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Pascal’s Principle (hydraulic systems)PowerPoint Presentation

Pascal’s Principle (hydraulic systems)

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Chapter 14 problems

- Joey Soprano tasks two underlings to dispose of a large heavy “package” in the bottom of a holding tank at a local water treatment plant. The pair take the package out to the middle of the (large) tank using a small boat, and drop it over the side (at which point it sinks to the bottom of the tank). Does the water level (as measured at the side of the tank) go up, go down, or stay the same after the package reaches the bottom and while the two hoods are moving back to the side of the tank (compared to its level when the package was still in the boat)?
(DOWN: 3 UP: 9 SAME: 13 NO ANSWER: 16 ??: 1 )

- When the package is still in the boat no water is displaced because the bouyant force is greater than the gravitational force so no part of the package is submerged. However when the package goes to the bottom it does displace water in the tank and so the water level will go up a little bit.
- The water level will be the smae. The requirement for the flotation of the boat is that the F(b)be equal to the weight of the water displaced. So the package would have made the boat sink in the water an amount to equal its mass in water.
- Goes down. When he is floating in the boat, he is displacing his own weight in water. When he sinks to the bottom, he is displacing his own volume in water. Since he sinks, it is safe to assume that he is more dense than the water. Since he is denser than water, his weight corresponds to a greater volume of water than his volume.

Equation of Continuity heavy “package” in the bottom of a holding tank at a local water treatment plant. The pair take the package out to the middle of the (large) tank using a small boat, and drop it over the side (at which point it sinks to the bottom of the tank). Does the water level (as measured at the side of the tank) go up, go down, or stay the same after the package reaches the bottom and while the two hoods are moving back to the side of the tank (compared to its level when the package was still in the boat)?(mass in must = mass out)

A1v1=A2v2

Assuming that r is constant

(i.e. an incompressible fluid)

E.G. with the Equation of Continuity heavy “package” in the bottom of a holding tank at a local water treatment plant. The pair take the package out to the middle of the (large) tank using a small boat, and drop it over the side (at which point it sinks to the bottom of the tank). Does the water level (as measured at the side of the tank) go up, go down, or stay the same after the package reaches the bottom and while the two hoods are moving back to the side of the tank (compared to its level when the package was still in the boat)?(mass in must = mass out)

What is the flow through the unmarked pipe?

E.G. with the Equation of Continuity heavy “package” in the bottom of a holding tank at a local water treatment plant. The pair take the package out to the middle of the (large) tank using a small boat, and drop it over the side (at which point it sinks to the bottom of the tank). Does the water level (as measured at the side of the tank) go up, go down, or stay the same after the package reaches the bottom and while the two hoods are moving back to the side of the tank (compared to its level when the package was still in the boat)?(mass in must = mass out)

13

What is the flow through the unmarked pipe?

(4+8+5+4 = 21 in; 13+2+6 = 21 out)

Bernoulli’s Principle heavy “package” in the bottom of a holding tank at a local water treatment plant. The pair take the package out to the middle of the (large) tank using a small boat, and drop it over the side (at which point it sinks to the bottom of the tank). Does the water level (as measured at the side of the tank) go up, go down, or stay the same after the package reaches the bottom and while the two hoods are moving back to the side of the tank (compared to its level when the package was still in the boat)?

P + ½rv2 + rgy = const

Rank the pressures at positions 1, 2, 3, 4 ?

- Why does the curtain crowd you in the shower? heavy “package” in the bottom of a holding tank at a local water treatment plant. The pair take the package out to the middle of the (large) tank using a small boat, and drop it over the side (at which point it sinks to the bottom of the tank). Does the water level (as measured at the side of the tank) go up, go down, or stay the same after the package reaches the bottom and while the two hoods are moving back to the side of the tank (compared to its level when the package was still in the boat)?
- (Bernoulli: 9; Water: 8; Temperature: 2; no answ: 17)
- The warm air (from the hot shower) rises out over the shower curtain, as the cooler air near the floor pushes in under the curtain to replace the rising air.
- Because the air inside the shower is hotter and has a lower pressure than the cold air outside the shower; therefore, the outside air forces the curtain inside because of the differences in pressure.(certainly, but what causes the pressure difference?)
- Bernoulli's principle says an increase in velocity results in a decrease in pressure, therefore, when the water flow is turned on, air flow moves the same direction as the water and the pressure inside drops.

