Discussion

The usual textbook depiction of the rolling ball with friction opposing the forward velocity.	Friction the other way? Friction in the direction of the velocity.

Some textbooks say that a ball or cylinder rolling on an infinite plane must come to a stop because "it is slowed by friction". As usually defined in textbooks, the force due to friction lies in the horizontal plane. If the friction force is in a direction to reduce the forward velocity, as shown in the first picture, it is in a direction to increase the angular velocity. If the friction force is in a direction to slow the rotation, as shown in the second pictre, it is in a direction to increase the forward velocity. Niether of these is consistent with a reduction in the cylinder's velocity. Clearly friction alone cannot stop the rolling body.

The resolution to this apparent paradox is to realize that perfectly rigid, non-deformable, bodies aren't possible in nature. All bodies deform somewhat on contact. This gives rise to non-uniform size reaction forces over the deformed surface of contact (and normal to it) and there can be a net torque from them. These reaction forces adjust size and distribution to allow slowing without slipping.

The normal forces due to surface deformation.

These figures show two cases. In the first case the ball deforms, but the surface doesn't. Normal forces are greater in front of the rotation axis, giving a net counterclockwise torque, the correct direction to slow the angular roation. In the second case, both the ball and the surface deform, but again the result is a net counterclockwise torque. Also, these forces normal to the deformed surface have components in the plane, whose sum is in a direction opposite to the forward velocity.

In fact, these forces also act to determine the exact deformation of the surface, and everything can adjust to slow the rolling body without allowing slipping.

Sliding.

(1) But what about the block's momentum? What happened to that?

(2) And what about the block's angular momentum? It doesn't have any, you say. Well, about its own center of mass it surely isn't rotating during this experiment. But why not? The only forces on the block are the forces due to friction at its bottom, the gravitational force, and the upward reaction force the table exerts on the block. The net (total) of the gravitational force and the reaction force must be zero, since the block isn't accelerating up or down. The force due to friction acts in the plane of the table and therefore has a lever arm about the center of mass, and therefore must exert a torque on the block about the center of mass. So why doesn't this set the block into rotation? There must be a counter-torque of equal size and opposite direction to the friction torque; but what is it?

Solutions

(1) Momentum of a system is conserved. The horizontal momentum lost by the block is given to the table through the continuous impulse of the force due to friction at the table-block interface.

(2) This one is more interesting. Take an axis through the block's center of mass, a distance H above the table, where H is the height of the block's center of mass (C) above the table. The force due to friction f supplies a clockwise torque fH about this axis. The gravitational force is balanced about the center of mass, and gives no net torque. We conclude that the normal forces along the bottom surface of the block cannot be balanced about the center of mass, but must be greater at the front of the block, providing the necessary counterclockwise torque to balance the torque due to friction, and resulting in no appreciable rotation of the block about its center of mass.

See the next item for an application of this insight.

Braking automobile.

Solution

When the automobile brakes, you notice that the front dips down. The front springs compress more than the rear ones, and the front tires do also. This indicates there's more vertical force on the front axle than on the rear.

The reason this happens is that the force of friction of the roadway on the tires produces a clockwise torque about the car's center of mass (see picture below). The vertical normal forces on the tires must adjust in size to produce an equal and opposite counterclockwise torque.

In the light of the previous problem of the sliding cube, the reason is clear. The center of mass of the vehicle is between the front and rear axles, and above them. Therefore the force due to friction between roadway and tires provides a torque that initiates a rotation, causing the downward dip of the front of the vehicle. This compresses the front springs more, causing the normal force to be greater at the front than at the rear. This torque is opposite to the torque due to friction, The forces on the tires due to friction f₁ and f₂ at the roadway are linked to the normal forces N₁ and N₂ through the relation f = µN where µ is the coefficient of friction. So, since N₂ is greater on the front tires than is N₂ on the rear tires, the friction at the roadway is also greater on the front tires. The brake drums or disks in the front also have an increased force due to friction, hence they must be more rugged.

That's the way the ball bounces.

This tells us that the ball's speed just before collision must be equal to its speed just after the collision. Considering momentum, we conclude that if the speed before collision was v, then the change of momentum was 2mv, since momentum is a vector quantity.

