How can a boat made out of a material more dense than water float?

Worksheet for Exploration 14.1: Floating and Density How can a boat made out of a material more dense than water float? The block has a mass of 0.185 kg (position is given in centimeters). If this block is a cube, what is the density of the block? Note that since it is greater than water (1000 kg/m3) the block sinks as shown in the animation. Restart.

Block Density=_________

We reshape the block so that it has the same depth into the screen, but is wider and taller with walls that are 0.21-cm thick.

a. When the animation runs, what is the volume of water displaced (the dimension of the water container into the screen that you cannot see is 10 cm)?

i. First sketch a free body force diagram for forces acting on the reshaped box. (One of these is the buoyant force).

Displaced water:

Length=_______ Width=_______ Depth=_______

Volume displaced=_______

b. Using the density of water (1000 kg/m3), find the mass of the water displaced. Show that it is equal to the mass of the reshaped block. Thus, the block floats. i. Think about how mass, density and volume all relate.

mass of water=_______

c. Another way to think about this is that in its new shape the block has an effective density (total mass/total volume) less than that of the water. Divide the mass (0.185 kg) by the new volume to find the new effective density of the block.

Effective Block Density=_______

d. How does the effective density compare to the density of water?

The weight (mass*9.8 m/s2) of the water displaced (even if the displaced water leaves the container) is equal to the buoyant force on the block. In the case of a floating object, the buoyant force is equal to the weight of the floating object.

Worksheet for Exploration 14.2: Buoyant Force

When an object is put into a liquid, it experiences a buoyant force that is equal to the weight of the liquid the object displaces. The force on the wire is given as the block is slowly lowered into the liquid (position is given in centimeters and force is given in newtons). You can change the mass of the block between 0.125 kg and 0.375 kg and the density of the liquid between 500 kg/m3 and 1000 kg/m3. The object is in static equilibrium when the clock stops. Restart.

a. What is the weight of the block and the tension in the string when the block is in the liquid? Therefore, what is the value of the buoyant force? The buoyant force and the tension in the string (the force on the support wire) act upward and the weight acts down.

i. First sketch a free body force diagram for forces acting on the reshaped box. (One of these is the buoyant force).

Fg block submersed=_______

FTension=_______

b. What is the volume of the block in the liquid—either the submerged part of the block if the block is partially submerged when you paused it or the entire block if it is completely submerged (the dimension of the block that is into the screen is 5 cm)?

Length=_______ Width=_______ Depth=_______

Volume Block=_______

c. What is the volume of the water that is displaced by the block (the dimension of both water containers into the screen is 10 cm)? Verify that this is equal to the answer in (b).

Length=_______ Width=_______ Depth=_______

Volume displaced water=_______

d. What is the mass of the liquid displaced? What is the weight of the liquid displaced? Check that this is equal to the buoyant force.

mass liquid=_______

e. Pick two different masses and densities and verify that the buoyant force is equal to the weight of the water displaced.

i. mass=_______ Density=_______

Fbuoyant=_______ Fg=_______

ii. mass=_______ Density=_______

Fbuoyant=_______ Fg=_______

Worksheet for Exploration 14.3: Buoyancy and Oil on Water

This Exploration will address the buoyant force in more depth (pun intended). Specifically, what happens if we put an object in two “layers” of fluids? Assume the brown block is a cube (position is given in meters and pressure is given in pascals). Restart.

Note: The format of the pressure is written in short hand. For example atmospheric pressure, 1.01×105 Pa, is written as 1.01e+005.

Move the pressure indicator and measure the pressure at the bottom of the wooden block and at the top of the block.

a. If the block is a cube, what is the force on the block due to the water (buoyant force)?

i. First sketch a free body force diagram for forces acting on the reshaped box. (One of these is the buoyant force).

Fbuoyant=_______

b. What, then, is the weight of the block? What is the density of the block? (Use the symbol Fg as the force due to gravity, ie., weight).

Fg=_______

density=_______

c. Another method: How much (what percentage) of the block is submerged? Check that the density of the block is that same percentage of the density of water (1000 kg/m3). i. When you determine density use the mass of the block from part b.

%Submerged=_______

density=_________

Now consider what would happen if we put the block in an oil with a different density.

d. Predict what you expect will happen if we put the block in an oil with a density of 700 kg/m3.

e. Try it. Was your prediction correct? Explain.

f. What is the pressure at the bottom of the block and at the top of the block? What is the buoyant force on the block in the oil?

Pbottom=_______

Ptop=_______

i. To determine the buoyant force you can use the mass of the block from above, and the density of the oil.

