14.1S Buoyancy, Ideal Fluid Flow solutions (1)

Physics/BioPhysics 150 Studio 14.1S: Buoyancy, Pressure and Fluid Flow i. Understand the relation between hydrostatic pressure in a fluid and depth ii. Calculate the buoyant force on an object in a fluid. iii. Use the equation of continuity to relate flow velocity to cross-sectional area . Warmup Problem A solid plastic sphere of density 350 kg/m 3 and radius 10.0 cm is attached by a string of negligible mass to the bottom of an aquarium filled with water as shown in the figure. The density of water is 1000 kg/m 3 . What is the tension in the string when the plastic sphere is half-submerged in the water? Be sure to: • Identify forces and draw a free-body diagram for the plastic sphere. • Write down the equation of motion for the sphere. • Solve the equation for the tension. (Note that all of the symbolic expressions below are useful in problem 1 and so should ideally be included in your solution in class.) First let’s identify the forces acting on the sphere. Since the ball isn’t moving, we know that the net force in the y direction must be 0. ∑𝐹𝐹 𝑦𝑦 = 0 = 𝐹𝐹 𝑏𝑏 − 𝑇𝑇 − 𝑚𝑚𝑚𝑚 We learned in class that the buoyant force is given by 𝐹𝐹 𝑏𝑏 = 𝜌𝜌 𝑤𝑤 𝑉𝑉 𝑤𝑤 𝑚𝑚 , where 𝜌𝜌 𝑤𝑤 is the density of the displaced fluid (here that’s water) and 𝑉𝑉 𝑤𝑤 is the volume of fluid that is displaced. Subsituting in for the buoyant force and solving for 𝑇𝑇 we find. 𝑇𝑇 = 𝜌𝜌 𝑤𝑤 𝑉𝑉 𝑠𝑠𝑠𝑠𝑏𝑏 𝑚𝑚 − 𝑚𝑚 𝑠𝑠 𝑚𝑚 𝐹𝐹 𝑏𝑏 𝑇𝑇 𝐹𝐹 𝑔𝑔

We don’t know the mass of the sphere, but we can find it using its density 𝑚𝑚 𝑠𝑠 = 𝜌𝜌 𝑠𝑠 𝑉𝑉 𝑠𝑠 . 𝑇𝑇 = 𝜌𝜌 𝑤𝑤 𝑉𝑉 𝑠𝑠𝑠𝑠𝑏𝑏 𝑚𝑚 − 𝜌𝜌 𝑠𝑠 𝑉𝑉 𝑠𝑠 𝑚𝑚 Since only half the ball is submerged, the volume of displaced fluid is half of the volume of the ball. With this last substitution we can go ahead and solve for the value of 𝑇𝑇 . 𝑇𝑇 = 𝜌𝜌 𝑤𝑤 � 𝑉𝑉 𝑠𝑠 2 � 𝑚𝑚 − 𝜌𝜌 𝑠𝑠 𝑉𝑉 𝑠𝑠 𝑚𝑚 = 𝑉𝑉 𝑠𝑠 � 𝜌𝜌 𝑤𝑤 2 − 𝜌𝜌 𝑠𝑠 � 𝑚𝑚 𝑇𝑇 = 6.16 N 1. In the warmup problem, would the tension in the string increase, decrease, or remain unchanged if each of the following changes were made? ( Hint : Using one of the symbolic expressions for the tension derived during the warmup, identify which variables are changing in each case, and then determine the effect of their change on the tension.) a. The water in the tank were replaced by heavy water (which has a higher