So we conclude that the collision changed the ball's kinetic energy hardly at all. Yet the ball's momentum was changed greatly. How can this be? Using conservation of energy and conservation of momentum we conclude that the floor gained no energy, but it gained momentum of 2mv. How can anything gain momentum but not energy?

Answer: You can't do work unless you move.

An important principle is often slighted in textbooks. The books tell you that work is force times displacement, but they seldom give examples to reinforce that and use it in an insightful way. The insight can be put this way: Work is not done on or by a body unless that body undergoes a displacement. And a displacement is indicated by the fact that the center of mass of the body moves to new position.

Now, we are talking about macroscopic mechanical work here, not about microscopic work on individual atoms and molecules, which can convert mechanical work to thermal energy.

An elastic impact is one in which no mecanical energy is lost, which means none is converted to thermal energy. An elastic impact with a very massive body causes very tiny displacement of that body's center of mass. An elastic impact with an infinitely massive body causes no displacement of that body's center of mass. Therefore an elastic collision of a body with one of infinite mass does no work on the infinite mass, and likewise, the infinite mass does no work on the smaller body. Therefore the smaller body's mechanical energy is conserved.

If this seems too "pat", you may be thinking that surely during the elastic deformation work is done at the deformed interface, and Newton's third law must apply there. True. But the amount of work done on the ball during the compression is exactly equal in size to the work done by it during expansion. Ditto for the infinite mass. All is right with the conservation laws.

Answer. The skater crashes into the pillar with the same speed she had when she grabbed the rope. And her velocity at the time she contacts the pillar is perpendicular to the pillar's surface. The rope does no work on the skater. How can we know this without complicated analysis?

The rope has negligible mass compared to the skater, so we treat it as massless. The force the pillar exerts on the rope does no work, for the portion of rope wrapped around the pillar doesn't have any component of motion along the rope's length. Therefore the other end of the rope does no work on the skater. Since no work is done on the skater, the skater's kinetic energy, and speed, remain constant.

It follows that at all times the rope is perpendicular to the skater's velocity. That's true, though not obvious, and not particularly easy to show directly. It's probably a calculus problem. But you can get a lot of insight by taking a round jar lid, fastening a string to it, and placing it on a sheet of paper. Put a loop in the free end, and with a pencil in that loop hold the string taut and draw the path on the paper as you swing around the round jar lid. The path will end up with the curve closing in on the lid, and the path will be perpendicular to the lid's rim where it hits. This picture was made in this way.

Consequences to ponder. The skater's speed remains constant. What about her angular velocity and angular momentum?

Answer: Taking the fixed point at the center of the pillar as a center of torques, we see that the rope's tension causes a torque on the pillar, giving the pillar and the Earth an angular momentum in the same sense as the skater's angular velocity. Therefore the skater's angular velocity and angular momentum decrease. (Her angular momentum reaches zero when she impacts the pillar.)

Another way to look at this: The rope exerts a torque on the skater. The lever arm of this force about the pillar's center is just the radius of the pillar, R. At any instant the torque, R X F is a vector opposite in direction to the skater's angular momentum, so this must decrease her angular momentum. [See Sutton, M-186]

At the moment of her impact with the pillar, her angular momentum has become zero and she impacts with velocity that is directly toward the center of the pillar.

One reason people sometimes get off on the wrong track when analyzing this problem is their previous familiarity with problems of central force motion, where the force is always directed toward a fixed point. The force of tension on the skater with rope around the pillar is not directed toward a fixed point. For an example of central force motion, consider a skater on a frictionless rink, holding a rope that is fastened at a fixed point. Now, as time goes on, the rope length remains constant, and the skater's motion is constant speed around a circle. Now suppose, due to some clever mechanism, the rope is continually shortened, perhaps by having it pass over a pulley at the center of the rink, and then to a motor winch. But we take care that the rope at the center always passes through a fixed point there. [One of these days maybe I'll do a diagram.] By insisting that the rope pass through a fixed point, we ensure that there's no torque on it about that point. Now the angular momentum must be constant, as it always is in central force motion. So as the radius decreases, the skater's speed must increase as she spirals toward the center. This tells us that work is being done on the skater. The skater's velocity is no longer perpendicular to the rope, but has a small radial component. This means there's a radial displacement toward the center, and the radial force is responsible for that displacement. That's where the work is done on the skater—work done by the motor that drives the winch that's shortening the rope.