Fbuoyant=_______

ii. OR you can use the pressure measurements at the top and bottom to help you determine the buoyant force on the block. The two results should agree.

Fbuoyant=_______

iii. How does the buoyant force compare to the weight of the object?

Now, suppose the wood block is put in a mixture of water on the bottom with oil on the top (the oil floats on the water and doesn’t mix with the water).

g. What do you expect will happen? Why?

h. Try it. Is more or less of the block submerged in water in this case compared with the block simply floating in water (without oil)? Why?

i. One way to look at what happened is to measure the pressures. Find the pressure at the bottom of the block and at the top of the block.

Pbottom=_______

Ptop=_______

j. What is the pressure difference and thus the TOTAL BUOYANT FORCE on the block?

∆P=_______

Fbuoyant=_______

k. In order for a block to float only in water (with air on top), to get the same pressure difference to support the block, why does the block need to be lower in the water? (Think about the density of air compared with the density of oil and, therefore, the change in pressure with depth in air and the oil.)

Another way to look at this is to compare the buoyant forces.

l. In comparison with the block floating in water only, has the buoyant force increased, decreased or stayed the same?

m. What is the volume of water that the block displaces? i. When in water/air.

Vwater displaced=_______

ii. When in water/oil

Vwater displaced=_______

n. What is the weight of that water? i. When in water/air.

Fg diplaced water (i)=_______

ii. When in water/oil.

Fg displaced water (ii)=_______

o. What is the volume of oil that the block displaces?

Voil displaced=_______

p. What is the weight of the displaced oil?

Fg displaced oil =_______

q. How do those two compare with the weight of the block? i. That is, compare the total weight of displaced material for each case.

Worksheet for Exploration 15.1: Blood Flow and the Continuity Equation

Blood flows from left to right in an artery with a partial blockage. A blood platelet is shown moving through the artery. How does the size of the constriction (variable from 1 mm to 8 mm from each wall) affect the speed of the blood flow?

Restart. Assume an ideal fluid (position is given in millimeters and pressure is given in torr = mm of Hg). We can use the continuity equation and Bernoulli’s equation to understand the motion:

Continuity: Av = constant Bernoulli: P + (1/2) ρv2 + ρgy = constant.

With a 2.0-mm constriction:

a. What is the platelet’s speed before and after it passes through the constriction?

vbefore=_______

vafter=_______

b. What is the platelet’s speed while it passes through the constriction?

vconstriction=_______

Set the constriction to 8.0 mm.

c. Does the speed of the platelet before it reaches the constriction increase, decrease, or not change?

d. With the 8-mm constriction, is the speed of the platelet in the constriction faster, slower, or the same as with the 2-mm constriction?

e. Assume that the blood vessel and the blockage are cylindrical (circular cross-sectional area for both). Measure the radius of the artery and the radius of the flow area where the blockage is. Verify the equation of continuity to compare the 2-mm and 8-mm cases.

Aout=_______ Ain=_______

Now compare the 2-mm and 8-mm cases.

f. What is the pressure inside of and outside the constriction (use the white box to measure pressure)?

Pin=_______ Pout=_______

g. Does the pressure decrease or increase in the region where the blockage is?

h. This result, (g), is surprising to many students so let’s figure out why: At the instant the platelet travels from the wide region to the narrower constricted region, what is the direction of acceleration? i. In addition to considering net acceleration of the platelet, also consider which side of the

platelet accelerates more (or first) as it enters the constriction region.

i. What, then, is the direction of the force that the platelet feels? i. For parts (i) and (h) you should consider what causes the force on the platelet. Sketch the

platelet and see if the force on the left side of it or right side of the platelet is greater.

ii. Also, as the platelet enters the constriction which side has a greater force acting on it? What does this do to the shape of the platelet (not shown in animation).

j. What region should have a larger pressure? i. To determine this you should carefully consider the answers above and note that the pressure

on the side of the platelet that has not yet entered the constriction remains constant. So what must happen on the constriction side to cause the acceleration you note?

k. Do the same analysis for the platelet as it leaves the constricted region and goes back to the unblocked artery (sketch a diagram to show the direction of acceleration and force).

l. Verify that Bernoulli’s equation holds inside and outside the constricted region for the 2-mm and 8- mm cases (760 Torr = 760 mm of Hg = 1.01 x 105 Pa). The density of blood is 1050 kg/m3.

Worksheet for Exploration 15.2: Bernoulli’s Equation

Bernoulli’s equation describes the conservation of energy in an ideal fluid system. Assume an ideal fluid (position is given in meters and pressure is given in pascals). The dark blue in the animation is a section of water as it flows into the region marked by the horizontal line and the corresponding water that must move out of the region in the top right. We will explore the connection between Bernoulli’s equation and conservation of energy. Restart.

