volume = 62.4 x \pi(5^2)dy

Potential Energy of the elementary disc of water =  ( π (52)dy)(62.4)g(1-y)

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

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