One might say that the central force circular motion case is the limit of the pillar case as the pillar radius goes to zero.

Follow-up question. Would you get the same result if the pillar wasn't round in cross section?

Answer: Yes. The analysis above in no way depended on the pillar being circular in cross-section. The next problem "Rope and Capstan" gives some additional insight into that.

Follow-up to the follow-up question. What if the pillar were square in cross section with slightly rounded corners. How does the time it takes to crash compare with the case of the round pillar? That time isn't hard to calculate in the square pillar case, since the speed is constant.

Rope and Capstan It is a well-known result in engineering statics that the tensions T_A and T_B at the ends of a rope or belt wrapped round a capstan are related by

T_B = T_A e ^{µ ß}

where µ is the coefficient of friction between rope and capstan, and ß is the angle of wrap.

This is generally derived for a cylindrical capstan. Prove that it also applies to other smooth cross-sectional shapes. Solution left for the reader.

The case of the skater problem with friction is much more involved. An excellent discussion of this problem can be found in:

Arons, Arnold. "An F=ma Analysis of the Spinning Skater and Decaying Satellite Orbit." The Physics Teacher, 37, 3 (March 1999), p. 154-160.

Mixtures

Liquid Transfer

Now, both beakers contain a mixture of water and alcohol, and both beakers contain 1 liter of liquid. Is there now more alcohol in the water than water in the alcohol?

Answer:

This is a classic example of a problem in which more information is given than is needed. It's also an example of insightful approaches versus detailed mathematical approaches. Many people will get out their calculators and start calculating mixture ratios with the numbers given.

It doesn't matter how much is initially in the beakers; they need not have equal volumes. It doesn't matter how much liquid is transferred each time. It isn't necessary to mix the liquid after a transfer. It doesn't even matter how many transfers are made, so long as each beaker ends up with the same volume it had initially.

And the result is... The water beaker will end up with as much alcohol as the alcohol beaker has water. The reasoning is simple. The water beaker ends up with some alcohol in it. That alcohol must have come from the alcohol beaker. Since the alcohol beaker now has as much liquid as it had initially, the amount of alcohol it lost has been replaced by that same amount of water. Q.E.D.

Martin Gardner discusses this in his chapter "Mathematical Card Tricks" in Hexaflexagons and Other Mathematical Diversions, a reprint of The First Scientific American Book of Mathematical Puzzles and Games, U. of Chicago Press, 1959, 1988.

The same principle may be applied to containers of marbles, beans, decks of cards, poker chips, coins, etc. I prefer this problem expressed in numbers of things rather than volumes. In the alcohol and water case, volumes are not preserved when the two materials are mixed, due to molecular interactions. This has been ignored in the answer given above. With discrete objects that problem doesn't arise.

With that in mind, here's a variation designed to baffle even physics students. Obtain some dry beans and some dry rice. Find a coarse screen or strainer that the rice can pass through but the beans can't. You'll also need three glass beakers or similar containers of equal weight and a beam balance for comparing their weights. Put a beaker on each balance pan, then add beans to one and rice to the other till they weigh the same. Now pour some beans from the bean container to the rice container. Shake or stir the mixture, and pour some back into the bean continer. Repeat this a few times, sometimes transfering a little (a spoonful), sometimes a lot (pour it from one container to the other).

Now ask someone in the audience to choose beans or rice. (This is a diversion that doesn't matter one bit.) Suppose they choose rice. Then ask them to choose one of the containers on the scales. (It doesn't matter which they choose, or we wouldn't ask.) Now transfer material from the heavier to the lighter container, pretending to very carefully select beans or rice, but actually it doesn't matter. (That's showmanship.) Do this until the two containers weigh the same. Announce that you have now made the rice (chosen by the audience) in the chosen container weigh exactly as much as the beans in the other container. You could even string the audience along by claiming that you could always judge the number of marbles in a container at carnivals, and win the prize. (That's irrelevant anyway, for we are using weight here, not numbers of things.)

Pour the contents of the chosen container through the sieve letting the rice fall into the empty beaker, and replace that beaker of beans on the scale. Now take the other container, with mixed beans and rice, and strain out the rice, replacing the remaining beans and container on the scale. The weights balance. It's a miracle! Finally, assign the students the task of explaining why it works every time.