Note: The format of the pressure is written in short hand. For example, atmospheric pressure, 1.01 x 105 Pa, is written as 1.01e+005.

The relationship between the speed and dimensions of the water going in compared with the water leaving is governed by the continuity equation (what flows in must flow out unless there is a leak in the pipes!): Av = constant, where A is the cross-sectional area and v is the speed of the liquid. Assume the pipes are cylindrical.

a. What is the volume of both (darker) blue regions (should be the same)?

Vleft=_______

Vright=_______

b. What is the speed of the water in the left pipe?

Speed left=_______

c. What is the cross-sectional area of the left pipe?

Aleft=_______

d. What is the speed of the water leaving the region (in the right pipe)?

Speed right=_______

e. What is the cross-sectional area of the right pipe?

Aright=_______

f. Does the continuity equation hold?

As the water travels through this pipe system, work must be done on the fluid to raise it up and to increase its speed. The work done must be equal to the change in kinetic plus potential energy.

The upper blue region is effectively the same as the lower blue region but at a later time (after traveling up the pipe). So when you make measurements on that region, you are determining how the properties of that mass of water change.

g. Given the pressure (you can move the red pressure indicators), find the force (from the water behind it) on the lower left dark blue region.

Pleft=_______

Fleft=_______

h. What is the work done by that force for the duration of the animation?

Wleft=_______

i. Similarly, find the force on the upper right dark blue region that opposes the motion.

Pright=_______ Fright=_______

j. What is the work done by that force for the duration of the animation (note that the displacement and force are in opposite directions, so this is negative work)?

Wright=_______

k. What is the net work done, then, during the duration of the animation on the water in the middle region? i. This really refers to the net work done to move the same volume of water that is in either dark

blue region from the bottom to the top.

Wnet=_______

l. Calculate the difference in kinetic energy of the dark blue regions. Note that since the volume is the same, the mass is the same. (The density of water is 1000 kg/m3.) i. This is the difference in kinetic energy between a dark blue volume of water at the low left

region and the kinetic energy of the same mass when it is up on the right.

∆KEright-left=_______

m. Calculate the difference in potential energy of the center of mass of the dark blue regions. Does the net work equal the change in kinetic energy plus the change in potential energy? i. Again this change is really the difference between the right and left dark blue masses of water.

∆PEright-left=_______

This is all described by Bernoulli’s equation.

n. Show that the net work is equal to (Pleft – Pright) Avt.

o. Show that the net change in kinetic energy is (1/2) ρAvt (vright 2 – vleft

2).

p. Show that the net change in potential energy is ρgAvt (yright – yleft).

P is the pressure, ρ is the density of the fluid, v is the speed of the fluid flow, A is the cross-sectional area, t is the time, and y is the height of the fluid. Combining these three terms, we get

(Pleft – Pright) = (1/2)ρ(vright 2 – vleft

2 ) + ρg(yright – yleft),

or Bernoulli’s equation, P + 1/2ρv2 + ρgy = constant, so that Bernoulli’s equation is simply another way to restate conservation of energy.

Worksheet for Exploration 15.3: Application of Bernoulli’s Equation

Adjust the height of the water in the reservoir and notice what happens to the water flow out of the opening. Assume an ideal fluid (position is given in meters). We can use Bernoulli’s equation (i.e., conservation of energy for fluids) to understand what happens, P + 1/2ρv2 + ρgy = constant, where P is the pressure, ρ is the density of the fluid, v is the speed of the fluid flow, and y is the height of the fluid (you can, of course, pick any point to be y = 0 m). Restart.

The amount of water leaking out is small during the animation. So the height effectively stays constant during the time this animation is running (to a good approximation).

a. Use Bernoulli’s equation to find the pressure at the bottom of the reservoir. Pick a height of water in the reservoir. The pressure above the water is atmospheric pressure (1.0 x 105 Pa). What is the pressure of the water at the bottom of the reservoir? (Note that for both of these cases, v = 0 m/s).

height=_______

Pbottom=_______

b. Use Bernoulli’s equation at the bottom of the reservoir to find the speed of water flow out of the reservoir. Equate Bernoulli’s equation somewhere in the middle of the bottom of the reservoir (where v = 0 m/s) to the water flowing out of the opening (where P is atmospheric pressure) and note that the value of y is the same for both cases.

vbottom=_______

c. Using this value for the initial x velocity of the water, calculate where the water will land and verify that it does land at the proper spot. Repeat this procedure for another value of the reservoir height.

First height=_________

Landing Coordinates=______________

Second height=_________

Landing Coordinates=______________