Question

vious Problem List Next (1 point) A rectangular tank that is 8 feet long, 3 feet wide and 9 feet dee is filled with a heavy iquid that weighs 120 pounds per cubio foot ch patteow, assume that the tank is initially full. Your answers must include the correct units. (You may enter lbfor lb f for ff-lb.) (a) How much work is done pumping all of the liquid out over the top of the tank? (b) How much work is done pumping all of the liquid out of a spout 6 feet above the top of the tank? (c) How much work is done pumping two-thirds of the liquid out over the top of the tank? (d) How much work is done pumping two-thirds of the liquid out of a spout 6 feet above the top of the tank? Note: You can earn partial credit on this problem. Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining. Page generated at 04/07/2017 at 10.14am CDT Weework o 1900-2015 l theme mathfiwa version 2.10 lpg vension: 210 The Weework Project

Answer 1

Given length l = 8ft, breadth b = 3ft and height h = 9ft

therefore Volume of the tank = 8 * 3* 9 = 216 lbm.

Given density = 120 lbs/cu.ft

Hence total mass = density * Volume

= 120 * 216 = 25920 lb

By following the linearity of the system, the center of gravity(CG) lies at half way he depth

Therefore CG lies at 4.5ft

(a) The work done in pumping the water from the top of the tank = 4.5 * 25920 = 116,640 ft - lb

(b) The total position of the CG = 4.5+10 = 10.5

therefore the work done in pumping the liquid at a height 6ft above the tank = 10.5 *25920 = 272,160 ft - lb

(c) The work done in pumping at a height 2/3 above the tank = (2/3)*25920 = 17,280 ft - lb

(d) The new CG is 3ft and above the tank it is 6ft, hence the resultant position of the CG will be (3/2) + 6 = 7.5ft

therefore the work done = (2/3) * 7.5 * 25920 = 129,600ft - lb