Make sure you are connecting and building the math with the physical situation.This week: Based on what you know about work, energy, motion, and what we covered last Tuesday, you should be able to answer most of the HW. We will talk more about fluid motion coming Tuesday. You should be reading Ch 15 (Knight) and Ch 13 (Tipler). Fluids (Understanding Physics) is posted on Blackboard.

1. (Birkett and Elby) Measuring pressure with a mercury barometerA mercury barometer is made by taking a tube completely filled with mercury, submerging it in a tub of more mercury, and then flipping the tube upside-down, while still leaving the open end of the tube submerged. The mercury no longer fills the tube completely. a) Is there air in the tube in the space above the mercury? Explain.There is no space and therefore no air between the mercury and the closed end of thetube. When the tube is tipped up, the mercury moves down, leaving space between it and the closed end of the tube, and forming a seal that prevents any air from entering that space. Hence that space is nearly a vacuum, empty except for tiny amounts of mercury vapor.b) Is the pressure at point A greater than, less than, or equal to the pressure at point B? Explain.Since points B and A are at the same height in the same fluid, their pressures are equal. By contrastif the pressure at B exceeded the pressure at A, then mercury would flow from B toward A, thus lowering the height of the mercury column. Or, if pA were greater than pB, then mercury would flow from A to B, adding fluid to the tube. Since the mercury level in the tube is stable, pB = pA.c) Why does some mercury remain in the tube rather than all draining into the tub?No air is in the tube that pushes down on the column of mercury, so the pressure at point B arises entirely from the weight of the mercury column above it. By contrast, the pressure at A arises entirely from the outside air: pA = pair. Since the pressure is the same at B and A, there must be mercury in the tube—just enough to create a pressure at A that is equal to the outside air pressure (at A).

d) What happens to the mercury level in the tube when the air pressure goes up? Explain why this would happen conceptually.

Increasing outside air pressure (pA) pushes mercury up the tube until the pressure that the weightof that higher column generates at B equals the new outside air pressure at A.

e) You make your own mercury barometer (in your spare time) and measure the height of the mercury in the tube, measured from the top surface of the mercury in the tub. Write your “barometer equation” that you can use to get the outside air pressure in terms of your measurement of the height of the mercury in the tube.

(e) We start with the insight that pA = pB, and use the hydrostatic equation: Since there is no airpressure in the tube, we have pB = ρgh, where ρ is the density of mercury and h is the height ofmercury in the tube. From part (b) we know pA = pair. HencepA = pB OR ρgh = pair

Which gives us pair in terms of hA little bit more: Converting the relevant quantities to S.I. units (pair = 1.00 atm = 1.01x 105 Pa, ρmercury = 13.58 g/cm3 = 13580 kg/m3, g = 9.80 m/s2), we find h = 0.759 m, which is 759 mm.You may hear this result expressed as “air pressure is about 760 mm of mercury,” since mercury barometers often measure atmospheric pressure in labs. Although we can interpret this statement correctly, it’s a sloppy use of units because “millimeter” is a unit of distance, not pressure. To talk about “760” as a pressure, scientists use a non-S.I. unit called a torr. One torr is thepressure that supports 1 mm of mercury in a barometer. So, atmospheric pressure is about 760 torr.2. (Birkett and Elby) The Hydraulic LeverA hydraulic lever can be used to lift a car easily. You push on a small piston (Asmall = 0.060 m2), which in turn pushes on the fluid in the container. The fluid then pushes up on the larger piston (Alarge = 6.0 m2),

SOLUTIONS TO Phys 260 Gupta HW 4 (Fluids)REQUIRED READING

which lifts the car. The car and the large piston together weigh 7200 N. The dotted line shows where the pistons were before you push on the small piston.a) Using Pascal’s principle, calculate the minimum force Fpush that you must apply to the

small piston to lift the car. Does this answer match with the fact that hydraulic levers make it easier to lift a car?

The car-and-piston push down on the fluid with 7200 N spread over area Alarge = 100Asmall = 6.0 m2.The resulting pressure on the fluid’s top surface is p = Flarge/Alarge = (7200 N)/(6.0 m2) = 1200 Pa.Since the small piston is at the same depth, the fluid pushes on it with the same pressure, 1200 Pa.The resulting upward force exerted on the small piston is only Fpush = pA = (1200 N)(0.060 m2) =72 N, or 1/100th (the ratio of the areas Asmall/Alarge) of the car’s weight. By pushing down on the small piston with a force slightly greater than 72 N, you can move it down, and move the large piston and car up.b) To understand more about how this system works, let’s look at energy considerations. Suppose you

push the small piston in by 0.50 m. How far up does the car move?The fluid pushed out of the small piston tube pushes an equal amount of fluid into the large piston tube: |ΔVsmall piston| = ΔVlarge piston OR AΔypush = (100A)Δycar