Hydrostatics

Floating Test Tube

An inverted glass test tube floats in water if there's enough air trapped inside. Obtain a large jar, half-full of water, and float an inverted test tube in the water with just enough water rising in the tube so that the tube floats as shown. Close the top of the jar with a flexible rubber sheet so it makes an airtight seal.

What will happen if we press the rubber sheet inward so that the volume inside the jar is reduced?

1. The test tube will rise up a bit with respect to the water level inside.
2. The test tube will sink a bit with respect to the water level inside.
3. The test tube will neither rise nor sink with respect to the water level inside.

Answer: Some books invoke Archimedes' principle, and say that the weight of the test tube including the water in it increases because water is forced up into it, making the average density of the tube (including its air and water) greater. This is true, but it seems a rather unsatisfying explanation in this case, since you seem to be arbitrarily defining the tube to have a "boundary" at its open end. You could take the boundary to include the glass and the air only, for a more plausible "explanation" based on the increase of density of the air trapped in it.

It helps to look closely at details and consider Pascal's law. If we adjust the amount of air in our inverted test tube so that the tube just barely floats, then it behaves in the same manner as the ancient toy called the "Cartesian diver". Slight added pressure on the walls of the outer container causes the test tube to sink all the way to the bottom. Discussions of this device usually don't explicitly consider the air in the top of the container. Does that air make a difference?

Pressing on the sides of the container reduces the total volume inside. Water doesn't compress much; it's the air inside the container that compresses to smaller volume. The air trapped in the test tube compresses, and, by the gas laws, its pressure increases. But the pressure of the air in the top of the container increases also. Pascal's principle tells us that the change of pressure is the same at all points in the liquid. Pascal's principle also applies to the air in the top of the container, and to the air trapped in the test-tube. We must now determine whether these pressure changes result in a net downward, or upward, force on the glass test-tube.

Suppose the pressure in the top of the container increases by amount P. It exerts a downward force on the glass. The pressure of the air trapped in the tube increases by the same amount. It exerts an upward force on the glass. The force due to gravity on the glass test tube is the same as before. Nothing else can exert a vertical component on the glass test tube. Therefore the net force on the test tube is still zero.

From this we conclude that the test tube is still in equilibrium after the pressure change. But do not conclude that it didn't move up or down during the pressure change! Remember that pressure doesn't change instantaneously throughout the container.

The pressure of the air trapped inside the tube has increased. This causes that air to occupy a smaller volume (from the Gas law). But the pressure difference between the water level outside and inside has increased by the same amount. That pressure change is due to the pressure change increasing everywhere in the water by the same amount (Pascal's principle). The height difference between the water level outside and inside the test tube is unchanged!

To see this crucial point, it's helpful to write an equation. The net force on the glass tube is the vector sum of the downward force due to pressure of air above, the upward force due to pressure of air below, and the weight of the glass. We can write an equation for the pressure at the water surface within the test tube.

P _a + dgh₁ = P_b + W/A

where P_a is the initial pressure in the top of the container, P_b is the initial pressure of the air trapped in the test tube, d is the density of water, g is the acceleration due to gravity, h₁ is the initial height difference of the water levels, W is the weight of the test tube, and A is the cross-sectional area of the inside of the test tube.

After the pressure change the two pressure terms have been increased by the same amount. W/A doesn't change. Therefore the dgh term must remain the same, and we conclude that the height between the water levels does not change.

Therefore, the test tube must have moved downward during the pressure change. Does anyone care to analyze the time-sequence of pressure and force changes during that brief time interval?

We conclude that the presence of air above the liquid in a Cartesian diver will not compromise its performance.

Hydrostatic Puzzles

• (a) Does the weight registered on the balance change?
• (b) Does the pressure on the bottom of the container change?

2. A glass beaker of water is on the balance. A metal ball is hung from the edge of the beaker but outside the beaker. The weight of everything is recorded. The ball is now put inside the beaker hanging from the string, not touching the bottom.

• (a) Does the weight registered on the balance change?
• (b) Does the pressure on the bottom of the container change?

3. Repeat experiment 2 with a cork, first outside the beaker on the balance pan, then floating in the water. 4. Do a free-body analysis of the ball and the cylinder in problems 2 and 3, accounting for everything.