density than water), while keeping the sphere exactly half-submerged. Let’s use our final expression for the tension, 𝑇𝑇 = 𝑉𝑉 𝑠𝑠 � 𝜌𝜌 𝑤𝑤 2 − 𝜌𝜌 𝑠𝑠 � 𝑚𝑚 . If we increase 𝜌𝜌 𝑤𝑤 , then our value of 𝑇𝑇 would also increase. b. The string were made longer, so that less than half of the sphere is submerged in the water. Again, for this part let’s use this expression from the warmup: 𝑇𝑇 = 𝜌𝜌 𝑤𝑤 𝑉𝑉 𝑠𝑠𝑠𝑠𝑏𝑏 𝑚𝑚 − 𝜌𝜌 𝑠𝑠 𝑉𝑉 𝑠𝑠 𝑚𝑚 . Lengthening the string and thereby decreasing 𝑉𝑉 𝑠𝑠𝑠𝑠𝑏𝑏 would thus decrease our value for 𝑇𝑇 . c. The mass of the sphere were increased, while keeping its volume constant and keeping it exactly half-submerged. For the mass of the sphere to increase without the volume changing, the density must have increased. We can see this from 𝑚𝑚 𝑠𝑠 = 𝜌𝜌 𝑠𝑠 𝑉𝑉 𝑠𝑠 . Now we can use our expression 𝑇𝑇 = 𝑉𝑉 𝑠𝑠 � 𝜌𝜌 𝑤𝑤 2 − 𝜌𝜌 𝑠𝑠 � 𝑚𝑚 to see that an increase in 𝜌𝜌 𝑠𝑠 results in a decrease in tension. d. The mass and volume of the sphere were increased, while keeping its density constant and keeping it exactly half-submerged.

See above to compare columns 1 and two at the line 𝑏𝑏 . Since line 𝑏𝑏 is in fluid A at the same height the pressure is the same in 1 and 2 at line b. In column 1, the pressure is 𝑝𝑝 𝑏𝑏 = 𝑝𝑝 𝑎𝑎𝑎𝑎𝑎𝑎 + 𝜌𝜌 𝐴𝐴 𝑚𝑚ℎ 1𝑏𝑏 . In column 2, the pressure difference is 𝑝𝑝 𝑏𝑏 = 𝑝𝑝 𝑎𝑎𝑎𝑎𝑎𝑎 + 𝜌𝜌 𝐵𝐵 𝑚𝑚ℎ 2𝑏𝑏 . So 𝜌𝜌 𝐴𝐴 ℎ 1𝑏𝑏 = 𝜌𝜌 𝐵𝐵 ℎ 2𝑏𝑏 and knowing that ℎ 2𝑏𝑏 > ℎ 1𝑏𝑏 → 𝜌𝜌 𝐴𝐴 > 𝜌𝜌 𝐵𝐵 . In other words, the greater height must have the lower density. Conclusion: A is denser than B. See above for comparison of columns 1 and 3 at the line a . Since line 𝑎𝑎 is in fluid A at the same height, the pressure is the same in 1 and 3 at line a . In column 1, the pressure is 𝑝𝑝 𝑎𝑎 = 𝑝𝑝 𝑎𝑎𝑎𝑎𝑎𝑎 + 𝜌𝜌 𝐴𝐴 𝑚𝑚ℎ 1𝑎𝑎 . In column 3, the pressure difference is 𝑝𝑝 𝑎𝑎 = 𝑝𝑝 𝑎𝑎𝑎𝑎𝑎𝑎 + 𝜌𝜌 𝐶𝐶 𝑚𝑚ℎ 3𝑎𝑎 . So 𝜌𝜌 𝐴𝐴 ℎ 1𝑎𝑎 = 𝜌𝜌 𝐵𝐵 ℎ 3𝑎𝑎 and knowing that ℎ 3𝑎𝑎 > ℎ 1𝑎𝑎 → 𝜌𝜌 𝐴𝐴 > 𝜌𝜌 𝐶𝐶 . Again, the greater height must have the lower density. Conclusion: A is denser than C. What we cannot definitely compare is the densities of fluids B and C, at least not without measuring the exact heights. Now let’s address the question about the pressure at x. ℎ 1𝑎𝑎 ℎ 2 ℎ 3𝑎𝑎 𝑎𝑎 ℎ 𝑏𝑏𝑏𝑏 𝑋𝑋 𝑏𝑏

