GP Chapter 5 - Solutions

Physics 131 – Spring 2024 Group Problems: Chapter 5 Learning Goals After working through this worksheet, students will be able to: 1. Identify which objects will float and which will sink. 2. Relate the magnitude of the buoyant force to the volume of fluid displaced by an object. 3. Analyze the forces on freely floating objects and relate volume displaced to ratio of the densities of the object and the fluid. 4. Analyze the forces on fully submerged objects 5. Analyze the motion of an object in a fluid while sinking under the influence of gravity. 6. Compute the pressure exerted by a fluid on an object. Problem 1: Floating and sinking You place a cork and a rock, both with the same size, in water: the first one floats, the second sinks. A. What can you conclude about the density 𝜌𝜌 𝑐𝑐 of cork in relation to the density 𝜌𝜌 𝑤𝑤 of water? Are they the same, is one greater than another, is there not enough information to tell, etc.? Since the cork floats, its density must be less than the density of the water. B. Similarly, what can you conclude about the density 𝜌𝜌 𝑟𝑟 of rock in relation to the density 𝜌𝜌 𝑤𝑤 of water? Since the rock sinks, its density must be greater than the density of water. For both the cork and the rock, there are two important forces acting on the object: the force of gravity on the object (pointing down) and a buoyant force (pointing up). C. The cork is floating stationary at the surface of the water. Draw a free-body diagram for the cork. How does the magnitude of the buoyant force compare with the weight of the cork? D. The rock sinks until it hits the bottom of the container, where it stays . Draw a free-body diagram for the rock. How does the magnitude of the buoyant force compare with the weight of the rock? Problem 2: Computing the buoyant force Consider three blocks of different sizes and densities. Block A floats partially submerged in water, block B floats fully submerged in water, and block C sinks . An object (wholly or partially) submerged in a fluid will feel a buoyant force, provided there is pulling the fluid downward. The magnitude of the buoyant force 𝐹𝐹 ⃗ 𝐵𝐵 from the fluid ( 𝑓𝑓 ) on the object ( 𝑜𝑜 ) is equal to the weight of the fluid displaced by the object. Since the weight of the displaced fluid is �𝐹𝐹 ⃗ 𝐺𝐺 � = 𝑚𝑚 𝑓𝑓 𝑔𝑔 , where 𝑚𝑚 𝑓𝑓 = 𝜌𝜌 𝑓𝑓 𝑉𝑉 𝑑𝑑 is the mass of the displaced fluid, 𝑉𝑉 𝑑𝑑 is the volume of the displaced fluid by the submerged object, and 𝜌𝜌 𝑓𝑓 is the density of the fluid, then: �𝐹𝐹 ⃗ 𝐵𝐵 � = 𝑚𝑚 𝑓𝑓 𝑔𝑔 = 𝜌𝜌 𝑓𝑓 𝑉𝑉 𝑑𝑑 𝑔𝑔 Since the cork is at rest, according to Newton’s 1 st law 𝐹𝐹 ⃗ net = 0 , therefore the buoyant force must have the same magnitude as the weight. 𝐹𝐹 ⃗ 𝐺𝐺 𝐸𝐸→𝑐𝑐 𝐹𝐹 ⃗ 𝐵𝐵 𝑤𝑤→𝑐𝑐 We use again Newton’s 1 st law, 𝐹𝐹 ⃗ net = 0 . In this case, there is also a normal force pointing vertically upward, that supports the rock at the bottom of the container. The upward forces (buoyant force plus normal force) must equal the downward forces (gravitational force), therefore the magnitude of the buoyant force alone must be smaller than the weight of the rock. 𝐹𝐹 ⃗ 𝐺𝐺 𝐸𝐸→𝑟𝑟 𝐹𝐹 ⃗ 𝐵𝐵 𝑤𝑤→𝑟𝑟 𝐹𝐹 ⃗ N 𝑆𝑆→𝑟𝑟