Answers:

Cylinder: Force P downward on bottom due to liquid pressure at bottom. Force N upward, due to the balance pan. Force T downward, due to tension in string holding the ball.

Ball: In the first case: Force T upward, due to string tension. Weight W downward. In second case one also has an upward buoyant force B on the ball equal to the weight of the displaced liquid.

T decreases by amount B to maintain the ball in equilibrium. Therefore, since N is constant, P must increase by that amount.

Sinking Boat

Answer:

The floating boat displaces its own weight of water (Archimedes' principle). When it is sunk it displaces its own volume of water (displacement principle). Its own weight of water is larger than its own volume of water, since its density is greater than that of water (after all, it did sink). Therefore, after sinking, the boat displaces less water, so the water level in the beaker drops. Therefore the pressure of the water on the bottom drops. The force of the boat sitting on the bottom compensates exactly for reduction in force due to pressure, so the total force on the bottom of the beaker doesn't change. After all, if the beaker were sitting on a pan of a balance, we wouldn't expect the balance reading to change.

Alternatively one could argue: The weight of the beaker and contents doesn't change. There's no vertical component of force due to the liquid pressure on the side walls of the beaker, only on the bottom. When the boat floats, the force on the bottom of the beaker is entirely due to liquid pressure. After the boat sinks, the force on the bottom is partly due to the weight of the boat sitting on the bottom. The net force on the bottom must be the same in both cases, since the total weight of the system didn't change. Therefore the pressure of the liquid must have decreased.

Physics on the Lake 1

You are in a boat on a perfectly calm lake. There's an anchor in the boat. You drop the anchor overboard and it sinks to the bottom of the lake. During this process, does the level of the lake rise, fall, or stay the same?

Answer:

We assume that there's a water level gauge of great precision somewhere, perhaps at the boat dock, to measure the lake level. This puzzle, and the next two puzzles, use Archimedes' principle: "The Buoyant force on an object submerged or floating is equal to the weight of the displaced liquid." They also use Newton's law: "The net force on a body in equilibrium is zero."

The anchor in the boat adds to the boat's weight, and the presence of the anchor there causes a displacement of an amount of water that weighs as much as the anchor. (The upward buoyant force due to the displaced water has the same size as the weight of the anchor, by Newton's law.) I.e., the anchor in the boat displaces a weight of water equal to its own weight. When the anchor is at the bottom of the lake it displaces an amount of water equal to its own volume. Its weight is greater than the weight of an equivalent volume of water (that's why it sank), so it displaces more water when in the boat than when at the bottom. The water level of the lake drops when the anchor is thrown overboard.

Physics on the Lake 2

Answer:

The floating log displaces its own weight of water. The log in the boat displaces its own weight of water. The level of the lake is unchanged.

Physics on the Lake 3

Answer:

The log in the boat displaces its own weight of water. The log at the bottom displaces its own volume of water. Its weight of water is less than its volume of water (that's why it normally floats), so the level of the lake rises. The anchor was at the bottom before and also after, so it causes no change in the lake level.

Milk Bottle

Note: Many materials, when mixed, occupy a volume different than their total volume when separate. This is generally a small effect. In this problem, this volume difference will be ignored. It would, in fact, have negligible contribution to the changes of pressure being considered here.

Answer:

The total force that the bottom of the bottle exerts on the table stays the same. After all, the total weight of bottle and contents hasn't changed. However, the pressure (force/area) of the milk on the bottom of the bottle decreases. Why?

First, we assume that a clever person made up this question, probably a physicist. So the wording of the problem should be examined with care. Why a milk bottle and not a beaker? Aha!, the shape of the bottle must play a role here. This suspicion is reinforced by the reference to the cream being only in the neck of the bottle.

Some students think that the problem has only to do with the liquid, forgetting that the liquid is in a container. They also imagine the bottle sitting on a scale, and since no mass has changed, the scale should read the same. That's correct. But then they incorrectly identify the scale reading with the pressure.

Some of you may not have ready access to Arnold B. Arons' A Guide to Introductory Physics Teaching, so I excerpt out the relevant parts of his discussion of this problem, in Section 11.3 HYDROSTATIC PRESSURE.

