=2

PHY 121/123 LECTURE NOTES -Week 3

Let me return to something discussed in lecture during Week 2, when it was especially topical. A home run champion, Mark McGwire or Sammy Sosa, hits another record, and jogs around the bases at 3 m/s. Ignoring any pauses at the bases, what is the mean velocity, the mean speed, and the total trip time? The baseball diamond he is circuiting is a square just under 30 m on a side. The results should be compared with those for the catcher, who stayes put next to home base.

As discussed earlier, mean velocity is defined by

$${\bf \vec{v}}_{av} = \Delta {\bf \vec{x}}/\Delta t.$$

Because he returns to his starting point, we have $\Delta {\bf \vec{x}}=0$, and therefore ${\bf \vec{v}}_{av} = 0$. The speed, on the other hand, is constant, and therefore gives the mean speed, s_av=s₀=3 m/s. The trip time is given by the relation distance traveled D=s_av t = 120 m, or t=40 s. The difference for the catcher who stands still is only that his average speed is zero - his zero average velocity is the same as for the runner! One can imagine his saying to the runner, ``You ran all that way and got out of breath, but now you are just back where you started.''

It's also interesting to ask about acceleration for the runner. As he starts from rest and ends at rest, the change in velocity is zero, and so is the average acceleration, but because his velocity changes during the circuit there must be accelerations at various times. Say he takes 1/2 s to get up to speed at the start, 1 s each to change direction at each base, and 1/2 s to stop when he gets back. This implies an average acceleration during the first half-second of ${\bf \vec{a}}_{av} = 6$ m/s² towards first base, acceleration of magnitude $3\sqrt{2}=4.2$ m/s² and direction towards the center of the diamond as he rounds first base, similarly for second and third base, and finally, if he stops in 1/2 s, an acceleration towards third base again of 6 m/s². If we add all the pieces together, we know they must give a total sum

$${\bf \vec{a}}_{av}^i\Delta t^i=0 ,$$

and it is easy to check that this is so. Note that if we add the average acceleration in the first half second and the last half second we get again a vector pointing towards the center, so at each of the four bases a vector velocity shift of the same magnitude occurs, always pointed towards the center of the diamond. This discussion should be a good introduction to the subject of circular motion, which we take up next week.

NEWTON'S THREE (ACTUALLY TWO) LAWS

Here is Newton's first law, which really amounts to a negative definition of force, by describing when no force is present. We shall see shortly that this is just a special case of Newton's second law, which is why I say there actually are only two laws.

1. If the net force on any object or system is zero, then the object does not accelerate: It moves with constant velocity ${\bf \vec{v}}$.Constant velocity means that the speed of motion does not change, and the direction of motion also does not change. If the velocity is written as a vector in terms of components, then each of the components is constant. If it is written in terms of a magnitude, and direction angles, then these do not change. The above statement works backwards also: If there is no acceleration, then the net force on the object is zero. Note that an object at rest has velocity ${\bf \vec{v}}=0$, which is just a special case of constant velocity.

Now we come to the second law, which includes the first.

2. For an object of fixed mass m, the net force is related to the acceleration by

$${\bf \vec{F}}=m {\bf \vec{a}},$$

where ${\bf \vec{F}}$is the net force vector and ${\bf \vec{a}}$ is the acceleration vector, i.e., the first time derivative of the velocity, ${\bf \vec{a}}=d{\bf \vec{v}}/dt$. In component notation, this means, for example, that the x component of the acceleration is equal to the time derivative of the x component of the velocity: a_x=dv_x/dt, and we know from before v_x=dx/dt, where x is the placement of the object along the x direction. Evidently, vanishing force implies vanishing acceleration, and vice versa, so the second law contains the first as a special case.

The third law is a very different statement, but essential for completeness and consistency.

3. If ${\bf \vec{F}}_{A \rightarrow B}$ is the force of an object A on an object B, and ${\bf \vec{F}}_{B \rightarrow A}$ is the force of B on A, then these two forces are equal and opposite:

$${\bf \vec{F}}_{A \rightarrow B}=-{\bf \vec{F}}_{B \rightarrow A} .$$

These laws and definitions only acquire meaning when we introduce specific forces. Let's start with one we already have discussed, gravity near the surface of the earth. In two space dimensions, horizontal x and vertical y, gravity points down,

$${\bf \vec{a}}=(a_x,a_y) = (0,-g).$$

We shall do a lot of two-dimensional problems, but it is good to realize that in some cases three dimensions all are important. With horizontal x and y directions and vertical z, gravitational acceleration is written

$${\bf \vec{a}}=(a_x,a_y,a_z)=(0,0,-g).$$

Going back to two dimensions, we have immediately the gravitational force on an object of mass m:

$${\bf \vec{F}}_{grav}=m {\bf \vec{a}}=(0,-mg).$$

If this were the only force on the object, it would fall with acceleration g, but often other forces are present, and we can use knowledge of the gravitational force to learn about others.

