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*4.A swimming pool is [0 fl wide and 36 f long and its bottom is an inclined plane; the shallow end having depth of 2 f and the deep end, 12 f. Ilthe pool is full o...

Question

*4.A swimming pool is [0 fl wide and 36 f long and its bottom is an inclined plane; the shallow end having depth of 2 f and the deep end, 12 f. Ilthe pool is full of' water, lind the hydrostatic force on the shallow end: (Use the fact that water weighs 62.5 Ib/f'

*4.A swimming pool is [0 fl wide and 36 f long and its bottom is an inclined plane; the shallow end having depth of 2 f and the deep end, 12 f. Ilthe pool is full of' water, lind the hydrostatic force on the shallow end: (Use the fact that water weighs 62.5 Ib/f'



Answers

A swimming pool is 20 ft wide and 40 ft long and its bottom is an inclined plane, the shallow end having a depth of 3 $\mathrm{ft}$ and the deep end, 9 $\mathrm{ft}$ . If the pool is full of water, estimate the hydrostatic force on (a) the shallow end, (b) the deep end, (c) one of the sides, and (d) the bottom of the pool.

So here for a party, we can say the weight of the water would be equaling the density times G times be and so the mass would be the density times of volume. And then, uh, the mass times acceleration due to gravity would give us the weight and so this would be equal to the density of water. 1000 kilograms per cubic meter multiplied by 9.80 meters per second squared. This would be multiplied by the volume 5.0 meters multiplied by 4.0 meters times 3.0 Brother being the same multiplied 3.0 meters and this would equal 5.9 times 10 to the fifth Newton's. This would be the weight of the water. Now for part B. Integration gives the expected result that the force is what it would be if the pressure were uniform and equal to the pressure at the midpoint. So essentially we could find the force. This would be equaling the density times acceleration due to gravity times. The area multiplied by D over too. And so we find that this is equaling 1000 kilograms per cubic meter multiplied by 9.8 zero meters per second. Squared multiplied. Rather, we can say for 0.0 meters multiplied by 3.0 meters multiplied by D over too 1.50 meters. And we find that the force is equaling 1.76 times 10 to the fifth Newton's. This would be your answer for part B. That is the end of the solution. Thank you for

And we have a man of mass M equals to 64 kg and density rho equals to 9 70. Program per meter cube. Standing in a shallow pool with 32% of its volumes. Volume is equals to 0.32 of the volume his of his body. Okay. Below the water. So calculate the normal force that the bottom of the pool exit on his feet. So I suppose this is the person and the weight of the person. This will be in the downward direction, so it will be M. G. And the person will feel a buoyancy force in the other direction and the normal force. This will be in the upward direction. So we can write from the free body diagram that and plus F B minus MG. This will be equal to zero. So from here we get normal force and this is equal to M G minus F B. Where FB is the buoyancy force. Okay, so this can be written as and that is equal to M G minus by and see force. It will be equal student city of flu, multiplied by volume displaced, multiplied by G. And it can be written as displaced volume. It can be written as displaced volume. It is given as 0.32 times of body volume of the body. So it can be written as M G minus density rho and displaced volume. It is 0.32 times of volume of body. Molecular biggie and volume of the body can be replaced in the term of mass M. So we can write that end. This will be equals two MG minus Romer player by 0.32 Market volume of the body can be written as must develop by density rho. Okay, this is the density of the flu. Okay, density of the flu. So we can write flew F. And this is the density of the body and G. So from here after rearranging we get MG. MG. And bracket so 1 -0.32 times of density of the flu. They were by density of the body. Okay, now substituting values in the situation. So since flurries here water. Okay, so substituting values for normal force and this will be equals two M. Mass, which is 64 K. G. And G, which is 9.81 And bracket one minus 0.32 and density of the flute. This is that is water. So this is 1000 program per meter cubed and density of the body, which is 9 70 Program permitted. So from here after solving we get normal force and equals to four .2 into 10 to the power to Newton. Okay, so this is the normal force exerted on his feet by the water. Okay?

However, where you are going to solve a problem numbered forced. So we have to given area we call 22 square feet on. We have to remember that there was a dynasty of border equals 62.4 found bear cubic feet so the force or the vicar force equal be pressure off by a that equal w what dynasty? Multiply edge, water height, multiply area. So that's equal 62 0.4 multiplied eight Multiple oy earlier. 22. So that equal 10,000 982 pound. Thanks for watching.

Over where you are going to solve a problem number to. So we have area equal 16 feet squirt on who we have to remember. But there were dynasty off further equal 62.4 of pounds per cubic feet so their force equal be multiply eight. That's equal w what s border? Deploy what head Multiply areas service that's called 62.4. Multiply eight multiply. 0 16. So that's what be equal. A seven mine 87 0.2 pounds. Thanks for


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