7.5 Fluid Pressure and Forces Greg Kelly, Hanford High School, Richland, Washington.

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7.5 Fluid Pressure and Forces Greg Kelly, Hanford High School, Richland, Washin

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If we had a 1 ft x 3 ft plate on the bottom of a 2 ft deep wading pool, the force on the plate is equal to the weight of the water above the plate. densitydepth pressure area All the other water in the pool doesn’t affect the answer!

Transcript of 7.5 Fluid Pressure and Forces Greg Kelly, Hanford High School, Richland, Washington.

Page 1: 7.5 Fluid Pressure and Forces Greg Kelly, Hanford High School, Richland, Washington.

7.5 Fluid Pressure and Forces

Greg Kelly, Hanford High School, Richland, Washington

Page 2: 7.5 Fluid Pressure and Forces Greg Kelly, Hanford High School, Richland, Washington.

What is the force on the bottom of the aquarium?

3 ft

2 ft

1 ft

Force weight of waterdensity volume

3

lb62.5 2 ft 3 ft 1 ftft

375 lb

Page 3: 7.5 Fluid Pressure and Forces Greg Kelly, Hanford High School, Richland, Washington.

If we had a 1 ft x 3 ft plate on the bottom of a 2 ft deep wading pool, the force on the plate is equal to the weight of the water above the plate.

3

lb62.5 ft

density

2 ft

depth

pressure

3 ft 1 ft

area

375 lb

All the other water in the pool doesn’t affect the answer!

Page 4: 7.5 Fluid Pressure and Forces Greg Kelly, Hanford High School, Richland, Washington.

What is the force on the front face of the aquarium?

3 ft

2 ft

1 ft

Depth (and pressure) are not constant.

If we consider a very thin horizontal strip, the depth doesn’t change much, and neither does the pressure.

3 ft

2 ft

2

0

ydy

62.5 3 yF y dy

densitydepth

area2

062.5 3 F y dy

22

0

187.52

F y 375 lb

It is just a coincidence that this matches the first answer!

Page 5: 7.5 Fluid Pressure and Forces Greg Kelly, Hanford High School, Richland, Washington.

6 ft

3 ft

2 ft

A flat plate is submerged vertically as shown. (It is a window in the shark pool at the city aquarium.)

Find the force on one side of the plate.

Depth of strip: 5 y

Length of strip:

y xx y

2 2x y

Area of strip: 2 y dy

62.5 5 2 yF y y dy

density depth area

3

062.5 5 2 F y y dy

3 2

0125 5 F y y dy

32 3

0

5 11252 3

F y y

1687.5 lbF

We could have put the origin at the surface, but the math was easier this way.

Page 6: 7.5 Fluid Pressure and Forces Greg Kelly, Hanford High School, Richland, Washington.

2 323

68%

95%

99.7%

34%

13.5%

2.35%

Normal Distribution:For many real-life events, a frequency distribution plot appears in the shape of a “normal curve”.

Examples:

heights of 18 yr. old men

standardized test scores

lengths of pregnancies

time for corn to pop

The mean (or ) is in the middle of the curve. The shape of the curve is determined by the standard deviation .

x

mu

x x-bar

sigma

“68, 95, 99.7 rule”

Page 7: 7.5 Fluid Pressure and Forces Greg Kelly, Hanford High School, Richland, Washington.

2 323

34%

13.5%

2.35%

Normal Distribution:

“68, 95, 99.7 rule”

The area under the curve from a to b represents the probability of an event occurring within that range.

In Algebra 2 we used z-scores and a table of values to determine probabilities. If we know the equation of the curve we can use calculus (and our calculator) to determine probabilities:

2 2/ 212

xf x e

Normal Probability Density Function:(Gaussian curve)

Page 8: 7.5 Fluid Pressure and Forces Greg Kelly, Hanford High School, Richland, Washington.

Normal Distribution:

2 2/ 212

xf x e

Normal Probability Density Function:(Gaussian curve)

The good news is that you do not have to memorize this equation!

Example 6 on page 406 shows how you could integrate this function to predict probabilities.

In real life, statisticians rarely see this function. They use computer programs or graphing calculators with statistics software to draw the curve or predict the probabilities.