General strategy for using Newton's second law to solve problems:

Draw a diagram; select a coördinate system1.Identify relevant objects/agents2.Sketch an interaction scheme3.Draw a free-body diagram, using the coördinate system chosen in Step 14.Apply Newton's second law to each relevant object, in each direction specified by the coördinate system chosen in Step 1

5.

Example: Static equilibrium

Determine the tension in the string.

Solution:

Chapter 4B: Applications of Newton's LawsTuesday, September 17, 2013 10:00 PM

Ch4B-F2015 Page 1

_________________________________________________________________

Example: Static equilibrium(a) Determine the normal force exerted by a table on a textbook of mass 1.20 kg resting on the table.(b) Repeat part (a) if a hand presses straight downwards on the book with a force of 5.00 N.(c.) Repeat part (a) if a string is attached to the book and pulls straight upwards on the book with a force of 4.00 N.

Solution:

Ch4B-F2015 Page 2

Note that the normal force in part (b) is greater than the normal force in part (a). Apparently the normal force adjusts depending on the applied forces. Every solid surface is, microscopically, a little bit like a trampoline. The atoms in the solid exert attractive electrical forces on each other, which is why the atoms in the solid hold together. You can picture the inter-atomic electrical forces as very tiny springs connecting each atom to nearby atoms.

When a force is applied to the surface of a solid, the atoms in the surface "flex." The "springs" stretch a little bit, and the result is that the surface exerts a force back on the object applying the force to the solid. This is called a normal force, because the force is perpendicular to the surface between the two objects.

If the applied force increases, the surface of the solid flexes more, and so the normal force increases. Similarly, if the applied force decreases, the surface of the solid flexes less, and so the normal force decreases. This explains how the normal force can adjust to different applied forces.

(c.) The applied lifting force reduces the net force acting on the table, and so reduces the normal force. The lifting force is not large enough to actually move the book, because the lifting force is less than the weight of the book.

Ch4B-F2015 Page 3

_________________________________________________________

Example: Static equilibrium

Calculate the tension in the strings if (assume the strings have no mass)the labelled angles are both 30°a.the labelled angles are both 10°b.one labelled angle is 40° and the other labelled angle is 60°c.

Ch4B-F2015 Page 4

Ch4B-F2015 Page 5

____________________________________________________________

We mentioned in class that you can test the previous example for yourself

Ch4B-F2015 Page 6

Mass and Weight

Apparently we don't "feel" gravitational forces directly; what we do feel is contact forces. This explains why we feel weightless when falling, and why astronauts feel weightless when they are in orbit around the Earth (in "free fall").

When you stand on your bathroom scale, you "feel" the normal force exerted by the scale on you; this is what the scale also measures. You feel a similar normal force when you sit on a chair. If you are just standing on your bathroom scale, the normal force from the scale balances your weight (i.e., the gravitational force that the Earth exerts on you), which is why the scale reading (which measures this normal force) equals your weight.

Make sure you understand the difference between mass and weight; your mass is a measure of the substances that you consist of, and is independent of your location. Your weight is a measure of the gravitational force that the Earth exerts on you; although it's proportional to your mass, it is not the same as your mass (W = mg). If you were on the Moon, your weight would be different, because what you feel as your weight would be the normal force that the surface of the Moon exerts on you, which is equal to the gravitational force that the Moon exerts on you.

When you are accelerating, your apparent weight (i.e., the normal force exerted on you by the surface you're standing on) depends on your acceleration.

http://courses2.cit.cornell.edu/physicsdemos/secondary.php?pfID=52

We mentioned in class that you can test the previous example for yourself by hanging a heavy picture frame that has a wire attached to the back. If you pull the wire extremely taut, the tension in the wire when it is hanging will be extremely high. Leaving some slack in the wire means that once the picture hangs from the hook, the angles of the wire relative to the horizontal will be a little larger, and so the tension will decrease.