2 1. For which of these blocks is the displaced volume 𝑉𝑉 𝑑𝑑 less than the object’s volume 𝑉𝑉 𝑜𝑜 . 2. For which of these blocks is the displaced volume 𝑉𝑉 𝑑𝑑 equal to the object’s volume 𝑉𝑉 𝑜𝑜 . 3. Can an object ever have 𝑉𝑉 𝑑𝑑 > 𝑉𝑉 𝑜𝑜 ? Explain why or why not. B. Consider a block floating at rest on the surface of a lake. The block has a mass of 3 kg and a volume of 4000 c m 3 . Using the FBD similar to what you drew in Problem 1C , determine the total volume displaced by the block as it floats on the water’s surface. The density of water is 1000 kg/m 3 . 1. Compute the fraction of the block which is submerged 𝑉𝑉 𝑑𝑑 𝑉𝑉 𝑜𝑜 . 2. Calculate the density 𝜌𝜌 of the block and the ratio 𝜌𝜌 𝜌𝜌 𝑓𝑓 . Compare this ratio to that from part B1. C. Now consider a denser block which has sunk and is at rest on the bottom of the lake. It has a mass of 2500 kg and a volume of 1 m 3 . The density of water is 1000 kg/m 3 . Using the FBD similar to what you drew in Problem 1D , compute the magnitude of: 1. the buoyant force on the block. For the block that floats partially submerged. For the block that floats fully submerged or the one that In general, this is impossible for an object with uniform density: the biggest the displaced volume can be is equal to the volume of the object. However, there is one case in which this is possible – object that is partially hollow, like a boat. The boat displaces more water than its volume, but only if the boat is floating on the surface. Since the buoyant force and the weight are the only two forces in the vertical direction and must add to a net force of zero, we have. 𝑉𝑉 𝑑𝑑 𝑖𝑖𝑖𝑖 𝑙𝑙𝑙𝑙𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑎𝑎𝑎𝑎 𝑡𝑡ℎ𝑙𝑙 𝑏𝑏𝑙𝑙𝑜𝑜𝑏𝑏𝑘𝑘 ′ 𝑖𝑖 𝑣𝑣𝑜𝑜𝑙𝑙𝑣𝑣𝑚𝑚𝑙𝑙 𝑉𝑉 𝑂𝑂 ! 𝐹𝐹 𝐵𝐵𝐵𝐵 + 𝐹𝐹 𝐺𝐺𝐵𝐵 = 0 ⇒ 𝜌𝜌 𝑓𝑓 𝑉𝑉 𝑑𝑑 𝑔𝑔 − 𝑚𝑚𝑔𝑔 = 0 ⇒ 𝑉𝑉 𝑑𝑑 = 𝑚𝑚𝑔𝑔 𝜌𝜌 𝑓𝑓 𝑔𝑔 = 𝑚𝑚 𝜌𝜌 𝑓𝑓 = 3 kg 1000 kg m 3 = .003 m 3 𝑉𝑉 𝑜𝑜 = 4000 cm 3 × � 1 m 100 cm � 3 = 0.004 m 3 𝑉𝑉 𝑑𝑑 𝑉𝑉 𝑜𝑜 = 0.003 m 3 0.004 m 3 = 0.75 𝜌𝜌 = 𝑚𝑚 𝑉𝑉 = 3 kg . 004 m 3 = 750 kg/m 3 Therefore, 𝜌𝜌 𝜌𝜌 𝑓𝑓 = 0.75, which is the same as what you got for 𝑉𝑉 𝑑𝑑 𝑉𝑉 𝑜𝑜 above. This is not a coincidence but follows directly when a floating object is acted upon by only the buoyant force and the force of gravity. In this case, �𝐹𝐹 ⃗ 𝐵𝐵 � = �𝐹𝐹 ⃗ 𝐺𝐺 � and hence 𝜌𝜌 𝑓𝑓 𝑉𝑉 𝑑𝑑 𝑔𝑔 = 𝜌𝜌𝑉𝑉 𝑜𝑜 𝑔𝑔 . Cancelling 𝑔𝑔 on both sides and rearranging gives us 𝑉𝑉 𝑑𝑑 𝑉𝑉 𝑜𝑜 = 𝜌𝜌 𝜌𝜌 𝑓𝑓 . Since the rock is entirely submersed, 𝑉𝑉 𝑑𝑑 = 𝑉𝑉 = 1 m 3 (the rock’s volume). Using the formula for the buoyant force, we get: �𝐹𝐹 ⃗ 𝐵𝐵 � = 𝜌𝜌 𝑓𝑓 𝑉𝑉 𝑑𝑑 𝑔𝑔 = 1000 kg m 3 × 1 m 3 × 10 m s 2 = 10,000 N = 10 kN.

