Answer to Question #212039 in Mechanical Engineering for Mary

Question

given, pool volume 86.9m^3

hydraulic pressure= .00028733N/m

F=PxA = .0166N

1.    If the pool is built with 15 cm concrete on the bottom and 10 cm concrete walls, calculate the net weight of the pool structure for each half. What are the resulting net forces on each half of the pool? You’ll need to show a free body diagram with the direction of the forces shown.

2.    Based on your results, can you comment on the photo and conclude why one half is higher than the other? And can you guess what is the water level around the pool? There is not one easy answer for this last question, use your judgment.

 

The density of concrete is 2320 kg/m³

 

Answer 1

Part 1

$W= mg$

But $m = \frac{\rho}{v}= \frac{2320}{86.9}=26.697 kg$

$W= mg =26.697* 9.8 = 261.63 N$

Net weight of the pool structure for each half

$W= \frac{mg}{3} =\frac{26.697}{3}* 9.8 = 87.2102 N$

Resultant force

$87.2102- 1*9.8*2=67.6102 N$

Part 2

One half is higher than the other to achieve pressure differences in the same pool.

The water level in the pool is 69240 ltrs.