Understanding the concept of "hydrostatic pressure" involves a good bit more than acquiring the definition "force per unit area." A major step toward understanding resides in appreciation of the full significance of Pascal's Law: that the pressure at any point in a fluid is the same in all directions. The usual formal "proof" of this idea is given by examining the static equilibrium of a small volume of fluid, for example, one having a parallelepiped shape and triangular cross section. This treatment is so abstract that, even if it is presented, very few students assimilate its physical implications. Many introductory courses now eschew such treatments entirely in order to save time for other subjects. This is a legitimate choice, but teachers must then remember that something very subtle and fundamental has been left out and must be prepared to help close the gap at some subsequent point.

The subtlety of the insight, and the fact that many individuals (including many active physicists) have not really acquired it, are indicated by the responses given to the following question: A container of the shape in Fig. 11.3.1 initially contains a uniformly dispersed mixture (or colloidal suspension) of two immiscible liquids of different densities (e.g., oil in water; cream in milk). As time goes by, the lower density fluid separates and collects in the throat of the container. How does the final pressure on the bottom of the container (after the separation is complete) compare with the initial pressure (when the fluids were uniformly dispersed)? Is it the same, greater, or smaller than the initial pressure?

Novices tend to say they do not know, but the majority of those who have had some physics but have never thought about such manifestations of fluid pressure tend to say that the pressure must remain the same. Those who suggest a line of reasoning rather than just making an intuitive guess say that, since the weight of the fluid is unchanged, the total force on the bottom, and therefore the pressure on the bottom, must remain unchanged.

This reasoning is incorrect because the pressure on the bottom of a container of any shape other than purely cylindrical is not equal to the weight of fluid divided by the area of the bottom. Since the fluid pressure is the same in all directions at any point, the fluid at a wall exerts a force on the wall, and the wall, in turn, exerts a force on the fluid. The sloping walls of the container exert a downward component of force that influences the pressure on the bottom of the container. Since, on separation, the average density of the fluid in the central column is less than it was initially, the pressure at the bottom of the central column has decreased. The pressure at the level of any point along the sloping walls has also decreased; the downward force exerted by the wall has decreased; the pressure has decreased all over the bottom.

This is a problem in which the answer is non-intuitive for most people. It is also a problem where ill-formed or wrong concepts can lead to a persuasive argument that is simply wrong.

Arons explicitly uses just part of Pascal's principle: the fact that pressure at a point acts equally in all directions. The other part, properly stated deals with changes in pressure:

When the pressure is changed (increased or decreased) at one point in a homogenous, incompressible fluid, all other points experience the same pressure change.

To answer this one must consider the free body diagram of the glass bottle. We are particularly interested in the vertical force components. There's the bottle's own weight, which doesn't change. There's the force of the fluid on the bottle's sloping shoulder. This has a vertical component. The horizontal components of pressure on the neck don't concern us, and also, they add to zero before and also after the cream separates. The vertical components of pressure on the neck add to an upward force that does change; it decreases as the cream separates. This exactly equals the size of the decrease of downward force of the fluid on the bottom of the bottle, since the glass bottle remains in equilibrium.

Why does the upward force component on the bottle neck decrease as the cream separates? Before separation the pressure at the neck was due to the milk-cream mixture above it. After separation, the liquid above had a greater ratio of cream to milk, and therefore a smaller average density.

Balance Puzzle.

1. The balance indicator will read zero. The scale is balanced.
2. The left pan (with the steel ball) moves down.
3. The left pan rises.

This diabolical problem has all sorts of distractions, tempting you to do it "the hard way". Let's look for the easy way. Break it down into logical steps.

1. The containers are identical, so their weights are equal.
2. The water volumes are equal, so their weights are equal.
3. The steel and plastic balls have equal volumes, so they displace the same amount of water.
4. Therefore the water rises to equal heights in both containers (with respect to their bottoms).
5. Therefore the water pressure on the bottoms of the two containers is equal.
6. Since the containers are identical in shape, those equal pressures cause equal net force on the bottom of the containers.
7. So far in our tally, the net force on each of the two balance pans is the same. But...
8. There's the additional upward force due to tension of the string that keeps the plastic ball submerged. This string is fastened to the bottom of the container.
9. Therefore the right pan (with the plastic ball) rises.
10. Therefore the left pan (with the steel ball) lowers.