Chapter 14 heavy “package” in the bottom of a holding tank at a local water treatment plant. The pair take the package out to the middle of the (large) tank using a small boat, and drop it over the side (at which point it sinks to the bottom of the tank). Does the water level (as measured at the side of the tank) go up, go down, or stay the same after the package reaches the bottom and while the two hoods are moving back to the side of the tank (compared to its level when the package was still in the boat)? problems

- A basketball is dropped onto a good hardwood floor and is observed to bounce several times. In your opinion, is such motion oscillatory? Is this motion simple harmonic motion? Please explain (briefly) why or why not in each case.
Most said yes to osc. But NO to SHM (changing amplitude [9] or freq. [7]

- Yes such a motion is oscillatory as well as simple harmonic motion. It passes through a minimum and maximum positon and moves periodically.
- The motion is oscillatory but does not exhibit a simple harmonic motion. It is oscillatory because it exhibits a sinusoidal pattern bouncing up and down. However, since there is a change in frequency because of a loss in height each time the ball bounces back up, it does not exhibit simple harmonic motion
Several said not oscillatory because there was “no fixed axis”?

- Explain as compactly as you can manage the essential requirement on a force if it is to be able to produce simple harmonic motion.
- When the restoring force (the force that opposes displacement) is opposite and directly proportional to the displacement,the oscillating object will exhibit Simple Harmonic Motion ( 17 Answered like this; but ONLY ONE one applied this to the previous Q!!).
- It must be elastic, such as a spring, where the force varies with distance, but it also must always be present, and not lose energy such as in the basketball. (Here, the basketball when it bounces, compresses like the spring during the impulse.) (2 answers)
- Energy conservation figured in a few answers.

Connection to Circular motion requirement on a force if it is to be able to produce simple harmonic motion.

Angular separation between

Jupiter and Callisto as viewed

from the Earth by Galileo in 1610!

Here there is a real angle associated with the

phase*; in most SHM case the phase angle

is just a way of keeping track of the motion.

* What is that angle in the case of the Callisto observations?

Chapter 15 requirement on a force if it is to be able to produce simple harmonic motion. problems

(The vertical scale mark is at

t=0.)

Chapter 15 requirement on a force if it is to be able to produce simple harmonic motion. problems

- A 30kg child is swinging on a swing whose seat is a distance of 5.0m from the pivot point. Estimate the optimal time between “pumps” that the child should execute to increase her swinging amplitude. How does this time change for a 15.0 kg child?
- (11 got this right, 3 made a slight error, 13 didn’t answer and 12 didn’t know what to do).
- My best guess is to use the equation T= 2(pi)*square root of (I/mgL). However, we dont have the rotational inertia of the pendulum (I). However, it is apparent that if you change the mass, it will affect the time between ˜pumps˜.
- use equation T=2piesqrootI/mgh so I=mLsqrd and the optimal time T=4.48sec so the finally answer for a 15kg child changes to T=4.49s i was not for sure how to plug in this data especially for I (for someone who did not know what to, this is VERY GOOD!)
- T = 2*pi*(L/g)^(1/2) The time does not change. The estimate time is 4.5 s.

Physical Pendulum of 5.0m from the pivot point. Estimate the optimal time between “pumps” that the child should execute to increase her swinging amplitude. How does this time change for a 15.0 kg child?

I a = mghsin(q) but if q<<1 then:

a= (mgh/I) q

Be careful to distinguish the physical

angle from the phase angle here!

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