PHY 171 Homework 9 solutions (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/m3 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 FB = 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 FB = 200 N By Archimedes’ Principle, FB = weight of water displaced = mass of water displaced g = ρw Vwd g where ρw is the density of water, and Vwd is the volume of water displaced. Substituting numbers, we get 200 N = 1000 kg/m3 Vwd 9.8 m/s2 so that Vwd = 200 = 2.04 × 10−2 m3 1000(9.8) Presumably, the object is completely immersed in water, so it is displacing a volume of water equal to its own volume, i.e., V = Vwd . Therefore, the volume of the object is V = 2.04 × 10−2 m3 Note: We are assuming that both zeros in 200 N are significant, in assessing the significant figures in the answer above. PHY 171 (Winter 2012) Homework 9 solutions 1. 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 o n 3 −2 3 W = ρiron V g = 7870 kg/m 2.04 × 10 m 9.8 m/s2 Therefore 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 ⇒ Wobject = weight of water displaced mass of wood block g = mass of water displaced g ⇒ ρwood Vwood g = ρwater Vwater displaced g ⇒ Vwater displaced ρwood = ρwater Vwood Now, since the wood floats with 2/3 of its volume submerged, Vwater displaced = 32 Vwood , so ρwood = ρwater 2 /3 Vwood Vwood = 2 2 3 10 kg/m3 = 6.7 × 102 kg/m3 ρwater = 3 3 Carrying out the same procedure as above, except with “water” replaced by “oil” we get ρwood Voil displaced 0.9 Vwood = = ρoil Vwood Vwood so that ρoil = ρwood = 7.4 × 102 kg/m3 0.9 Page 2 of 7 PHY 171 (Winter 2012) Homework 9 solutions 3. A block of wood has a mass of 3.67 kg and a density of 600 kg/m3 . It is to be loaded with lead so that it will float in water with 0.90 of its volume submerged. The density of lead is 1.13 × 104 kg/m3 . (a) Calculate the mass of lead needed if the lead is attached to the top of the wood. Solution: Given mwood = 3.67 kg, ρwood = 600 kg/m3 , and ρlead = 1.13 × 104 kg/m3 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 mlead . So we can write Wobject = weight of water displaced ⇒ mwood + mlead g = mass of water displaced g ⇒ 3.67 kg + mlead g = ρwater Vwater displaced g (H9.3.1) In this part, the lead is attached to the top of the wood block, which means that it is above the 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 0.90 Vwood . We can find Vwood from the mass and density of the wood block by writing Vwood = mwood /ρwood . So n o 3.67 kg + mlead S g = ρwater 0.9Vwood S g from which we obtain mlead 3.67 kg = 1000 kg/m 0.9 − 3.67 kg = 1.84 kg 600 kg/m3 3 (b) Calculate the mass of lead needed if the lead is attached to the bottom of the wood. Solution: In this part, the lead is attached to the bottom of the wood and is fully submerged, so in addition to the water displaced by the wood (= 0.9Vwood ), the lead will displace an additional amount of water equal to its own volume, Vlead . So, (H9.3.1) must be changed to n o 3.67 kg + mlead S g = ρwater 0.9Vwood + ρwater Vlead S g so that mlead 3.67 kg 3 + 1000 kg/m 3.67 kg + mlead = 1000 kg/m 0.9 600 kg/m3 1.13 × 104 kg/m3 1000 3.67 3 − 3.67 ⇒ mlead = 2.01 kg ⇒ mlead 1 − = 1000 kg/m 0.9 1.13 × 104 600 3 Page 3 of 7 PHY 171 (Winter 2012) Homework 9 solutions 4. Three children, each of weight 356 N, make a log raft by lashing together logs of diameter 0.300 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 800.0 kg/m3 . 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 h i Weight of 3 children + Weight of X logs = ρwater Vwater 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 Vlog and the mass of a log as mlog = ρlog Vlog , we get from the equation above that 3 356 N + X mlog g = ρwater X Vlog g or 3 356 N + X ρlog Vlog g = ρwater X Vlog g Moving terms containing X to one side, we get XVlog ρwater − ρlog g = 3 356 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 dlog = 0.300 m and height as hlog = 1.80 m, so that Vlog = π dlog 2 2 hlog = π 0.300 m 2 2 2 1.80 m = π 0.150 m 1.80 m Substituting this and other values into equation (H9.4.1), we get h 2 i 2 3 3 X π 0.150 m 1.80 m 1000 kg/m − 800 kg/m 9.8 m/s = 3 356 N which reduces to so that X 249.3796 = 3 356 N X= 3(356 N) = 4.3 249.3796 Therefore, in order to keep the children afloat, the raft must have at least 5 logs Page 4 of 7 PHY 171 (Winter 2012) Homework 9 solutions 5. 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 5000.0 m3 . The density of helium gas is 0.160 kg/m3 . Instead, if you filled the blimp with hydrogen gas (density = 0.081 kg/m3 ), 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 so that mpayload + mballoon g = mair displaced g Since m = ρ V , we get when the gas is filled with He that mpayload + ρHe Vballoon g = ρair Vair displaced g (H9.5.1) Instead, if the balloon were filled with hydrogen, we would be able to carry an additional amount of payload mnew payload = mpayload + madditional payload so that mpayload + madditional payload + ρhydrogen Vballoon g = ρair Vair 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 mpayload + ρHe Vballoon S g = mpayload + madditional payload + ρhydrogen Vballoon S g Canceling mpayload from both sides, we get ρHe Vballoon = madditional payload + ρhydrogen Vballoon so that madditional payload = ρHe − ρhydrogen Vballoon = 0.16 kg/m3 − 0.081 kg/m3 5000 m3 which gives madditional payload = 395 kg We don’t usually use hydrogen because it is flammable, of course — remember the Hindenburg? Page 5 of 7 PHY 171 (Winter 2012) Homework 9 solutions 6. What (minimum) force must you exert to lift a car of mass 1500 kg, if Aout = 1.6 m2 and Ain = 0.2 m2 ? 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 Pin = Pout Using the definition of pressure as the force per unit area perpendicular to the surface, we can therefore write Fin Fout = Ain Aout This gives Fin = Ain Aout = 0.2 m2 1.6 m2 Fout h i 1500 kg 9.80 m/s2 = 1.8 × 103 N Therefore, the minimum input force required to lift the car is 1.8 × 103 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 Ain ∆yin = Aout ∆yout Therefore, the required ∆yin = Aout Ain ∆yout = 1.6 0.2 0.25 m = 2 m Therefore, we will have to move the input piston through a distance of 2 m Page 6 of 7 PHY 171 (Winter 2012) Homework 9 solutions 7. 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, vD = 2.44 m/s, and asked to find the speed of the ambulance, vs . Use the speed of sound in air, vair = 340 m/s. The relation for the Doppler effect is f′=f vair ± vD vair ± vs 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 ′ f =f vair + vD vair + vs Substituting numbers 340 + 2.44 1590 = 1600 340 + vs so that or 1590 340 + vs = 1600 342.44 1590 vs = 547, 840 − 540, 600 which gives vs = 7240 = 4.55 m/s 1590 Therefore, the ambulance is moving at a speed of 4.55 m/s Page 7 of 7