(1) A all is floating in. If the density of the all is 0.95x10 kg/m, what percentage of the all is aove the? (2) An with a density of 19.x10 kg/m and a mass m50 kg is placed into a cylinder which contains a fluid. (a) If the fluid is, how much does the weigh when it is in the? () Suppose the fluid has a density of 1x10 kg/m. What is the apparent weight in this fluid? () An floats with 50% of its volume in fluid 1 while it floats with 75% of its volume in fluid 2. If fluid 1 is, what is the density of fluid 2? (4) The density of air is 1.29 kg/m. Gold has a density of 1x10 kg/m. Suppose you have a piece of gold which weighs 1 kg. What is the radius of a sphere which could e made from gold which would float in air? (5) Suppose helium has a density of 1.79x10-1 kg/m. A alloon with a volume 0.5m has a mass m50g. When the alloon is filled with helium, how much additional weight will the alloon e capale of lifting? (6) A cylindrical ody of volume v A :( a+ + c) and density 250 kg/m is floating in an oil- mixture as shown with a rod to stailize the cylinder. The oil layer has a thickness given y:η : 0 η 1. If the densities of the fluids are 1000 kg/m and oil 800 kg/m, what are the ratios a and c?
(1) A all is floating in. If the density of the all is 0.95x10 kg/m, what percentage of the all is aove the? Solution: The principle involved here is Archimedes principle which states that the uoyant force is equal to the weight of the fluid displaced. The will float if the uoyant force is equal to the weight of the. Thus: F wt. of fluid displaced g The second part is this: the floats if the uoyant force is equal to the weight of the.. Thus: Wt. of g And since it is floating, we must then have: g g We can thus find the fraction of the : We are, however, asked for the percentage of the aove the. This is given from: + Thus, The density of is find: not not not 1 1 kg 1 10 (you should rememer this value). Thus we m not. 95x10 1 1 0. 05 1x10 This is the fraction of the all which is aove the. In order to determine the percentage of the all aove the, we multiply y 100 which gives 5% aove the.
(2) An with a density of 19.x10 kg/m and a mass m50 kg is placed into a cylinder which contains a fluid. (a) If the fluid is, how much does the weigh when it is in the? () Suppose the fluid has a density of 1x10 kg/m. What is the apparent weight in this fluid? Solution: (a) Since the has a density higher than that of, the will not float in. Thus it sinks. When it sinks, it will still displace and there will e a uoyant force on the which decreases its weight elow the weight that the has in air. The change in weight is given y the uoyant force. Thus: ( weight) F We can find out the volume of the fluid displaced if we know the volume of the. This is given y: m 50kg m. x 2 59 10 m 19. x10 kg/m The change in weight is given y: F g 1x10 x 9. 8 x 2. 59x 10 25. 4 N fluid ( ) ( ) ( ) The weight of the is given y mg. Thus: weight mg 50 x 9. 8 490 N ( ) ( ) The apparent weight of the in is then 490N-25.4N465N. () This fluid is most likely something like mercury. We can repeat the calculation for this new fluid. The will still sink in this fluid since the density of the is greater than that of the fluid. The uoyant force is now given y: F g 1x10 x 9. 8 x 2. 59x10 0 N fluid ( ) ( ) ( ) which is the change in apparent weight of the. Thus the apparent weight of the in this new fluid will e 490N-0N160N. The has the same density as gold.