We also mentioned that the same idea can be used to pull a car out of a ditch; you just have to attach one end of a rope to the car, attach the other end to an object that will not move (a very sturdy tree, for example), and then tighten the rope as much as you can. Pressing on the middle of the rope sideways will exert a very large tension in it, which will pull the car out a little bit. Then tighten the rope and repeat. (Caution: I've never actually pulled a car out of a ditch like this, so one should try this out with less massive objects before trying it with a real car. It's quite possible that the tree might be uprooted, as the tension will be very high!) Check out the following link:

__________________________________________________________

Ch4B-F2015 Page 7

acceleration. _______________________________________________________

Example: Apparent weight

Calculate the apparent weight of a 70 kg person in an elevator that isaccelerating upward at 4 m/s2.1.moving upward at a constant speed of 5 m/s.2.accelerating downward at 2 m/s2.3.

_________________________________________________________

What happens if the cable snaps?

Ch4B-F2015 Page 8

Thus, if the cable snaps, then the normal force becomes zero, which means the apparent weight of the passenger becomes zero; the passenger feels weightless.

_________________________________________________________

Friction

You may have noticed that it is difficult to push a refrigerator. The reason for this is that the bottom of the refrigerator is rough, and so is the floor.

You may also have noticed that it's more difficult to move the refrigerator over a carpet than over a smooth floor, because the carpet is rougher than the floor. Thus, the frictional force between two surfaces depends on the surfaces. This dependence is so complicated that we don't have any good theory for predicting the friction between two surfaces; the best we can do is just measure the frictional forces in experiments.

Further reflection will reveal other facts about friction. You may have noticed that when you begin to push on the refrigerator, it doesn't move. However, as you gradually increase your pushing force, eventually the refrigerator will move. Once the refrigerator is moving, it takes less pushing force to keep it moving than the force needed to get it moving in the first place.

Conclusion: the coefficient of static friction is greater than the coefficient of kinetic friction. Check out the following link:

http://courses2.cit.cornell.edu/physicsdemos/secondary.php?pfID=41

Furthermore, if you stack another heavy object on the refrigerator, it will be harder to move it. Similarly, if someone presses down on the refrigerator, that also makes it harder to move. Thus, frictional forces depend on the normal force acting between the two surfaces in contact.

Conclusions:

fs ≤ µsn

fk = µkn

Typically, the coefficient of kinetic friction is less than the coefficient of static

Ch4B-F2015 Page 9

Typically, the coefficient of kinetic friction is less than the coefficient of static friction for the same pair of surfaces; the coefficient of rolling friction is much less than the coefficient of static friction for the same surfaces.

µr << µk < µs

Sample coefficients of friction are found on Page 101 of your textbook. For example, for rubber on concrete, approximate typical values are

µs = 1.00, µk = 0.80, µr = 0.02_____________________________________________________________

Example: A refrigerator of mass 100 kg is resting on a horizontal floor. The coefficients of friction between the refrigerator and the floor are µs = 1.00 and µk = 0.73. Determine the force needed to get the refrigerator moving.a.the force needed to keep the refrigerator moving at a constant speed once it has already started moving.

b.

Ch4B-F2015 Page 10

In the previous problem, what happens if the original 980 N applied force is maintained once the refrigerator is moving? That is, the kinetic friction force opposing the motion is 715 N, but the applied force is greater than the frictional force, so the net horizontal force Ch4B-F2015 Page 11

_____________________________________________________________

Example: A boat of mass 62,000 kg is being pulled by two tugboats. The first tugboat has a mass of 12,000 kg and pulls with a force of 41,000 N in a direction 23 degrees North of East, and the second tugboat has a mass of 10,000 kg and pulls with a force of 48,000 N in a direction 14 degrees South of East. The water provides a resistive force of 31,000 N directly opposing the boat's motion. Determine the acceleration of the boat.