A first example is a 10 kg box sitting on a floor. The box is not moving, and therefore certainly is not changing velocity, i.e., accelerating, so the force of gravity must be balanced against something else. What is that? The answer is rigidity, or impenetrability, or resistance to compression of solid objects like the box or the floor. Because the box cannot go through the floor, we know that the floor must be pushing on the box with a force (0,mg)= (0,100 N). This means that the net force on the box is zero, permitting it to remain stationary.

Now there is an important difference between the force of gravity and the new force of rigidity. If you know the mass of an object, then you know the force of gravity on it. However, the upward force of the floor on the box depends on the mass of the box, not on particular characteristics of the floor. Instead, the material properties of the floor (and the box) determine the maximum force which could be sustained before one or the other or both would collapse under the weight. Thus a particular floor-box combination has a maximum capacity for resisting compression, and above that limit there will be a catastrophic collapse, the force will drop suddenly to zero, and, if it is the floor that collapses, the box will fall with the full acceleration of gravity. Thus, in a plot of F_up versus weight (i.e., the force of gravity on the box), the magnitudes of the two will be equal up to a certain limit (giving a 45^o line in the plot), after which F_up will drop rapidly to zero.

A little apology and reminder: In the above discussion I described, but did not draw, a plot. Perhaps by the latter part of the semester I'll be doing drawings. However, meanwhile, the lack of drawings may actually be helpful in using these notes. If you sketch what should be in a drawing for yourself, that could allow you to fix the ideas in your mind better than if you just glanced at a drawing of mine.

Note that F_down always is equal in magnitude to F_up, i.e., the downward force of box on floor is always equal to the upward force of floor on box, in obedience to Newton's third law. Only if the box is stationary is F_down equal to the gravitational force on the box, also known as the weight of the box. Once collapse occurs, F_up goes to zero, and therefore so does F_down.

This discussion brings to mind a definition in the text, a free-body diagram: Draw an imaginary circle around the box, and look at all the forces on the box. The first force is gravity pushing the box down, which may be considered as acting at the center of the box. The second force is the floor pushing the box up, and as just indicated this exactly balances gravity, so the box stays still.

We have seen already that the force of the box on the floor, which is just the weight of the box, is equal and opposite to the force of the floor on the box, in obedience to Newton's third law. However, we haven't yet discussed the question, ``What is the third-law `back-force' to the force of gravity on the box?'' This is tricky, because we shall be learning to think of gravity as a force from the whole earth acting on the box. Therefore, the reaction force balancing the attraction of the box to the earth is an equal and opposite force attracting the earth to the box. We don't notice this attraction because the great mass of the earth means that its acceleration due to the attraction is too small to see. Nevertheless, in principle, this force should be there. Of course, as long as the box is resting on the floor, the upward attraction of the box on the earth is balanced by the downward force of the box's weight on the earth. Only if the box is falling towards the earth is the earth also falling (very slowly!) towards the box.

Let's make the box problem more interesting by tilting the floor, assuming the box can slide freely along the floor. The floor surface becomes an inclined plane, at angle ${\theta}$ to the horizontal. The component of gravity perpendicular to the plane is balanced by the upward force of the floor, but the component of gravity parallel to the plane pushes the box with no compensating resistance. This gives $F_{slide} = mg {\rm sin} \theta ,$ and therefore acceleration $a_{slide} = g {\rm sin} \theta .$ For $\theta = 30^o$,and the box 10 m from the bottom of the ramp, how long does it take to get to the bottom, starting from rest? We have $\Delta x = (1/2)at^2$, where x now is placement along the ramp, not along the horizontal direction. This gives $10 \ {\rm m} = (1/2) \times (1/2) \times 10 \ {\rm (m/s^2)} \times t^2$, or $t=2 \ {\rm s}$.