() An floats with 50% of its volume in fluid 1 while it floats with 75% of its volume in fluid 2. If fluid 1 is, what is the density of fluid 2? Solution: As in previous prolems, the principle involved here is Archimedes principle. Thus:, for fluid 1: The floats with F g.. 0 5 Thus, we can find the uoyant force at this point as: F 0. 5 g Now since the is floating, this is also equal to the weight of the, which is given y: We then have: weight g 0. 5 g g 0. 5 Now that we know the density of the, we ll e ale to find the density of fluid 2. We have that in fluid 2: 0. 75 Thus, the uoyant force at this point is given y: F 0. 75 fluid 2g And since the is floating, this will e equal to the weight of the. But the weight of the is given y Equating the two, we then find: weight g 0. 5 g 0. 75 g 0. 5 g 667 0. 5 kg fluid2 fluid2 075. m
(4) The density of air is 1.29 kg/m. Gold has a density of 19x10 kg/m. Suppose you have a piece of gold which weighs 1 kg. What is the radius of a sphere which could e made from this gold which would float in air? Solution: Although this prolem may e technically impractical, let s see if it is. The type of sphere which would need to e made from the gold is one which weighs 1 kg, ut is hollow and evacuated inside. Essentially, the average density of the sphere would have to e less than that of air. (This is the same idea that allows aircraft carriers to float on ). The density of the spherical would thus need to e the same as that of air. The mass is given then y: m We can thus find: Now, the volume of a sphere is given y air m 1kg air 1. 29kg/ m 0 775. m π 4 4 0 57 The interior of the sphere needs to e a vacuum. 4 (. ) sphere R R 0 775 π π. m The rest of this prolem is for your interest only. We can get an idea of how thin the gold shell would e for this sphere. The equation we need involves sutracting the volume of two spheres and equating this difference to the volume of gold involved. It s proaly easier to see the math here than to interpret the words. The volume of our gold is given y: mgold 1kg 5 gold gold x kg/ m 7 69 10 1 10. x m This volume of gold needs to e squished into a spherical shell of outer radius R 1 and inner radius R 2 where R 1 we ve already found to e 0.57m. Thus: 4 π R R We ll solve this for R 2 : ( 1 2) gold R R 0. 57 7. 69x 10 0. 56980 m 5 2 1 4π gold 4π A more precise value for R 1 is: 0.56982m The thickness of the gold film would e: 0.56982-0.569802x10-5 m. This film would need to hold ack the enormous pressure of the Earth s atmosphere. Gold won t stand up to this type of pressure at this thickness so the sphere proaly could not e made of gold. In fact, I am not aware of any material which would work here.
(5) Suppose helium has a density of He 1 kg 1. 79 10. A alloon with a volume m 0. 5m has a mass m50g. When the alloon is filled with helium, how much additional weight will the alloon e capale of lifting? Solution: The uoyant force is given y the weight of the fluid displaced. In this prolem, the displaced fluid is air with a density of 1.29 kg/m. The alloon has a mass of 50g or 0.05 kg. 0.5 m of helium has a mass of 1 kg m 1. 79x 10 0. 5m 0. 0895 kg ( )( ) helium helium alloon m The uoyant force is given y: kg m ( )( )( 2) F g 1. 29 0. 5m 9. 8 6. 21 N air alloon m s The weight of the alloon and helium is then: w + h ( 0. 05+ 0. 0895) 9. 8 1. 67 N The alloon can thus lift an additional weight of lift 6. 2N 1. 7N 4. 59 N This is aout ½ kg. The value is proaly larger than what will actually happen ecause helium in the alloon may not e pure. You might, however, want to do the same prolem with hydrogen (8.99x10-2 kg/m ) and think aout the Hindenurg. In any event, this does seem like quite a it of lift.
(6) A cylindrical ody of volume v A :( a+ + c) and density 250 kg/m is floating in an oil- mixture as shown with a rod to stailize the cylinder. The oil layer has a thickness given y:η : 0 η 1. If the densities of the fluids are 1000 kg/m and oil 800 kg/m, what are the ratios a and c? Solution: According to Archimedes principle the weight of the fluid displaced is equal to the uoyant force. et s assume the cylinder has a constant cross sectional area A. Then the uoyant force is given y: F aa g+ A g A a + g [ ] oil oil Since the is floating, the uoyant force is equal to the weight of the. Thus: F Ag The two forces are equal. Thus: A a + g Ag a + [ ] [ ] oil oil Now we know how this is related to the thickness of the oil layer (look at the wording of the prolem to see this). Thus: [ a ] {} + oil η It is now easy to solve this for a/: a oil a {} oil oil η η η η oil η It is also easy now to solve for c y replacing a in this expressing y --c: c c c oil c oil 1 1 η η 1 +η 1 It s reasonale to check these answers against: a + + c 1. a c oil oil + + η +η+ 1 +η 1 1 For the particular prolem at hand, we then have: a η 0. 25 0. 80η : 0. 75+ 0. 20 η oil a c As a limiting cases: suppose η 0 0. 25: 0. 75. Notice that there is a certain point at which this formulation of the prolem will not work ecause I have assumed that the displaces an amount of oil equal in volume to: ηa. The smallest value of a for which this prolem will work is then given y: a oil oil 250 η η η. oil 800 0 0 0 125 After this point, the floats completely in the oil regardless of how much deeper the oil layer goes. In fact, this is exactly the ratio / that would e otained if the floated only in oil. c