PHY 171. Homework 9 solutions. (Due by beginning of class on Thursday, March 8, 2012)

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1 PHY 171 Due by beginning of class on Thursday, March 8, 2012 Submit neat work, with answers or solutions clearly marked by the question number. Unstapled, untidy work will be charged a handling fee of 20% penalty. Writing only an answer without showing the steps used to get to that answer will fetch very few points, even if the answer is correct. Late homework will not be accepted. 1. An iron anchor of density 7870 kg/m 3 appears 200 N lighter in water than in air. a What is the volume of the anchor? Solution: Since the buoyant force is given by F B = weight in air weight in water and we are told that the object appears 200 N lighter in water than in air, this means that By Archimedes Principle, F B = 200 N F B = weight of water displaced = mass of water displaced g = ρ w V wd g where ρ w is the density of water, and V wd is the volume of water displaced. Substituting numbers, we get 200 N = 1000 kg/m 3 V wd 9.8 m/s 2 V wd = = m 3 Presumably, the object is completely immersed in water, so it is displacing a volume of water equal to its own volume, i.e., V = V wd. Therefore, the volume of the object is V = m 3 Note: We are assuming that both zeros in 200 N are significant, in assessing the significant figures in the answer above.

2 PHY 171 Winter Continued from previous page... b How much does the anchor weigh in air? Solution: To find the weight of the anchor in air, we need to find its mass, which from the definition of density is given by density of iron, the material of which the anchor is made times the volume of the anchor which we determined in part a. So, weight of the anchor is W = mass of anchor g which gives Therefore W = ρ iron V g = {7870 kg/m 3} m m/s 2 W = 1570 N 2. A block of wood floats in fresh water with two thirds of its volume submerged. In oil, the block floats with 0.90 of its volume submerged. Find the density of the wood and the oil. Solution: An object will float if its weight is equal to the buoyant force on the object. Meanwhile, we have from Archimedes Principle that the buoyant force on an object is equal to the weight of fluid displaced by the object. So, we can write W object = weight of water displaced mass of wood block g = mass of water displaced g ρ wood V wood g = ρ water V water displaced g ρ wood ρ water = V water displaced V wood Now, since the wood floats with 2/3 of its volume submerged, V water displaced = 2 3 V wood, so 2 / 3 V wood ρ wood = ρ water = 2 V wood 3 ρ water = kg/m 3 = kg/m 3 3 Carrying out the same procedure as above, except with water replaced by oil we get ρ wood ρ oil = V oil displaced V wood = 0.9V wood V wood ρ oil = ρ wood 0.9 = kg/m 3 Page 2 of 7

3 PHY 171 Winter A block of wood has a mass of 3.67 kg and a density of 600 kg/m 3. It is to be loaded with lead it will float in water with 0.90 of its volume submerged. The density of lead is kg/m 3. a Calculate the mass of lead needed if the lead is attached to the top of the wood. Solution: Given m wood = 3.67 kg, ρ wood = 600 kg/m 3, and ρ lead = kg/m 3 Once again, an object will float if its weight is equal to the buoyant force on the object, while we have from Archimedes Principle that the buoyant force on an object is equal to the weight of fluid displaced by the object. In this case, the weight of the object is the weight of the block of wood and the lead attached to the top of the block of wood; let us denote the mass of lead as m lead. So we can write W object = weight of water displaced m wood +m lead g = mass of water displaced g 3.67 kg+m lead g = ρ water V water displaced g H9.3.1 Inthispart,theleadisattachedtothetopofthewoodblock, whichmeansthatitisabovethe water surface and so it does not displace any water. Therefore, the volume of water displaced is equal to the volume of the wood block that is submerged, namely 0.90V wood. We can find V wood from the mass and density of the wood block by writing V wood = m wood /ρ wood. So 3.67 kg+m lead g = ρ water {0.9V wood } g from which we obtain m lead = 1000 kg/m [ ] 3.67 kg 600 kg/m kg = 1.84 kg b Calculate the mass of lead needed if the lead is attached to the bottom of the wood. Solution: Inthispart,theleadisattachedtothebottomofthewoodandisfullysubmerged, so in addition to the water displaced by the wood = 0.9V wood, the lead will displace an additional amount of water equal to its own volume, V lead. So, H9.3.1 must be changed to 3.67 kg+m lead g = ρ water {0.9V wood }+ρ water V lead g [ 3.67 kg 3.67 kg+m lead = 1000 kg/m 0.9 ] kg/m 3 kg/m 3 m lead kg/m 3 m lead [ ] = 1000 kg/m Page 3 of 7 [ ] m lead = 2.01 kg 600

