The mechanical advantage of a gear train is its gear ratio. The gear ratio (V_r) is the product of the gear ratios between the pairs of meshed gears. Let N represent the number of teeth for each gear:

V_r = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) \( \frac{N_3}{N_4} \) ... \( \frac{N_n}{N_{n+1}} \)

In this problem, we have three gears so the equation becomes:

V_r = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) = \( \frac{30}{12} \) \( \frac{12}{4} \) = \( \frac{30}{4} \) = 7.5

Normal force arises on a flat horizontal surface in response to an object's weight pressing it down. Consequently, normal force is generally equal to the object's weight.

Coefficient of friction (μ) represents how much two materials resist sliding across each other.  Smooth surfaces like ice have low coefficients of friction while rough surfaces like concrete have high μ.

The mechanical advantage of a wheel and axle is the input radius divided by the output radius:

MA = \( \frac{r_i}{r_o} \)

In this case, the input radius (where the effort force is being applied) is 8 and the output radius (where the resistance is being applied) is 3 for a mechanical advantage of \( \frac{8}{3} \) = 2.67

MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{55 lbs.}{2.67} \) = 20.6 lbs.

As an object falls, its potential energy is converted into kinetic energy. The principle of conservation of mechanical energy states that, as long as no other forces are applied, total mechanical energy (PE + KE) of the object will remain constant at all points in its descent.