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 = \( \frac{R_ad_a}{d_b} \) = \( \frac{50 lbs. \times 7 ft.}{9 ft.} \) = \( \frac{350 ft⋅lb}{9 ft.} \) = 38.89 lbs.

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 = \( \frac{R_ad_a}{d_b} \) = \( \frac{75 lbs. \times 1 ft.}{8 ft.} \) = \( \frac{75 ft⋅lb}{8 ft.} \) = 9.38 lbs.

f_Ad_A = f_Bd_B + f_Cd_C

For this problem, this equation becomes:

30 lbs. x 8 ft. = 45 lbs. x 3 ft. + f_C x 6 ft.

240 ft. lbs. = 135 ft. lbs. + f_C x 6 ft.

f_C = \( \frac{240 ft. lbs. - 135 ft. lbs.}{6 ft.} \) = \( \frac{105 ft. lbs.}{6 ft.} \) = 17.5 lbs.

The mechanical advantage of a wheel and axle is the input radius divided by the output radius:

MA = \( \frac{r_i}{r_o} \)

In this case, the input radius (where the effort force is being applied) is 5 and the output radius (where the resistance is being applied) is 6 for a mechanical advantage of \( \frac{5}{6} \) = 0.83

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 19 ft. = 370 lbs. x 3 ft.
F_e = \( \frac{1110 ft⋅lb}{19 ft.} \) = 58.4 lbs.