4 C. Calculate the magnitude of the drag force once the rock is at terminal speed, with and without including the buoyant force. Does it change much? For zero acceleration, Newton’s second law gives us 𝐹𝐹 𝑛𝑛𝑛𝑛𝑛𝑛 , 𝐵𝐵 = 0 ⇒ �𝐹𝐹 ⃗ 𝐵𝐵 � + �𝐹𝐹 ⃗ 𝑑𝑑𝑟𝑟𝑑𝑑𝑑𝑑 � − �𝐹𝐹 ⃗ 𝐺𝐺 � = 0 . Therefore: �𝐹𝐹 ⃗ 𝑑𝑑𝑟𝑟𝑑𝑑𝑑𝑑 � = �𝐹𝐹 ⃗ 𝐺𝐺 � − �𝐹𝐹 ⃗ 𝐵𝐵 � = 35 N − 0.024 N = 34.976 N If we had ignored the buoyant force, we would have obtained �𝐹𝐹 ⃗ 𝑑𝑑𝑟𝑟𝑑𝑑𝑑𝑑 � = 35 N . The two answers are the same to within 4 significant digits. Therefore, if the density of the fluid is much, much smaller than the density of the object, we can safely ignore the buoyant force. D. In this case the drag force is inertial , so �𝐹𝐹 ⃗ drag � = �𝐹𝐹 ⃗ 𝐷𝐷 � = 1 2 𝐶𝐶 𝐷𝐷 𝜌𝜌 𝑓𝑓 𝐴𝐴𝑣𝑣 2 . Assuming the rock has a drag coefficient 𝐶𝐶 𝐷𝐷 = 0.6, and cross-sectional area 𝐴𝐴 = 0 .02 m 2 , what is the rock’s terminal speed in m s ? From the drag-force equation above: �𝐹𝐹 ⃗ 𝐷𝐷 � = 1 2 𝐶𝐶 𝐷𝐷 𝜌𝜌 𝑓𝑓 𝐴𝐴𝑣𝑣 𝑛𝑛 2 ⇒ 𝑣𝑣 𝑛𝑛 = � 2�𝐹𝐹 ⃗ 𝐷𝐷 � 𝐶𝐶 𝐷𝐷 𝜌𝜌 𝑓𝑓 𝐴𝐴 𝑣𝑣 𝑛𝑛 = � 2(34.976 N) (0.6) × � 1.2 kg m 3 � × (0 .02 m 2 ) = 69.7 m s E. Does your answer above seem reasonable? Compare it with some speed you are familiar with (e.g. a car speeding on the highway or the speed of a baseball after a pitch). Recall: 1 m s is about 2.2 mph . The speed above is about 153 mph. That’s a high speed. For comparison, a speeding car may be going at 80-90 mph and the fastest baseball pitch recorded was 105 mph. It seems reasonable that a rock falling from a high place like the top of a skyscraper could end up going at these high speeds if the rock has enough time to accelerate to its terminal velocity. Check your answers with an instructor before continuing. Problem 5: Sedimentation of red blood cell through blood plasma (viscous drag force) Consider a red blood cell of mass 45 pg (p = p ico = 10 −12 ) and volume 40 μ m 3 ( 𝜇𝜇 = m ic r o = 10 −6 ) falling through blood plasma (of density 1025 kg m 3 ). A. First we need all the units given in S.I. units. The density is already in S.I. units, while the mass is: 𝑚𝑚 = 45 pg = 45 × 10 −12 g = 45 × 10 −15 kg Convert the