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 18 ft. = 220 lbs. x 1 ft.
F_e = \( \frac{220 ft⋅lb}{18 ft.} \) = 12.2 lbs.

Newton's Law of Univeral Gravitation defines the general formula for the attraction of gravity between two objects:  \(\vec{F_{g}} = { Gm_{1}m_{2} \over r^2}\) . In the specific case of an object falling toward Earth, the acceleration due to gravity (g) is approximately 9.8 m/s². 

This problem describes a second-class lever and, for a second class lever, 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 3 ft. = 270 lbs x 0.5 ft
F_e = 135 ft-lb. / 3 ft 
F_e = 45 lbs

A second-class lever is used to increase force on an object in the same direction as the force is applied. This lever requires a smaller force to lift a larger load but the force must be applied over a greater distance. The fulcrum is placed at one end of the lever and mechanical advantage increases as the object being lifted is moved closer to the fulcrum or the length of the lever is increased. An example of a second-class lever is a wheelbarrow.

The Joule (J) is the standard unit of energy and has the unit \({kg \times m^2} \over s^2\).