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

R_a = \( \frac{R_bd_b}{d_a} \) = \( \frac{50 lbs. \times 5 ft.}{2 ft.} \) = \( \frac{250 ft⋅lb}{2 ft.} \) = 125 lbs.

f_Ad_A = f_Bd_B + f_Cd_C

For this problem, this equation becomes:

40 lbs. x 10 ft. = 50 lbs. x 2 ft. + f_C x 7 ft.

400 ft. lbs. = 100 ft. lbs. + f_C x 7 ft.

f_C = \( \frac{400 ft. lbs. - 100 ft. lbs.}{7 ft.} \) = \( \frac{300 ft. lbs.}{7 ft.} \) = 42.86 lbs.

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.

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. = 470 lbs. x 4 ft.
F_e = \( \frac{1880 ft⋅lb}{5 ft.} \) = 376 lbs.

The work-energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle. Simply put, work imparts kinetic energy to the matter upon which the work is being done.