volume into units of m 3 . Convert micrometers to meters and then cube it : 40 μ m 3 × � 1 m 1 × 10 6 𝜇𝜇m � 3 = 4 × 10 −17 m 3 B. Determine the magnitude of the buoyant force on the blood cell. �𝐹𝐹 ⃗ 𝐵𝐵 � = 𝜌𝜌 𝑓𝑓 𝑉𝑉 𝑑𝑑 𝑔𝑔 = � 1025 kg m 3 � × (4 × 10 −17 m 3 ) × � 10 m s 2 � = 4.1 × 10 −13 N = 0.41 pN Because the buoyant force due to the air is typically much smaller than the weight of an object, it can almost always be neglected. However, if the object has a very tiny weight (like a balloon filled with helium), the buoyant force from air must be considered.

5 C. How does the buoyant force compare with the weight of the blood cell? �𝐹𝐹 ⃗ 𝐺𝐺 � = 𝑚𝑚𝑔𝑔 = 4.5 × 10 −13 N = 0.45 pN . In this case, the buoyant force is only slightly smaller than the weight, because the density of the fluid is only slightly smaller than the density of the blood cell: 𝜌𝜌 = 𝑚𝑚 𝑉𝑉 = 1125 kg/m 3 compared to 𝜌𝜌 𝑓𝑓 = 1025 kg/m 3 D. Using Newton’s laws determine the magnitude of the drag force once the cell is at terminal speed. We again have �𝐹𝐹 ⃗ 𝐵𝐵 � + �𝐹𝐹 ⃗ 𝑑𝑑𝑟𝑟𝑑𝑑𝑑𝑑 � − �𝐹𝐹 ⃗ 𝐺𝐺 � = 0 . Therefore: �𝐹𝐹 ⃗ 𝑑𝑑𝑟𝑟𝑑𝑑𝑑𝑑 � = �𝐹𝐹 ⃗ 𝐺𝐺 � − �𝐹𝐹 ⃗ 𝐵𝐵 � = 0.45 pN − 0.41 pN = 0.04 pN = 4 × 10 −14 N E. In this case the drag force is viscous , so �𝐹𝐹 ⃗ drag � = �𝐹𝐹 ⃗ 𝑉𝑉 � = 6 𝜋𝜋𝜋𝜋𝜋𝜋𝑣𝑣 . If the cell has a radius 𝜋𝜋 = 3.5 μ m and the viscosity of blood is 𝜋𝜋 = 0.001 Pa ⋅ s = 0.001 N⋅s m 2 , what is the cell’s terminal speed (don’t forget to convert to S.I. units)? We can solve for the terminal velocity: �𝐹𝐹 ⃗ 𝑉𝑉 � = 6 𝜋𝜋𝜋𝜋𝜋𝜋𝑣𝑣 𝑛𝑛 ⇒ 𝑣𝑣 𝑛𝑛 = �𝐹𝐹 ⃗ 𝑉𝑉 � 6𝜋𝜋𝜋𝜋𝑟𝑟 𝑣𝑣 𝑛𝑛 = 4.09 × 10 −14 N 6 𝜋𝜋 × � 0.001 N ⋅ s m 2 � × (3.5 × 10 −6 m) = 6.06 × 10 −7 m s = 606 nm s F. At this terminal speed, how many hours would it take the cell to fall 5 cm (about the length of a test tube)? We know that speed is distance divided by time: 𝑣𝑣 = 𝐿𝐿 𝑡𝑡 ⇒ 𝑡𝑡 = 𝐿𝐿 𝑣𝑣 = 0.05 m 6.06 × 10 −7 m s = 82,508 s × � 1 hr 3600 s � = 22.9 hours Your result in part (F) above illustrates one of the reasons why we need a centrifuge to sediment cells to the bottom of a test-tube: spinning in a centrifuge increases the apparent weight of the cells.

