F = k h

^{x}v

^{y}r

^{z })

Solution: Use the table for dimensions to write them out for each factor:

For Force : F has dimensions M L T

^{-2}

^{}

^{}Velocity v has dimensions L

^{ }T

^{- 1}

^{}

^{}h has dimensions M L

^{-1}T

^{-1}

^{}

^{}r has dimension L

Then, equating dimensions on both sides:

M L

^{ }T

^{– 2}= [M L

^{-1}T

^{-1}]

^{x}[LT

^{- 1}]

^{y}[L]

^{z}

^{}

^{}

We next equate

**indices**for M, L, and T on both sides:
For M: 1 = x

For L: 1 = -x - y + z

For T: -2 = -x – y

Next, solve for each of the indices:

a) x = 1

b) y = 1

c) z = 3

Finally: F = k h v r

^{ 3}
2) Repeat the exercise above to obtain the equation for the period (T) of a pendulum’s swing. (Hint: The pendulum's

*period*T should depend on its length L and the acceleration of gravity, g.)
Solution: Write out the provisional eqn.:

t = g

^{x }M^{ }^{y}r^{z}
Write out dimensions for each side:

[T] = [ L

^{ }T^{- 2 }] [M]y {L]^{z}
Solving for the indices:

1 = - 2x or x = - 1/2

0 = y

0 = x + z or x = - z = - (-1/2) = 1/2

Then:

t = k g - 1/2

^{ }ℓ 1/2
(k will be found by separate analysis to be: 2 p)

So that:

t = 2 p

**Ö**ℓ_{ }/**Ö**g_{ }
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