Drag is friction that opposes movement through a fluid like liquid or air. The amount of drag depends on the shape and speed of the object with slower objects experiencing less drag than faster objects and more aerodynamic objects experiencing less drag than those with a large leading surface area.

This problem describes an inclined plane and, for an inclined plane, the effort force multiplied by the effort distance equals the resistance force multipied by the resistance distance:

F_ed_e = F_rd_r

Plugging in the variables from this problem yields:

F_e x 5 ft. = 310 lbs. x 2 ft.
F_e = \( \frac{620 ft⋅lb}{5 ft.} \) = 124 lbs.

Mechanical advantage is resistance force divided by effort force:

MA = \( \frac{F_r}{F_e} \) = \( \frac{150 lbs.}{30 lbs.} \) = 5

Mechanical advantage (MA) is the ratio by which effort force relates to resistance force. If both forces are known, calculating MA is simply a matter of dividing resistance force by effort force:

MA = \( \frac{F_r}{F_e} \) = \( \frac{7 ft.}{7.78 ft.} \) = 0.9

In this case, the mechanical advantage is less than one meaning that each unit of effort force results in just 0.9 units of resistance force. However, a third class lever like this isn't designed to multiply force like a first class lever. A third class lever is designed to multiply distance and speed at the resistance by sacrificing force at the resistance. Different lever styles have different purposes and multiply forces in different ways.

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 = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) = \( \frac{22}{12} \) \( \frac{12}{8} \) = \( \frac{22}{8} \) = 2.8