Solution: Focus your attention on the boat being pulled. Draw a free-body diagram for the boat being pulled. Note, therefore, that the masses of the tugboats are irrelevant. They would only be relevant if we had to draw free-body diagrams for the tugboats, which it turns out is unnecessary.

The resistive force of 31,000 N opposes the motion, so its direction is along the line of the net applied force. The acceleration takes place along the same line, so let's first focus attention on determining the magnitude and direction of the sum of the two applied forces. As usual, the easiest way to do this is to separate each of the applied forces into components. Therefore, the net applied force acting on the boat in the x-direction is

The net applied force in the y-direction is

force is greater than the frictional force, so the net horizontal force on the refrigerator is 980 - 715 = 265 N. Thus, the refrigerator accelerates to the right.

Ch4B-F2015 Page 12

The magnitude F of the net applied force is

The direction of the applied force is (see the diagram above)

The net force acting on the boat, including the resistive force, is

Thus, the magnitude of the acceleration of the boat is

Thus, the acceleration of the boat is 0.86 m/s2, directed at 3 degrees north of east._____________________________________________________

Ch4B-F2015 Page 13

Motion on a Ramp

Determine the acceleration of the block down the ramp if there is no friction between the block and the ramp.

a.

Determine the minimum coefficient of static friction needed to keep the block at rest.

b.

Example: Consider a block of mass 5 kg that is sitting at rest on a ramp inclined at an angle of 30 degrees relative to the horizontal.

Solution: (a) Draw a diagram, and then draw a free-body diagram of the block.

Now choose a coördinate system wisely, and separate any forces that are not on the axes into components:

There are various ways to separate the weight vector mg into components; the following diagram hints at this (for example, you could use the black pair of components, or the blue pair, or one of each):

_____________________________________________________

Ch4B-F2015 Page 14

Now write down Newton's second law of motion in the y-direction:

Now write down Newton's second law of motion in the x-direction:

Equation 2 is enough to solve part (a) of this problem, as the only unknown is the acceleration. Equation 1 is not relevant for this problem; it gives us the strength of the normal force, which may be interesting, but we weren't specifically asked about this.

Solving equation 2, we get:

Note that the result is independent of the mass of the block. Is this reasonable?

(b) Now draw a new free-body diagram, including the frictional force.

of components, or the blue pair, or one of each):

Ch4B-F2015 Page 15

Once again, a wise choice of coördinate system is helpful:

Once again, write down Newton's second law for each component; first in the y-direction:

Newton's second law in the x-direction is:

We have two equations for two unknowns, which is fine, but we were asked about the coefficient of static friction, which doesn't appear in our equations so far. OK, so we need to include the connection between the normal force and the friction force, which will bring in the coefficient of static friction:

Now solve equation 1 for the normal force, solve equation 2 for the frictional force, and substitute the values into equation 3 to determine the coefficient of static friction:

Ch4B-F2015 Page 16

The following material on air drag and terminal speed is OPTIONAL, and will NOT be included in tests and the final exam for this course.

____________________________________________________

Thus, the minimum value of the coefficient of static friction between the block and the ramp to keep the block from moving is 0.58.

It's interesting that the value is independent of the mass of the block, and also independent of the acceleration due to gravity. This means that the same values would solve the problem on the Moon or some other planet.

A related problem, that we'll examine later involves banking of curves on highways. In the absence of friction (wet or icy conditions), banking the road helps to provide the turning force that friction would otherwise provide. It turns out that the best banking angle is also independent of the mass, which is great, because once you get the banking angle ideal, the same angle is ideal for transport trucks, motorcycles, and everything in between.

