Class Notes, Fall 2001

Week of	Aug 22	Sep 5	Oct 1	Nov 5	Phy 205 page
	Aug 27	Sep 10	Oct 8	Nov 12
		Sep 17	Oct 15	Nov 19
		Sep 24	Oct 22	Nov 26
			Oct 29		Notes for Fall, 2000

Phy 205, Week 6

24 Sep Look at the solution to the problem that I was working out last time. The key concept to keep in mind is that there are only two types of forces that we will encounter this semester:

1. A force can come from direct contact with another physical object.
2. Gravity.

The forces on the cart in this case were from the track and gravity. (Air too is in contact, but I neglected that.)

I again briefly set up the equations to show how the acceleration comes from the total force applied: a = F_total/m. After breaking the component equations apart into the separate algebraic equations, I solved them. The final part of this process is to look at the results to see if they make sense.

24 Sep Emphasize the structure of solving mechanics problems. Look again at the single mass sliding on the track and repeat the setup. Then take the case in which this mass is attached to a string and that string is attached to another mass that is hung over a pulley. I emphasize the orderly approach to all these problems:

Remember the list written above? Only (1) the gravitational pull (by the Earth in this case) and (2) something in contact with it will contribute.

Let m₁ be the mass on the track.
Let m₂ be the mass hanging down from the string.
The things that act on m₁ are
1. gravity
2. track
3. string
4. air

There were other things that various people suggested for this list:
5. table
6. friction
7. m₂
The table is not in contact with m₁, so it does not exert a force on it.
Friction is not a thing, it a property of the force that the track exerts on m₁. It is then already present under item 2.
m₂ is not in contact with m₁, so it exerts no force on it.

Repeat this for the the second mass. What acts on it are
1. gravity
2. string
3. air
and that's it.

In both of these cases I'll neglect the force exerted by the air; it's pretty small. Next I choose coordinate system and basis in order to describe the vectors that I am using. I'll use the obvious one, with i to the right and j up.

and the acceleration of this mass is

Note: I didn't assume that I know the value of F_track. I let the equations tell me. I didn't assume that i know the valus of F_string. I let the equations tell me. If you try to guess the value of these ahead of time, you may get one right but you will certainly get the other one wrong.

The basic dynamical equation is F_total = m a. This is a vector equation that has two components. Break these into separate equations so that I can do the algebraic manipulations.

These are two equations for the three unknowns, F_string, a, and F_track.

Next, repeat the process for m₂.

28 Sep Finish the problem of the two masses tied together, with a string run over a pulley. One mass moves horizontally, one vertically. I had done the mas m₁, the horizontally moving mass last time. I repeated that and went on to the other mass.

The things acting on m₂ are gravity and the string. At this point I made the approximation (a very good one) that the string is so light that I can neglect its mass. This means that the total force on each piece of it will have to be zero and that the tension in the string is the same all through it, in particular it is the same on the two ends.

Here I used the fact that the string doesn't break. This says that as one end moves so does the other. The magnitudes of the two accelerations are then the same; all that is needed is to straighten out the directions. That's what the "- j" is for.

This equation has only one component,

With the two equations for the first mass this is enough to solve the problem. The unknowns, a and F_string are now known.

The final step is to analyze the solution. First check the dimensions, then look at special cases of the parameters that occur in the problem. Here the only thing that I can vary are the two masses: the two extremes are that m₁ >> m₂ and the reverse.