Tension is a force that stretches or elongates something. When a cable or rope is used to pull an object, for example, it stretches internally as it accepts the weight that it's moving. Although tension is often treated as applying equally to all parts of a material, it's greater at the places where the material is under the most stress.

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 = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) = \( \frac{28}{16} \) \( \frac{16}{8} \) = \( \frac{28}{8} \) = 3.5

Gravitational potential energy is energy by virtue of gravity. The higher an object is raised above a surface the greater the distance it must fall to reach that surface and the more velocity it will build as it falls. For gravitational potential energy, PE = mgh where m is mass (kilograms), h is height (meters), and g is acceleration due to gravity which is a constant (9.8 m/s²).

To balance this lever the torques on each side of the fulcrum must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:

R_ad_a = R_bd_b

Solving for d_a, our missing value, and plugging in our variables yields:

d_a = \( \frac{R_bd_b}{R_a} \) = \( \frac{20 lbs. \times 9 ft.}{20 lbs.} \) = \( \frac{180 ft⋅lb}{20 lbs.} \) = 9 ft.

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 20 ft. = 370 lbs. x 5 ft.
F_e = \( \frac{1850 ft⋅lb}{20 ft.} \) = 92.5 lbs.