4 PHY 171 Winter Three children, each of weight 356 N, make a log raft by lashing together logs of diameter m and length 1.80 m. How many logs will be needed to keep them afloat in fresh water? Take the density of the logs to be kg/m 3. Solution: Yet again, an object will float if its weight is equal to the buoyant force on the object, while we have from Archimedes Principle that the buoyant force on an object is equal to the weight of fluid displaced by the object. In this case, if we assume that X logs are required, the weight of the object will be the weight of the 3 children plus the weight of these X logs. So [ ] Weight of 3 children +Weight of X logs = ρ water V water displaced g Now, we certainly don t want the children displacing any water because that would imply part of them was in the water instead of being on the raft above the water level, so the volume of water displaced is simply the volume of the X logs. In other words, we are assuming that the logs are just under the water surface in order to find the minimum number of logs needed to keep the children afloat. Writing the volume of a log as V log and the mass of a log as m log = ρ log V log, we get from the equation above that N +X m log g = ρ water XV log g or N +X ρ log V log g = ρ water X V log g Moving terms containing X to one side, we get XV log ρ water ρ log g = N H9.4.1 The volume of a log, of course, can be found by assuming that each log is a cylinder whose diameter is given as d log = m and height as h log = 1.80 m, V log = π dlog m 2 h log = π 1.80 m = π m 1.80 m 2 Substituting this and other values into equation H9.4.1, we get [ 2 ] X π m 1.80 m 1000 kg/m kg/m m/s 2 = N which reduces to X = N X = 3356 N = 4.3 Therefore, in order to keep the children afloat, the raft must have at least 5 logs Page 4 of 7

5 PHY 171 Winter We often see blimps cruising at low altitudes over Chicago, either during game time or some other shows. Suppose such a blimp, filled as usual with helium, has a maximum useful payload, including crew and cargo, of 1280 kg. The volume of the helium filled interior space is m 3. The density of helium gas is kg/m 3. Instead, if you filled the blimp with hydrogen gas density = kg/m 3, how much more payload could the blimp carry? Aside: Can you think of why blimps are usually filled with helium, even though hydrogen would allow you to carry a heavier payload? Solution: We remind ourselves again that an object will float if its weight is equal to the buoyant force on the object, while we have from Archimedes Principle that the buoyant force on an object is equal to the weight of fluid displaced by the object. In the present case, the fluid displaced is air, while the contributing factors to the weight are the payload including crew and cargo, and the weight of whatever gas fills the balloon. This gives Weight of payload+weight of gas filling balloon = Weight of air displaced m payload +m balloon g = m air displaced g Since m = ρv, we get when the gas is filled with He that m payload +ρ He V balloon g = ρ air V air displaced g H9.5.1 Instead, if the balloon were filled with hydrogen, we would be able to carry an additional amount of payload m new payload = m payload +m additional payload mpayload +madditional payload +ρhydrogenvballoon g = ρ air V air displaced g H9.5.2 Since the right hand side of both equations H9.5.1 and H9.5.2 above must be the same if the balloon is to float under the same conditions, we can set their left hand sides equal to get m payload +ρ He V balloon g = m payload +m additional payload +ρ hydrogen V balloon g Canceling m payload from both sides, we get m additional payload = ρ He V balloon = m additional payload +ρ hydrogen V balloon ρ He ρ hydrogen V balloon = 0.16kg/m kg/m m 3 which gives m additional payload = 395 kg We don t usually use hydrogen because it is flammable, of course remember the Hindenburg? Page 5 of 7

6 PHY 171 Winter What minimum force must you exert to lift a car of mass 1500 kg, if A out = 1.6 m 2 and A in = 0.2 m 2? If you want to raise the car by 25 cm, how far will you have to move the input piston? If you have difficulty understanding what the symbols mean, please see the figures on page 11 of the notes for Lecture 18 Feb 29 posted on the course website. Solution: This problem uses Pascal s law, which states that the pressure exerted at any point in a fluid is transmitted undiminished throughout the fluid. Therefore, Pascal s law gives for our present problem that P in = P out Using the definition of pressure as the force per unit area perpendicular to the surface, we can therefore write F in = F out A in A out This gives F in = Ain A out F out 0.2 m 2 [ = 1500 kg9.80 m/s 2] 1.6 m 2 = N Therefore, the minimum input force required to lift the car is N Meanwhile, to find how far we will have to move the input piston, consider that the volume displaced on the input side must be the same as the volume displaced on the output side, so A in y in = A out y out Therefore, the required y in = Aout A in y out = m = 2 m 0.2 Therefore, we will have to move the input piston through a distance of 2 m Page 6 of 7

7 PHY 171 Winter An ambulance with a siren emitting at 1600 Hz overtakes and passes a cyclist pedaling a bike at 2.44 m/s. After being passed, the cyclist hears a frequency of 1590 Hz. How fast is the ambulance moving? Solution: Given that the frequency emitted by the source ambulance is f = 1600 Hz, and frequency received by the detector is f = 1590 Hz. We are given the speed of the cyclist, v D = 2.44 m/s, and asked to find the speed of the ambulance, v s. Use the speed of sound in air, v air = 340 m/s. The relation for the Doppler effect is f = f v air ±v D v air ±v s where the choice of signs in the numerator and denominator must be made by taking the movement toward or away into account. This is a case when both source and detector are in motion. Since we are looking at the situation after the ambulance has passed the cyclist, the source ambulance is moving away from the detector cyclist, so we expect a lower frequency, hence we use the + sign in the denominator. Meanwhile, since the detector is still moving toward the source, and we expect a higher frequency, we should use the + sign in the numerator. Therefore, we get Substituting numbers or which gives f = f [ ] vair +v D v air +v s [ ] = v s v s = v s = 547, ,600 v s = 7240 = 4.55 m/s 1590 Therefore, the ambulance is moving at a speed of 4.55 m/s Page 7 of 7

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