W_in = W_out
F_effort x d_effort = F_resistance x d_resistance

In this problem, the effort work is 260 ft⋅lb and the resistance force is 130 lbs. and we need to calculate the resistance distance:

W_in = F_resistance x d_resistance
260 ft⋅lb = 130 lbs. x d_resistance
d_resistance = \( \frac{260ft⋅lb}{130 lbs.} \) = 2 ft.

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. 

A concentrated load acts on a relatively small area of a structure, a static uniformly distributed load doesn't create specific stress points or vary with time, a dynamic load varies with time or affects a structure that experiences a high degree of movement, an impact load is sudden and for a relatively short duration and a non-uniformly distributed load creates different stresses at different locations on a structure.

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.

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 R_b, our missing value, and plugging in our variables yields:

R_b = \( \frac{R_ad_a}{d_b} \) = \( \frac{30 lbs. \times 9 ft.}{8 ft.} \) = \( \frac{270 ft⋅lb}{8 ft.} \) = 33.75 lbs.