Air drag

very complicated, much like friction; we don't have good theories of air drag, so we can't predict very well the amount of air drag that will be present in some situation; the best we can do is to experiment, sometimes with scale models (wind tunnels, for example, are used for experimentation)

•

As an approximation, that is reasonably good in certain circumstances, the force of air drag (i.e., air resistance) in newtons is about:

D = 0.3Av2

where A is the cross-sectional area of the object that is moving through

Ch4B-F2015 Page 17

where A is the cross-sectional area of the object that is moving through the air (in square metres), and the speed is measured in m/s. Notice the quadratic dependence on speed; this means if the speed doubles, the air drag quadruples. This makes it very difficult indeed to get things moving through the air at high speeds. A lot of engineering design work goes into reducing air drag by designing shapes and using materials for objects that will reduce the coefficient "0.3" to a smaller number. A more detailed formula is

D = 0.25ρAv2

where ρ represents the density of air; a typical value for the density of air at sea level is ρ = 1.22 kg/m3.

Remember that these formulas are very rough approximations, and experts in various fields will have their own better approximations that deal specifically with their own situations.

A consequence of air drag is terminal speed. If you drop an object from rest, the gravitational force on the object is approximately constant for the entire fall, but the air drag starts off very small (because the speed is small) and then increases rapidly with increasing speed. At some point, if the object falls for long enough, the air drag will grow so large that it balances the gravitational force. After this point, the net force acting on the falling object will be zero, so the object's speed will be constant; this is called the object's terminal speed.

Example: Terminal speed

Calculate the terminal speed of a human being of mass 70 kg.1.Calculate the area of a parachute needed to reduce the terminal speed of a human being of mass 70 kg to (i) 10 km/h, and (ii) 5 km/h.

2.

Assume that the person falls feet-first, so that we can estimate his or her cross section as approximately 0.1 m2. (We can probably stand on a 1-foot square kitchen tile without "sticking out", right?) Thus, the drag force on the person is about

Ch4B-F2015 Page 18

This is very fast. You can minimize the terminal speed by increasing the cross-sectional area, which increases the drag force. One way to do this is to fall "belly-flop" style instead of feet-first. Another way is to use a parachute, which is explored in Part (b).

Remember that such calculations are only approximate. The actual drag force depends on many factors that are not included in the formula, such as the shape of the object, the material on the surface of the object, and so on.

When the drag force is equal to the force of gravity on the person, then the net force on the person is zero, and the person's speed is constant. Thus, the terminal speed satisfies

Ch4B-F2015 Page 19

This seems awfully fast. Would you want to be hitting the ground at this speed? We can compare this terminal speed with speeds attained by jumping from various heights. Or, better yet, let's use the kinematics equations to determine the height from which you would have to drop a ball so that it hits the ground at a speed of 6.8 m/s (assuming no air resistance):

Do the results seem reasonable? Not really, because typical sky-diving parachutes have cross-sectional areas in the range of 10 m2 to 50 m2. Evidently, sky divers hit the ground at speeds significantly greater than what we assumed.

Suppose we take a typical parachute cross-sectional area of 50 m2. Here is the resulting terminal speed:

Ch4B-F2015 Page 20

Sliding blocks

This seems reasonable, as the height is less than the height of a typical one-storey roof. I've seen athletic teenagers jump off one-storey roofs with no ill effects (I'd probably break my legs), so jumping from a lesser height is probably safe. It's reasonable to expect a parachute to slow a novice sky-diver to similar speeds.

(End of optional material.)

__________________________________________________________

Imagine pushing a heavy refrigerator across the floor. If there is a fragile sculpture, made of thin glass, on the far side of the fridge, it's conceivable that the sculpture might slide along, being pushed by the fridge, without being damaged. However, if you pushed from the other direction, the sculpture would surely crumble as you pushed it against the fridge.

We can understand this situation in the context of Newton's laws as follows.

Example: Calculate the acceleration of the system and the magnitudes of the inter-block forces ifa 50 N force acts on the left-most block towards the right.a.a 50 N force acts on the right-most block towards the left.b.Assume there is no friction between the blocks and the surface they rest on.

