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

R_a = \( \frac{R_bd_b}{d_a} \) = \( \frac{30 lbs. \times 3 ft.}{8 ft.} \) = \( \frac{90 ft⋅lb}{8 ft.} \) = 11.25 lbs.

Mechanical advantage is resistance force divided by effort force:

MA = \( \frac{F_r}{F_e} \) = \( \frac{540 lbs.}{90 lbs.} \) = 6

For any given surface, the coefficient of static friction is higher than the coefficient of kinetic friction. More force is required to initally get an object moving than is required to keep it moving. Additionally, static friction only arises in response to an attempt to move an object (overcome the normal force between it and the surface).

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 = \( \frac{R_ad_a}{d_b} \) = \( \frac{75 lbs. \times 4 ft.}{6 ft.} \) = \( \frac{300 ft⋅lb}{6 ft.} \) = 50 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.