Let’s say that the height from the 𝑏𝑏 line to the point X is ℎ 𝑏𝑏𝑏𝑏 . Ascending from the height of 𝑝𝑝 𝑏𝑏 and subtracting “ 𝜌𝜌𝑚𝑚ℎ ” in each column until we reach the height X gives the pressure at X. Column 1 ascends through fluid A to point X. 𝑝𝑝 1𝑥𝑥 = 𝑝𝑝 𝑏𝑏 − 𝜌𝜌 𝐴𝐴 𝑚𝑚ℎ 𝑏𝑏 Column 2 ascends through fluid B to point X. 𝑝𝑝 2𝑥𝑥 = 𝑝𝑝 𝑏𝑏 − 𝜌𝜌 𝐵𝐵 𝑚𝑚ℎ 𝑏𝑏 . Since the density of B is smaller, 𝑝𝑝 2 , 𝑏𝑏 > 𝑝𝑝 1 , 𝑥𝑥 Column 3 ascends through fluid A only to point X. 𝑝𝑝 3𝑥𝑥 = 𝑝𝑝 𝑏𝑏 − 𝜌𝜌 𝐴𝐴 𝑚𝑚ℎ 𝑏𝑏 = 𝑝𝑝 1 , 𝑥𝑥 Now let’s read the answer choices. I is true. II is definitely true: 𝑝𝑝 1 , 𝑏𝑏 < 𝑝𝑝 2 , 𝑥𝑥 . III is definitely false, 𝑝𝑝 3𝑥𝑥 = 𝑝𝑝 1 , 𝑥𝑥 . Answer C. 3. As blood flows out from the heart into the circulatory system, it first passes through the aorta, which then branches into arteries, which in turn branch into arterioles and then into capillaries, as indicated schematically in the diagram. The aorta in a typical person has a radius of about 1 cm (and there is only one aorta), whereas there are about 6 × 10 9 capillaries, each with a radius of about 5 × 10 −4 cm. a. If blood flows through the aorta with an average speed of 0.26 m/s, what is a typical flow speed in a capillary? ( Hint: What is the cross-sectional area of the aorta? The same blood that was in the aorta splits up and flows through many capillaries at once. What’s the total cross-sectional area of capillary that it is passing through at any instant?) We know that for that the blood flow should be constant throughout the body, that is the value of 𝑄𝑄 = 𝑣𝑣𝑣𝑣 does not change. Thus, 𝑄𝑄 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝑄𝑄 𝑐𝑐𝑎𝑎𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐𝑠𝑠 , and from this we find. 𝑣𝑣 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑣𝑣 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝑣𝑣 𝑐𝑐𝑎𝑎𝑐𝑐 , 𝑎𝑎𝑎𝑎𝑎𝑎 𝑣𝑣 𝑐𝑐𝑎𝑎𝑐𝑐 = ( 𝑁𝑁 𝑐𝑐𝑎𝑎𝑐𝑐 𝑣𝑣 𝑐𝑐𝑎𝑎𝑐𝑐 ) 𝑣𝑣 𝑐𝑐𝑎𝑎𝑐𝑐 Now we just solve for 𝑣𝑣 𝑐𝑐𝑎𝑎𝑐𝑐 , 𝑣𝑣 𝑐𝑐𝑎𝑎𝑐𝑐 = 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑣𝑣 𝑎𝑎𝑎𝑎𝑎𝑎 𝑁𝑁 𝑐𝑐𝑎𝑎𝑐𝑐 𝐴𝐴 𝑐𝑐𝑎𝑎𝑐𝑐 = 𝜋𝜋 ( 0 . 01 ) 2 ( 0 . 26 ) ( 6 × 10 9 ) 𝜋𝜋 ( 5 × 10 −6 ) 2 = 1.7 × 10 −4 m/s b. A typical capillary has length of 1 mm. How long does any one blood cell spend in a capillary? (Does this seem like a long or a short time to you? Note that a red blood cell has to be in a capillary long enough for most of the oxygen it’s carrying to diffuse into the surrounding tissue.)