Solution: (a) Sketch a free-body diagram for each block.

Ch4B-F2015 Page 21

Solution: (a) Sketch a free-body diagram for each block.

Notice that we have five unknowns, so we need five independent equations to be able to solve this problem. Applying Newton's second law to each component direction for each block produces 4 equations; where are we going to get one more equation? After thinking for a while, you may realize that the two inter-block forces have equal magnitudes, according to Newton's third law:

OK, let's write down Newton's second law for each component direction of each block. (Note that because the blocks move together, they have the same acceleration, so we can use one single symbol to represent the acceleration of each block.)

Combining equations 1 and 5 gives us

Ch4B-F2015 Page 22

Example: Calculate the acceleration of the system and the magnitudes of the inter-block forces ifa 10 N force acts on the left-most block towards the right.a.a 10 N force acts on the right-most block towards the left.b.Assume there is no friction between the blocks and the surface they rest on.

and substituting the result into equation 3, and solving for a, gives us:

Using equation 5, we can calculate the inter-block force as

(b) Now I invite you to complete part (b) for yourself. If you compare your solution for part (b) to the solution for part (a), you will see that the solution processes are exactly the same; the only difference is that the roles of the two blocks are interchanged. Thus, you will find that the acceleration in part (b) is the same as in part (a) (does this make sense?), but the inter-block forces in part (b) are 100 times greater than in part (a). Does this make sense? Does this explain to you what we intuitively understand, that the delicate glass sculpture gets smashed in part (b) but not in part (a)?

_____________________________________________________________

The next example is much like the previous one, but with three blocks instead of two.

Ch4B-F2015 Page 23

Ch4B-F2015 Page 24

Ch4B-F2015 Page 25

Ch4B-F2015 Page 26

Example: Calculate the tension in the string and the acceleration of each block. Assume that the strings are massless and don't stretch, and that the pulley is massless and there is no friction at the bearing. Further assume that the string slides on the pulley without friction (or, more accurately, turns the pulley without slipping, so that friction need not be considered).

Blocks and pulleys

______________________________________________________________

Ch4B-F2015 Page 27

Ch4B-F2015 Page 28

Ch4B-F2015 Page 29

Ch4B-F2015 Page 30

Example: Calculate the tension in the string and the acceleration of each block. Assume that the strings are massless and don't stretch, and that the pulley is massless and there is no friction at the bearing. Further assume that the string slides on the pulley without friction (or turns the pulley without friction needing to be considered). Suppose that the coefficient of static friction between the 2 kg block and the horizontal surface is 0.8, and that the coefficient of kinetic friction between the same surfaces is 0.5.

Ch4B-F2015 Page 31

Now for an important bit of strategy: Note that if the tension force is less than the maximum value of the static friction force, then the blocks will not move. The static friction force will adjust itself to match the applied force, up to a point; if the applied force exceeds the maximum static friction force, then the blocks move, and then kinetic friction will apply.

Thus, let's first assume that the blocks do not move, and we'll calculate the tension force. Then we'll compare the tension force to the maximum force of static friction. If the tension force is greater than the maximum static friction force, then we know the blocks will move, and we'll re-solve a part of the problem using the kinetic friction force to determine the actual accelerations of the blocks.

Here goes the first step, using static friction:

Ch4B-F2015 Page 32

The tension force is greater than the maximum static friction force, so the blocks actually move. Thus, we can go back to equation (*) and replace f by the force of kinetic friction to determine the actual tension in the string.

Ch4B-F2015 Page 33

Example: Calculate the tension in the string and the acceleration of each block. Assume that the strings are massless and don't stretch, and that the pulley is massless and there is no friction at the bearing. Further assume that the string slides on the pulley without friction (or turns the pulley without friction needing to be considered).(a) Suppose that the surface is frictionless.(b) Suppose that the coefficient of static friction between the 2 kg block and the slanted surface is 0.8, and that the coefficient of kinetic friction between the same surfaces is 0.5.