7 Problem 7: Pressure For an object partially or entirely submersed in a fluid, it suffices to think about just the buoyant force. However, some objects are in contact with a fluid on a single side, and in such a case one needs to think about the pressure exerted by the fluid. Pressure is the magnitude of force per unit area: 𝑃𝑃 = �𝐹𝐹 ⃗ � 𝐴𝐴 A. Draw a free-body diagram for the piston. You can denote the pressure force from the gas by 𝐹𝐹 ⃗ 𝑝𝑝 . B. The weight of the piston is supported by the pressure force from the gas such that the piston is at rest. The cross-sectional area at the top and bottom of the piston is 80 cm 2 . Determine the pressure of the gas in S.I. units (the S.I. unit for pressure is the Pascal, Pa = N/m 2 ) Since the piston is at rest, the force from the gas pressure must exactly cancel the piston’s weight (i.e. the net force must add to zero, since 𝐹𝐹 ⃗ net = 𝑚𝑚𝑎𝑎 ⃗ = 0 �⃗ ). Hence �𝐹𝐹 ⃗ 𝑝𝑝 � = �𝐹𝐹 ⃗ 𝑑𝑑 � = 100 N Then 𝑃𝑃 = �𝐹𝐹 ⃗ 𝑝𝑝 � 𝐴𝐴 = 100 N 80 cm 2 × (100 cm) 2 (1 m) 2 = 12500 Pa C. Consider the same container and piston, but now moved to a room with normal atmosphere (pressure of 10 5 Pa ) outside the piston, instead of a vacuum. Redraw the free body diagram. Are there any additional forces on the piston that were not there in part (B)? Now the free-body diagram has another force – the pressure force from the atmosphere pushing down. D. What is the piston’s acceleration? From Newton’s second law in the y-direction, we have 𝑚𝑚𝑎𝑎 𝐵𝐵 = 𝐹𝐹 𝑛𝑛𝑛𝑛𝑛𝑛 , 𝐵𝐵 = 𝐹𝐹 𝑝𝑝𝐵𝐵 𝑑𝑑𝑑𝑑𝑔𝑔 + 𝐹𝐹 𝑝𝑝𝐵𝐵 𝑑𝑑𝑛𝑛𝑚𝑚𝑜𝑜 + 𝐹𝐹 𝐺𝐺𝐵𝐵 = 𝑃𝑃 𝑑𝑑𝑑𝑑𝑔𝑔 𝐴𝐴 − 𝑃𝑃 𝑑𝑑𝑛𝑛𝑚𝑚 𝐴𝐴 − 𝑚𝑚𝑔𝑔 = − 800 N So 𝑎𝑎 𝐵𝐵 = − 800 N 10 kg = − 80 m/s 2 Turns out that atmospheric pressure is strong (it works out to 14 lbs. for every square inch!) Gas Vacuum 𝐹𝐹 ⃗ 𝐺𝐺 𝐸𝐸→𝑃𝑃 𝐹𝐹 ⃗ 𝑝𝑝 𝐺𝐺𝑑𝑑𝑔𝑔→𝑃𝑃 𝐹𝐹 ⃗ 𝐺𝐺 𝐸𝐸→𝑃𝑃 𝐹𝐹 ⃗ 𝑝𝑝 𝐺𝐺𝑑𝑑𝑔𝑔→𝑃𝑃 𝐹𝐹 ⃗ 𝑝𝑝 𝑑𝑑𝑛𝑛𝑚𝑚𝑜𝑜→𝑃𝑃