Question

Lecture 1: Challenge Problem #1:

Could someone explain to me the steps in solving this problem, namely these:

Therefore we have the following set of algebraic relations
0 = -3V +W + X +2(Y +Z)
0 =V
1= -2X

Thanks.

Answer 1

Adi, I think I understand what youre asking, so here goes: These three equations were derived from eqn 1.3. The goal here is to make all the units match up. Because the left side of (1.3) has no units of length or mass, the right side must have all units of length and mass disappear. Im going to add in units of mass and length on the left side for clarity, but they dont matter since they are raised to the zero power, which is equivalent to one. \[M^0*L^0*T^{1}=[M^V*L^{-3V}]*[L^W]*[L^X*T^{-2X}]*[L^{2Y}]*[L^{2Z}]\] By the commutative law of multiplication:
\[M^0*L^0*T^1=M^V*L^{-3V}*L^W*L^X*L^{2Y}*L^{2Z}*T^{-2Z}\]Combining exponents of like terms gives\[M^0*L^0*T^1=M^V*L^{-3V+W+X+2Y+2Z}*T^{-2X}\]
In order to make the right side match up with the left, we get three equations:
\[M: 0=V\]
\[L: 0=-3V+W+X+2Y+2Z\]
\[T: 1=-2X\]
These three equations are the constraints that will make eqn 1.1 true. Using algebra and some further assumptions about the physical situation, you will be able to solve this problem! Let me know if this helps! Good luck!