One point easy to miss is the important principle that when tallying the net force on an object, we must include only things that touch it, or that exert forces on it through fields, such as gravitational attraction. Gravitational forces between the parts of this system are negligible, so we only tally contact forces. The plastic ball does not contact the container, so resist any temptation to include the plastic ball's weight as adding to that of the pan. The plastic ball's effect on the system is accounted for by the tension it produces in the string.

Quite a few variations may be concocted to perplex students. Actually some of these are easier.

• Two balls of equal volume but different density, both suspended from the ceiling and completely immersed in the water.
• Two identical hollow balls, one floating on the water surface, one tethered to the bottom and completely submerged.
• Two identical steel balls, one in the bottom of one can, the other suspended from the ceiling, completely immersed in the water in the other can.

Optics

Mirror Reversal?

The figure below shows that a pair of plane mirrors at right angles behaves differently than does a single plane mirror. Have two mirrors hinged (for ease of storage) with a precise arrangement to hold them at perfect right angle alignment for the demo. Curved cylindrical mirrors can be made by forcing thin plastic mirrors into a slight curve. This may not seem like physics, but it is certainly a good exercise in geometric thinking.

Answer:

The plane mirror doesn't reverse up/down, all agree. It doesn't reverse left/right either. If you point a finger up, the image finger points up. If you point a finger to your right, the image points its finger to your right. When you keep all right/left references consistently in a fixed reference frame there's no problem. But there's more that's worth investigating here.

There's considerable literature on this question, some of it misses the mark. A good summary can be found in: Martin Gardner's The New Ambidextrous Universe, Freeman, 1990 ... 1979, pp. 3-6 and 19-22. Much of the confusion rests on three things:

1. The ambiguity of the terms up/down and left/right.
2. Other semantic difficulties.
3. Shifting reference coordinates during the discussion.
4. Failure to explain the matter and still address the fact that the plane mirror is symmetric about any of its normals.

What I've forgotten is that my clone had to rotate 180 degrees about a vertical axis after leaving my body and then walking around to face me. That is a reversal of the clone's left/right coordinate reference. (It also reverses forward/backward, with respect to a line from my nose perpendicular to the mirror, a fact whose importance will be clearer in explanation 2.)

Explanation 2. The plane mirror, having symmetry about any normal, can be rotated about that axis without any change in the image. For simplicity, imagine the axis from your nose and normal to the mirror. All image transformations must be symmetric with respect to that axis. The mirror does not invert up/down or right/left. It inverts along the mirror's axis. If you mentally walked forward through the mirror, till your right ear was at the position of the image's right ear, and your left ear at its left ear, your nose would be at the back of its head, and the back of its head would be at your nose. The top of your head would be at the top of the image's head, and both are still "up." Your reference for right/left and up/down hasn't changed, for you walked along the symmetry axis of the mirror. The mirror inverts the image along its symmetry axis, inverting the nose to the back of the head and vice versa. It is a well-known, and easily provable fact, that any point on the image is exactly as far from the mirror as the corresponding object point. Thus your image's nose is nearer to you than the back of its head.

Further experiments.

Hold a printed page facing the mirror. The writing on the image is 'backwards' just as it would be if the printing were on a transparent sheet and you were trying to read it from the back side.

For a new slant on this problem, hold a large plane mirror at arm's length and in a horizontal plane facing up and look at the reflection of a poster on the wall that has printed text. The image is upside down and the text reads backwards. Now swing your arm to the right and rotate the mirror until it is in a vertical plane, keeping the poster's image in your view. The image rotates through 180 degrees, and is now upside down. Swing your arm up and high, rotating the mirror another 90 degrees till it is facing down. The image has rotated another 180 degrees, and is now right side up. The image rotates through twice the angle that the mirror rotates. Rotate your arm another 90 degrees until... Whoops. You'd better seek medical help for that arm.

The 'simple' magnifier

"It's really quite simple, Watson"

Answer:

This is a nice essay question to foil the plug-and-chug crowd—those students who do physics problems by blindly plugging data into a formula and grinding out an answer without involving the higher functions of the brain.

The wisest thing to do is to derive the magnification from first principles for the two cases, rather than to try to get one case from the other. They are fundamentally different cases.

