In the challenge question for "1. Units and Dimensional Analysis" step 1.4 of the solution asserts: 0 = -3V +W + X + 2(Y + Z). I understand -3V corresponds to L^-3 in [M*L^-3], but don't understand why X is just 1X, why not -2X because of [L*T^-2]. Also, how are the ratios determined if there are more than one dimension? Maybe I'm not understanding how they are getting the algebraic relations in 1.4 - see this document http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/introduction-to-mechanics/units-and-dimensional-analysis/MIT8_01SC_problems01_soln.pdf

Question

In the challenge question for "1. Units and Dimensional Analysis" step 1.4 of the solution asserts: 0 = -3V +W + X + 2(Y + Z).  I understand  -3V corresponds to L^-3 in [M*L^-3], but don't understand why X is just 1X, why not -2X because of [L*T^-2].  Also, how are the ratios determined if there are more than one dimension?  Maybe I'm not understanding how they are getting the algebraic relations in 1.4 - see this document http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/introduction-to-mechanics/units-and-dimensional-analysis/MIT8_01SC_problems01_soln.pdf

anonymous · Answer

see,to solve for X they used the time dimension which was -2X on RHS and since it was the only time dimension in RHS,therefore [T] = [T^-2]^X
                                                  or     1=-2X