Hence Δycar = Δypush/100 = 0.0050 m, or half a centimeter. Generally, the car rises a hundredth of the distance over which you push the small piston, since the ratio of the areas is Asmall/Alarge = 1/100.c) Calculate (i) the work you do on the piston, and (ii) the work the fluid does on the car. Which, if

either, is larger? Explain.Work is given by the product of the (average) force and the distance over which that force acts. The work you do on the piston is thus Wby you = FpushΔypush = (72 N)(0.50 m) = 36 J. The work done lifting the car is Won car = Fon carΔycar= (7200 N)(0.0050 m) = 36 J. Hence the energy you put into the system has gone directly into lifting the car.d) In class, when I asked the question (a), most of you intuitively thought that you would have to apply a

force equal to the weight of the car -- Roughly speaking, “you can only get out what you put in.” Reconcile that expectation with the answer that you get in part (a) - are you getting more than what you put in?

Although you get more than you give in terms of force, there is a trade-off in distances in order forenergy to be conserved: the force applied to the car is one-hundred times larger than your pushforce, but the car moves only one-hundredth of the distance through which you push. Thus whileyou push with much less force, you must do it over a much greater distance than that which the carrises.3. (Birkett and Elby) Soap FloatsA soap bar of volume 25.0 cm3 and mass 20.0 g floats in a large bucket of water.a) What percentage of the soap is beneath the surface of the water?b) Suppose you seal the top of the bucket with a lid and pump put almost all the air out of the bucket.

Does the percentage of the soap that’s submerged increase, decrease or stay the same? Explain.c) After answer part (b), a thoughtful student says: “I thought the soap would come higher out of the

water when the air was pumped out because air pressure pushes down on the top of the soap. By removing most of that air pressure, we lessen that downward force, allowing the soap to rise a little.” If that reasoning is correct, show how it is consistent with your reasoning from part (b). If that reasoning is incorrect, explain why.

d) The bucket (not sealed) and soap are somehow relocated to Mars, where gravity is weaker. Does the percentage of the soap that’s submerged increase, decrease, or stay the same? Explain.

SOLUTIONS TO Phys 260 Gupta HW 4 (Fluids)REQUIRED READING

4. The Force on a Dam, part 1

A rectangular dam is 20 m high from the bottom to the water’s surface. The shaded section represents the part of the dam in contact with the water it traps. The dam is 50 m wide and 2.5 m thick.a) A student, asked to find the overall force exerted by the trapped water on the dam,

reasons as follows: “The force is the pressure times the dam’s area, A = 1000 m2. And the pressure is just p = pair + ρwatergh, with h = 20 meters. So, I’ll calculate that p and plug it into F = pA.” Does that approach overestimate, underestimate, or correctly estimate the force? Explain.

The approach overestimates the force on the dam. The top of the dam has much lesser pressure than at the bottom; so the force on the top is also less. Multiplying the entire area of the dam with the pressure at the very bottom of the dam assumes that everywhere on the dam the pressure is equal to the maximum pressure (at the bottom) - so, it would give me a much larger number than the actual pressure on the dam.b) Another student suggests modifying the strategy from part (a). “To make that strategy work, you just

need to use the average pressure on the dam, not the pressure at the bottom. So, plug h = 10 meters instead of h = 20 meters into p = pair + ρwatergh, and then use that average pressure in F = pA.” Does this modified approach work? If not, does it overestimate or underestimate the actual force on the dam? Explain.

I think this would work! See the top part of the dam has less pressure than at the middle, the bottom has more. In fact, for each point above the middle-depth, I can find a corresponding point below the mid-depth, such that the average of the pressures at those points would be equal to the pressure at the mid-height. (E.g. [ph=5m + ph=15m]/2 = ph=10m) So, I can imagine that instead of the pressure varying with depth, it is the same everywhere - BUT equal to that at h = 10m!c) If the dam were trapezoidal rather than rectangular (See picture on the right), would using the average

pressure work? [Figure B: The dam is 50 meters wide at the water’s surface and 40 meters wide at the bottom, and it’s 2.5 meters thick.]. If so, which h would you use, and why? If not, does it overestimate or underestimate the actual force on the dam. Explain.

