Class Notes for Fall, 2000

Week of	Aug 23	Sep 6	Oct 2	Nov 6	Phy 205 page
	Aug 28	Sep 11	Oct 9	Nov 13
		Sep 18	Oct 16	Nov 20	Notes for Fall, 1999
		Sep 25	Oct 23	Nov 27
			Oct 30

Phy 205, Week 6

25 Sep State Newton's laws of motion, following Tipler's statement about the first law, as that's better than most. In order to see what an "inertial system" is, show some examples that aren't. In this course we won't be dealing with any such complications.

State in outline what has been discovered about the fundamental forces in nature. In this course, we will be using only two of them, and a special case at that. For us, the only forces will be either (1) gravity, or (2) "contact forces" and this last one is really a special case of electromagnetism, though that fact is not needed in order to use it.

Do in great detail the problem of a cart sliding down a ramp without friction. The demo really has negligible friction. Despite the simplicity of this first example, it contains most of the structure of more complicated problems.

First, list the things that act on the cart:
(1) The gravitational pull of the Earth
(2) the ramp the cart is sliding down.

And that's IT.

I pointed out several possible confusions and false starts that you can make in even this first step, but in the end it came down to these two things.

Next, translate this into mathematics. The forces are vectors, so you should pick a basis with which to describe them. (In the 11:00 class, I used a set tilted along the ramp. At 4:00 people wanted them horizontal and vertical. No matter.) Whichever basis you pick, the gravitational force is down. Because the force from the ramp has negligible friction, the force that it exerts will be perpendicular to the ramp itself. For gravity, the force has magnitude mg. For the ramp, the magnitude is unknown, so I make up a symbol for it and call its magnitude F_ramp.

The acceleration is unknown, but at least I know its direction. It points down the length of the ramp. Again, use a symbol for this unknown magnitude (a, of course).

Now write the vectors in terms of components and equate the two vectors F_total and ma. For the tilted basis, these equations were

For the basis that is horizontal and vertical, the corresponding equations are

In either case, you break the vector equations into components to get two scalar equations in the two unknowns F_ramp and a. Solve them and then interpret them to see if they make sense.

27 Sep If the basic equations of Newtonian mechanics are taken at face value, and if you don't take any short cuts, then there are only a few basic tools that you need use to set up even complex problems. (Solving the resulting equations for complex problems can still be hard.)

To show the beginnings of this, take two masses, a book and an eraser in contact with each other and resting on the table. If I push horizontally on one of them I can cause them both to start moving. Figure out the acceleration and the force of contact between them.

Treat each mass separately.

Acting on m1	Acting on m2
1. gravity	1. gravity
2. table	2. table
3. other mass (m2)	3. other mass (m1)
4. ME

For this problem today, I'll assume that friction with the table is negligibly small, even though it obviously is not. This just makes the problem easier. Use a standard basis, with i horizontal to the right and j up. Then

	Fon 1	=	- m1 g j + Ftable1 j - F12 i + FME i
		=	m1 ax i
and
	Fon 2	=	- m2 g j + Ftable2 j + F12 i
		=	m2 ax i

To set these equations up, I used Newton's third law, saying that the force of mass#2 on mass#1 is the same magnitude as #1 on #2, but in the opposite direction. I also used the fact that the two masses are moving together, so that their respective velocities are the same and so are their accelerations.

Now break these vectors into components and get four equations:

1:	- F12 + FME	=	m1 ax
2:	- m1 g + Ftable1	=	0
3:	+ F12	=	m2 ax
4:	- m2 g + Ftable2	=	0

I can now solve these simultaneously. Equations 2 and 4 aren't very interesting. The other two are easy to solve. Add equations 1 and 3, to get

This says a_x = F_ME / ( m₁ + m₂ ).
Substitute this back into equation 3 (or 1) to get F₁₂ = m₂ F_ME / ( m₁ + m₂ ).

The first of these says that the two masses move together as is they were one (in this case).
The second equation isn't so obvious, but it's easy to check in a couple of special cases. If m₁ is much larger than m₂, then it implies that the contact force is much smaller than the force I apply.
If the masses are reversed, it says that the contact force is almost the same as what I apply. You can check this if you feel the contact force with your finger. Use a book and a pencil or other light object. Place your finger between them and push on one or the other.

29 Sep Try to sort out some confusion about what is and is not a force:
In this course, the only two classes of force will be either
(1) gravity (usually from the Earth) or
(2) contact from another object.

If it isn't on the above list, it isn't.

Is ma a force? Is it gravity? No. Is it in contact with something and pushing it around? No. So... It isn't a force.

Does it have the units of force? Yes, but that doesn't make it a force. Newton's second law says that once you've added up all the forces on a mass, m, then you can predict its acceleration by dividing,

Do another detailed example of the application of Newton's laws, two masses tied together by a (light) string and the string run over a pulley. The lists of what acts on the two masses are

Acting on m1	Acting on m2
1. gravity	1. gravity
2. track	2. string
3. string

I spent some time explaining why it is reasonable to neglect friction with this apparatus (the bearings are very good). Also the approximation that the mass of the string is very small and the mass of the pulley is very small leads to the approximation that the tension in the string is constant throughout it, though in this case the real derivation will have to wait for another couple of chapters.

At this point, I pick a basis for the forces on m₁ and write the total force. The acceleration is along the direction of the track, so I wrote that out too. Then apply F_total = m a. Break the vector equation into components, getting two equations.

Now repeat the process for the second mass. Here I also know that (unless the string breaks) the magnitude of the acceleration will be the same as that for m₁, only with a different direction.

Finally solve the resulting scalar equations and analyze them. The results are

Check the dimensions of the results and then see what happens in a special case or two, letting one mass be a lot bigger than the other.