Ch4B-F2015 Page 34

(a) If the surface is frictionless, then f = 0, and the equations above simplify to (also noting that the sine of 30 degrees is 1/2):

Because we are only interested in calculating the tension in the string and the acceleration, we can ignore equation (1). Solve equation (2) for T and substitute the result in equation (3), to obtain:

Substituting the value of the acceleration into equation (2), we can solve for the tension in the string:

Ch4B-F2015 Page 35

(b) Now for an important bit of strategy: Note that if the tension force is less than the maximum value of the static friction force, then the blocks will not move. The static friction force will adjust itself to match the applied force, up to a point; if the applied force exceeds the maximum static friction force, then the blocks move, and then kinetic friction will apply.

Thus, let's first assume that the blocks do not move, and we'll calculate the tension force. Then we'll compare the tension force to the maximum force of static friction. If the tension force is greater than the maximum static friction force, then we know the blocks will move, and we'll re-solve a part of the problem using the kinetic friction force to determine the actual accelerations of the blocks.

Here goes the first step, using static friction:

Ch4B-F2015 Page 36

Note that the acceleration is less in this problem than in the previous problem, which is not surprising, since (because of the slope of the slanted surface) gravity works against the acceleration of the 2-kg block.

__________________________________________________________________

Next consider a pulley problem that is a little more sophisticated.

Example: Determine the accelerations of each block and the tension in the rope. (Because we assume that the rope is massless and does not stretch, and the pulleys are massless and frictionless, the tension is the same in each part of the rope.)

The tension force is greater than the maximum static friction force, so the blocks actually move. Thus, we can go back to equation (*) and replace f by the force of kinetic friction to determine the actual tension in the string.

Ch4B-F2015 Page 37

Solution: Draw three free-body diagrams; one each for the two outside masses, and another one for the central pulley. As usual, we'll choose positive directions for each free-body diagram so that the accelerations will all be compatible.

We have three equations for four unknowns, so we need another equation. Because the rope does not stretch, the accelerations are all connected, but in a way that is a little more complex than the previous problems in which there were only two blocks. Note that if the middle block is displaced downward so that its displacement is x2, the sum of the displacements of the other two blocks is twice as large as x2, because two lengths of ropes are attached to the middle pulley.

Ch4B-F2015 Page 38

Because

if the pulleys are released from rest, then

Thus,

(Calculus lovers could derive equation 4 in an alternative way by simply starting with the displacement relation above and differentiating each term twice.)

So now we have four equations for four unknowns, and the problem is solvable. One way to complete the solution is to solve each of the first three equations for the accelerations and substitute the resulting expressions into equation 4:

Ch4B-F2015 Page 39

We can now back-substitute the tension into the first three equations to determine the accelerations:

Are the results reasonable? Well, the total mass of the outside blocks is 9 kg, and the mass of the central block is 6 kg, so it seems reasonable that all three blocks accelerate in the negative directions.

_______________________________________________________________

To get a better feel for what can happen in general, it's worth solving the problem in general. This is done in the following example.

Example: Determine expressions for the accelerations of each block and the tension in the rope.

Ch4B-F2015 Page 40

Solution: As in the previous example, we'll draw a free-body diagram for each of the two outer blocks and one for the middle pulley. The relations among the accelerations is the same as in the previous example, by the same reasoning used there.

We can solve the problem by using the same solution process as in the previous example; solve each of the first three equations for the accelerations, then substitute the resulting expressions into equation 4 to obtain an equation involving just the tension. After solving this equation for the tension, we can then back-substitute into each of the first three equations to obtain expressions for the accelerations.

