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 13 ft. = 330 lbs. x 4 ft.
F_e = \( \frac{1320 ft⋅lb}{13 ft.} \) = 101.5 lbs.

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 = \( \frac{R_bd_b}{d_a} \) = \( \frac{60 lbs. \times 5 ft.}{1 ft.} \) = \( \frac{300 ft⋅lb}{1 ft.} \) = 300 lbs.

The mechanical advantage (amount of change in speed or torque) of connected gears is proportional to the number of teeth each gear has. Called gear ratio, it's the ratio of the number of teeth on the larger gear to the number of teeth on the smaller gear.  For example, a gear with 12 teeth connected to a gear with 9 teeth would have a gear ratio of 4:3.

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.

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