Now the game changes again! Of course, approach in (a) does not work - for the same reason (the pressure is varying with depth). I think the approach of (b) would also not work for the trapezoidal dam. This is because now the width of the dam is changing as we go down in the water. So, the pressure at the bottom acts on a smaller area, leading to smaller force than before. So yes, the average of pressure at 5 m and 15 m is still equal to the pressure at 10 m; but at 5m now, the pressure acts on a larger area than at 15 m. This breaks the nice balance that we talked about in (b), where the increased pressure (and force) below the mid-point exactly compensated for the decreased pressure (and force) above the mid-point. Because the dam is narrowing with depth, using the approach in (b) would overestimate the actual force if you used 50m as the width of the dam; and underestimate the actual force if you used 40m for the width of the dam.5. The Force on a Dam, part 2Now, let’s actually do some calculations and see if the math matches our reasoning.a) Use calculus to find the force exerted by the trapped water on the rectangular dam. Explain why you

set up the integral in the way you did. [Hint (on the hints page- see course website)]b) Is your answer here consistent with your response for the rectangular dam in parts (a) and (b) of the

previous problem?Parts 2c) and 2d) are only required for the Honors section.c) Use calculus to find the force exerted by the trapped water on the trapezoidal dam. Explain why you

set up the integral in the way that you did. d) Is your answer consistent with your answer in part (c) of the previous question?

SOLUTIONS TO Phys 260 Gupta HW 4 (Fluids)REQUIRED READING

6. (adapted from Birkett and Elby) Flow Rates in a PipeCompressed air is pumped into the pipe at point C and flows rightward towards point D. Transitioning from the thinner to the thicker segment of pipe, the gas becomes less compressed, i.e., its density goes down: ρD < ρC (but density is uniform within each segment)a) Explain why it’s not true that vC AC = vD AD. [hint: think about the amount of air that passes each point

in 1 second.]

In every second, the volume of air that goes past point C is vC AC *(1 sec) = vC AC meters3. Similarly the volume of air that goes past point D is vD AD. The equation in part (a) is saying that the volume of air passing point C in one second is equal to the volume of air passing point D in one second. Since the density of air at point D is less than the density of air at point C, the same volume implies that fewer air molecules are leaving this section of the pipe at point D than are entering at point C. If this is true, then the air molecules would build up in the pipe and the increased pressure would cause the pipe to burst. Common Sense!: this equation can’t be true, since we know that you can pump air into a pipe without it bursting!

b) At which point is vA greater? Why?

Based on my reasoning in part (a), the volume of air passing point D must be greater than passing point C, so vD AD > vC AC. If a certain number of air molecules go into the pipe at point C, the same number of air molecules has to come out at point D, to prevent the pipe bursting. Since the density of air at D is less than density at C, then the volume of air passing point D must be greater for the number of air molecules to be the same.

c) c) Modify the equation from part (a) to make it true. [hint: what rate must be the same at both points?Try to connect this equation to your reasoning in part (a)]

The equation in part (a) conserves the amount of volume of air that goes through the pipe. The thing we actually want to conserve here is the rate at which the Mass of the air flows through the pipe (This distinction did not matter when the density was same everywhere, but now it does). Think about the mass of air that flows past one point every second: the volume that flows past the point every second is vA. If we multiply it by the density (mass of air / volume), then we get the mass of air that flows past the point every second, ρvA. The equation must express the fact that we want the mass of air that flows past points C and D in a certain time to be equal, so ρCvCAC = ρDvDAD.

Now, instead of pumping air into the tube, we fill the thicker segment with oil and the thinner segment is filled with water and water is being pumped into the pipe at point C. The oil and the water won’t mix, oil is less dense than water, and neither oil or water are compressible, i.e., they don’t change their density in this process.d) In this case, which equation is true: “vC AC = vD AD” or your modified equation in part (c)? Explain.[Hint: what rate must be the same at both points? Which equation conserves this rate?]

The question here is: should the flow rate being conserved be the volume per second or the mass per second. If the conserved rate is mass per second, then because the oil and water are incompressible (density doesn’t change) and the oil is less dense then water, this would mean that more volume of oil had moved past point D than volume of water had moved past point C. This would mean that as you pump the water in at point C, more volume of fluid would leave the pipe. In this case, pumping water into this section would cause it to empty out, which doesn’t make sense.

SOLUTIONS TO Phys 260 Gupta HW 4 (Fluids)REQUIRED READING

Then the conserved rate would have to be volume per second. Does this make sense? Well, imagine sticking a plunger (like in a syringe) into the end of the pipe at point C. Because the water and oil aren’t going to compress, it makes sense that the volume of fluid that leaves the tube at point D would be equal to the volume displaced by the plunger at point C (just like in a syringe). Therefore, the correct equation should be the one that conserves the volume flow rate, ρCvCAC = ρDvDAD. (A third line of reasoning: Incompressibility would imply that the space occupied is conserved: thus conserving volume, as long as we are not losing stuff or adding stuff to the system)

The reason this analogy would fail with the air is that once you push the air into the larger section of pipe, the air expands and takes up more volume (because I said the density decreases), so you would actually expect a larger volume to exit at point D than had entered at point C.

SOLUTIONS TO Phys 260 Gupta HW 4 (Fluids)REQUIRED READING