Ch4B-F2015 Page 41

Thus,

Similarly (do the algebra!),

and

Ch4B-F2015 Page 42

_________________________________________________________________

It's possible for the first block to have zero acceleration. One way to do this is to arrange for the first block and the third block to have the same mass and the middle block to have twice the mass of one of the others. But this solution is a bit boring (although it makes sense, doesn't it?), because then all three blocks have zero acceleration. A more interesting solution is to arrange for the first block to have twice the mass of the third block, and middle block to have four times the mass of the first block. Then the first block has zero acceleration, but the third block flies up with an acceleration of g, and the middle block moves down with an acceleration of g/2.

•

You can arrange for the third block to have an acceleration of zero by just following the same choices as in the previous item, but interchanging the roles of the first and third blocks.

•

You can arrange for the middle block to have an acceleration of zero by choosing the masses of the middle block and the third block to be equal, and the mass of the first block to be a third of the mass of the middle block. Then the middle block does not accelerate, but the third block accelerates downward with magnitude g/2, and the first block accelerates upward with magnitude g/2.

•

Now it would be interesting to play with these expressions to see some examples of possible motions. For example:

There are many other interesting possibilities, and the right sort of person would have fun playing with this on a rainy day. If you're good at programming, you might like to program a simulation, so that you could see the results of changing the values of the masses in various ways. That would also be fun, would it not?

Finally, notice that in all of the problems in this chapter, time has been ignored. This is the usual tactic in first-year physics: We simplify as much as possible at the beginning, sometimes at the cost of oversimplification. As an example of a situation where time must be included in the analysis, consider the following massive block suspended by a thread, and having another thread attached below it:

Ch4B-F2015 Page 43

Suppose that you pull down on the bottom thread; what happens? Well, it depends on how gradually you increase the force as you pull down. If you give the bottom thread a sudden jerk, then the bottom thread will snap. If you give the bottom thread a slow, gradual increase in force, then the top thread will snap.

You can explain this for yourself if you draw free-body diagrams for each thread (that's right, the threads, not the block), and notice that the tension in the top thread is greater, because it also has to support the weight of the block. However, if you give the lower thread a sudden jerk, there is not enough time for the force to be transmitted to the upper thread, and so the lower thread will break.

It's a bit like a train engine pulling on a long train of cars; not every car begins moving at once. Instead, the first car begins moving, and there is a slight time delay until the coupling between the first car and the second car engages, which gets the second car moving. Then, after another short time delay, the third car begins moving, and so on. You can see the same phenomenon (for a different reason) when a line of cars gets moving when the traffic light changes from red to green; there is a time delay for each car in the line to get moving.

You can experience this phenomenon (with the block and threads) in every day life when you break a paper towel from its roll, or break toilet paper from its roll. If you "snap" the paper, then it will break along one of the perforations, but if you pull steadily the paper will unroll without breaking off.

The same thing happens at the grocery store when you break off a plastic bag from its roll in the produce department. A sudden snap breaks a bag from the roll, but a gradual pull just unrolls the bags.

There are other practical consequences to the tension experienced by different members of hanging structures. Consider the following hanging walkway.

Ch4B-F2015 Page 44

http://en.wikipedia.org/wiki/Hyatt_Regency_walkway_collapse

Moral: Critical thinking involves actively searching for errors. It's unreasonable to expect that no errors will be made; however, one needs systematic searches for errors.

This is one of the great strengths of science: Nothing is taken for granted, and nothing is taken by authority. Each idea is always tested, over and over again. The upside is that errors are found and corrected; the system of science is "self-correcting." The downside is that new ideas are often subject to vicious attacks, and sensitive people are often turned away from scientific research because of its rough-and-tumble nature.

Even the very greats suffer the slings and arrows of attack (to be fair, this is common in every field of creative endeavour, not just science), and sometimes they succumb; witness Boltzmann's suicide, which may have been driven by the vicious attacks on his revolutionary ideas in statistical mechanics.

Resilience is helpful in all fields of endeavour, so perhaps it's not surprising that it is helpful for survival in scientific research too.

Ch4B-F2015 Page 45