Mathematical Tools for Physics - Department of Physics - University

2—Infinite Series 39

mistake; go back and find it. You can do this sort of analysis everywhere, and it is one technique that
provides an automatic error finding mechanism. If an equation is dimensionally inconsistent, backtrack
a few lines and see whether the units are wrong there too. If they are correct then you know that your
error occurred between those two lines; then further narrow the region where the mistake happened by
looking for the place at which the dimensions changed from consistent to inconsistent and that’s where
the mistake happened.
The second tool in your analysis is to examine all the parameters that occur in the result and
to see what happens when you vary them. Especially see what happens when you push them to an
extreme value. This is best explained by some examples. Start with some simple mechanics to see the
procedure.

m 1 ax

M

m 2

Two masses are attached by a string of negligible mass and that is wrapped around a pulley

of massM so that it can’t slip on the pulley. Analyze them to determine what is wrong with each.

Assume that there is no friction betweenm 1 and the table and that the string does not slip on the

pulley.

(a)ax=

m 2 −m 1

m 2 +m 1

g (b)ax=

m 2

m 2 +m 1 −M/ 2

g (c)ax=

m 2 −M/ 2

m 2 +m 1 +M/ 2

g

(a) Ifm 1 m 2 , this is negative, meaning that the motion ofm 1 is being slowed down. But

there’s no friction or other such force to do this.

OR Ifm 1 =m 2 , this is zero, but there are still unbalanced forces causing these masses to accelerate.

(b) If the combination of masses is just right, for examplem 1 = 1kg,m 2 = 1kg, andM= 2kg,

the denominator is zero. The expression foraxblows up — a very serious problem.

OR IfM is very large compared to the other masses, the denominator is negative, meaning thataxis

negative and the acceleration is a braking. Without friction, this is impossible.

(c) IfMm 1 andm 2 , the numerator is mostly−M/ 2 and the denominator is mostly+M/ 2.

This makes the whole expression negative, meaning thatm 1 andm 2 are slowing down. There is no

friction to do this, and all the forces are the direction to cause acceleration toward positivex.

OR Ifm 2 =M/ 2 , this equals zero, saying that there is no acceleration, but in this system,axwill

always be positive.

The same picture, butwithfrictionμkbetweenm 1 and the table.

(a)ax=

m 2

m 2 +μkm 1 +M/ 2

g (b)ax=

m 2 −μkm 1

m 2 −M/ 2

g (c)ax=

m 2

m 2 +μkm 1 −M/ 2

g

(a) Ifμkis very large, this approaches zero. Large friction should causem 1 to brake to a halt

quickly with very large negativeax.

OR If there is no friction,μk= 0, thenm 1 plays no role in this result but if it is big then you know

that it will decrease the downward acceleration ofm 2.

(b) The denominator can vanish. Ifm 2 =M/ 2 this is nonsense.

(c) This suffers from both of the difficulties of (a) and (b).

Trajectory Example

When you toss an object straight up with an initial speedv 0 , you may expect an answer for the motion

as a function of time to be something like