Another force which depends on the detailed properties of materials is friction. A good rule of thumb is that the frictional force resisting sliding of an object along a surface is given by $F_{friction} \le \mu F_{normal}$, where F_normal, also called $F_{\perp}$ is the magnitude of the force with which the object pushes into the surface, which also is the magnitude of the force with which the surface pushes out on the object. Here the coefficient of friction is given by the Greek letter $\mu$, pronounced `mu'. We can combine our first three forces by considering a box resting on an inclined plane, slanted at angle ${\theta}$ to the horizontal. The box does not move in the direction normal to the plane surface, because the component of its weight in that direction is exactly balanced by the upward force of the plane on the box. This implies

$$\vert F\vert _{friction} \le \mu F_{\perp}=\mu mg {\rm cos} \theta .$$

First question: If the inclined plane is horizontal ($\theta=0$), obviously gravity doesn't try to make the box slide. How big does ${\theta}$ have to be for gravity to just balance the maximum friction, so that the box can slide down the ramp? Of course, for still larger angles than this, gravity overcomes friction, and the box actually accelerates, while for smaller angles friction is bigger than gravity (actually the component of the gravity force along the plane), and so the box, if moving initially, decelerates until it stops. The component of gravity along the plane is $F_{slide} = mg {\rm sin} \theta$, and so the angle for exact balance is given by $F_{slide} = \mu F_{\perp}$, or

$$mg{\rm sin} \theta = \mu mg {\cos} \theta,$$

meaning

$${\rm tan} \theta = \mu .$$

Suppose $\mu=1$; usually the value is smaller than this. Thus, for $\theta \gt 45^o$, the box will slide. On the other hand, suppose the surface is very smooth, so that friction is negligible ($\mu \approx 0)$. Then the box accelerates with $ma=F_{slide}=mg{\rm sin} \theta$, or $a_{slide}=g{\rm sin} \theta$, as we saw above. For F_slide bigger than the maximum frictional force, we have $a_{slide}=g({\rm sin \theta - \mu cos \theta})$. Take as an example a box starting from rest 5 m from the end of a ramp with $\theta=60^o$ and $\mu = 1/\sqrt{3} = .577$. Because we have ${\rm sin \theta =\sqrt{3}/2}$ and ${\rm cos \theta = 1/2}$, we get $a = 5.77 {\rm m/s^2}$, and $t= \sqrt{2\vert\Delta x/a}=1.32$ s.

There is an important difference between friction and tension - While both have limits depending on the properties of the materials involved, friction simply reaches its maximum value and stays constant after the sliding force exceeds this value. Friction does not show a catastrophic collapse. There is still another force which does collapse above a maximum strain, and that is tension. A string or rope or cable or chain does not resist compression as a table or floor would, but it does resist stretching, and, below a maximum tension which would rip it apart, simply acts as a force transmitter, even transmitting tension or pulling force around curves.

Let's look at some simple examples. Suppose a string is looped over a pulley, a light, low-friction wheel, and you are pulling on one end with a force of 50 N, while the other end, on the other side of the pulley, is hooked onto the top of our famous 10 kg box. Gravity pulls down on the box with 100 N, and the string pulls up with 50 N, so that the floor under the box only supplies 50 N upward normal force to balance that part of the weight of the box not supported by the string. In a free-body diagram, there are two upward forces of 50 N each, one from the floor and one from the string, and one downward force of 100 N applied at the center of the box, namely, gravity.

Now let's get some movement by making the box 5 kg and pulling on the other end of the string with a force of 100 N. Clearly this means the box will rise with an acceleration of 10 m/s². If the box goes over the pulley and starts coming down, its downward acceleration will be 30 m/s² (Can you explain why?). Getting still more complicated, suppose that instead of pulling with 100 N we put a 10 kg box on the other side of the pulley. Now we expect the string to remain taut, so the two boxes will accelerate at the same rate, one up and one down. Therefore, for the two boxes, the free-body diagrams and Newton's second law give

$$ma = T-mg, \ Ma = -T + Mg,$$

where T is the tension in the string. Adding these up, we get

a=(M-m)g/(M+m) =g/3.

Note that this is less than the acceleration of the smaller box when a force equal to the weight of the large box was applied by hand. The reason is that now that same force has to move three times as much mass, so gives one-third the acceleration.

Now suppose the boxes are both falling, just connected by a straight vertical piece of string. Common sense tells us that they will both have acceleration g, but let's do it the long way with free-body diagrams. This time call down positive. Then for the lighter box which we may assume is above, we have ma = mg + T, and for the lower, Ma= Mg - T, as tension always pulls in towards the center of the string. Adding up these two equations gives immediately the expected result a=g. Plugging that into either equation gives T=0, so indeed the two fall at the same rate, and removing the string would make no difference. Compare this with the case of pulling down on the light box with 100 N, where we found a to be three times as big. Again, the reason is that only one-third as much mass is being moved by the same total force.