Also, remember that the magnifier formulas are derived under the assumption that the magnifier is held near the eye. Students often forget this and assume that the formulas also apply when the magnifier is held near the page, much nearer than to the eye. They don't. The magnification is greatest when the lens is near the eye, and the object brought close. Also the field is larger and flatter when the magnifier is held near the eye. However, when Sherlock must examine something that can't be brought close, obviously the lens isn't very near the eye.

Angular magnification is M = tan(ß)/tan(ß₀) where ß is the angular size of the image seen with the optical instrument and ß₀ is the angular size of the object seen without the optical instrument.

In the case where the image is at the near point, d is the distance to the image, when the image is at the near point. The formula M = d/f + 1 is not valid for an image placed anywhere else.

In the case where the image is at infinity, d has only one meaning, the distance at which the object can be seen most clearly with the unaided eye. The formula M = d/f is not valid for an image placed anywhere else.

So, the meaning of d is different in the two cases. But one doesn't appreciate that fully until one derives both cases from first principles, with appropriate diagrams. Many textbooks do this derivation carelessly or incorrectly, though always arriving at the "right" formula.

What's the best image?

Look through a telescope at a distant object. Focus the telescope by moving the eyelens in its tube until the image is "best". Now (and this isn't easy) determine where the virtual image is (infinity, near point, somewhere in between). Explain this in comparison with the simple magnifier case.

Look through a microscope at a specimen. Focus the microscope until the image is "best." Where is the image? Why did you put the image there to get the "best" image?

Is the operational meaning of "best" the same in all of these cases?

Answers:

When using the simple magnifier close to the eye, an image may be seen clearly if it is located between the near point and far point of the eye. Consider looking through the lens at an object held in your hand. Most people bring the hand closer (and the image comes closer as well) until the image is at the near point of the eye. Bringing it even closer makes it "out of focus". There's a significant change in the image size, the image being larger as it is brought closer. Your brain easily senses this, and, desiring to see more detail, guides the hand holding the object to a position that makes the image as large as possible. This is not the most relaxed condition for the eye, but the brain weights clarity over comfort in this case.

When using a telescope, the image can be seen if it is anywhere between the near point and far point of the eye. But there's very little change in size of the image over this range—hardly enough to notice.So your brain guides your focusing hand to put the image where the eye will be most relaxed (the muscles controlling the lens will be under the least tension). This position is at the far point of the eye.

The microscope eyelens functions as a simple magnifier, so it's no surprise that the greatest clarity of image occurs when it is at the near point of the eye. However, for long term use, the user will often place the image at the far point for most relaxed and comfortable viewing.

Microscope magnification

M = -(d/f_o)(25/f_e)

In this formula d is the distance from the objective lens, of focal length f_o, to the real image near the eyelens. f_e is the focal length of the eyelens. The first factor is the linear magnification of the objective, the second is the angular magnification of the eyelens, used as a simple magnifier with virtual image at infinity. The result, M, is angular magnification, a point that should be stressed, because one is looking at a virtual image. The 25/f_e looks like the formula for the angular magnification of a simple magnifier used with the image at infinity. But the formula for the angular magnification of a simple magnifier of focal length f is given in most books is M = d/f + 1, where d is the distance from object to lens, usually taken as 25 cm or 10 inches. Yet those same books show a diagram of the microscope with the virtual image somewhere in the vicinity of the objective lens, that is, a finite distance not much different from 25 cm. Isn't there a contradiction here?

But we are passing up what folks in the ed-biz call a "teachable moment" if we don't raise a question here. When using a simple magnifier (or a microscope) one can use the instrument in such a way as to place the image anywhere from 25 cm (the near point of the eye) to infinity (the far point). Then why do we usually assume 25 cm?

When using a telescope, we can likewise place the image anywhere from 25 cm to infinity, depending on how we "focus" the instrument. Then why do we assume the image is at infinity when calculating the telescope's magnifying power?

The fact that textbooks give the angular magnification of a telescope as f_o/f_e tells us that the authors have assumed that the telescope is adjusted so that the focal point of the eyelens and objective lens are coincident, and that an object at infinity has its image at infinity. Yet some texts that make that point have a diagram showing a telescope with an image at a finite distance somewhere near the objective lens. Seems contradictory, doesn't it? See What's The "Best" Image above.

Additional insight problems with longer expositions.