W_in = W_out
F_effort x d_effort = F_resistance x d_resistance

In this problem, the effort work is 800 ft⋅lb and the resistance force is 200 lbs. and we need to calculate the resistance distance:

W_in = F_resistance x d_resistance
800 ft⋅lb = 200 lbs. x d_resistance
d_resistance = \( \frac{800ft⋅lb}{200 lbs.} \) = 4 ft.

A third-class lever is used to increase distance traveled by an object in the same direction as the force applied. The fulcrum is at one end of the lever, the object at the other, and the force is applied between them. This lever does not impart a mechanical advantage as the effort force must be greater than the load but does impart extra speed to the load. Examples of third-class levers are shovels and tweezers.

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.

Dead load is the weight of the building and materials, live load is additional weight due to occupancy or use, snow load is the weight of accumulated snow on a structure and wind load is the force of wind pressures against structure surfaces.

To balance this lever the torques at the green box and the blue arrow 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 R_b, our missing value, and plugging in our variables yields:

R_b = \( \frac{R_ad_a}{d_b} \) = \( \frac{70 lbs. \times 7 ft.}{8 ft.} \) = \( \frac{490 ft⋅lb}{8 ft.} \) = 61.25 lbs.