Question

A tank is full of water. Find the work (in ft-lb) required to pump the water out of the spout. Use the fact that water weighs 62.5 lb/ft3. 3 ft4 ft5 ft A right triangular prism tank with a spout is given. The legs of the right triangle are 3 ft and 5 ft. The depth of the prism is 4 ft.

Answer 1

The Work required to pump water out of the spout  would be equal to 464,062.5 π

How can we calculate the work done?

Given data :

3 ft, 4 ft, 5 ft A right triangular prism tank with a spout is given.

weight of water = 62.5 Ib/ft³

inner radius ( r ) = 3 x 2 =6 ft

Outer radius ( R ) = 3 x 4 = 12 ft

Height ( h ) = 3 x 5 = 15 ft

To determine the work required to pump water out of the spout

y = Vertical distance

dv = cross-section volume

dw = weight

The radius of the tank

Radius = ( r - R ) y/h + R

Radius = ( 6 - 12 ) y/15 + 12

So, Radius = 12 - 0.4y

dv ( cross section volume )

$\pi r^2\\\\\pi (12 - 0.4y)^2dy$

Weight of cross-section ( dw ) = 62.5 ( 12.5 - 0.4y )² π dy

The work done to pump water out of the spout will be

Work done =  ∫¹⁵₀ 62.5 ( 12.5 - 0.4y )² π dy

∴ Work done ( W ) = 464,062.5

Hence we can conclude that the Work required to pump water out of the spout  would be 464,062.5 π

Learn more about work here;

brainly.com/question/13148947

#SPJ1