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

d_b = \( \frac{R_ad_a}{R_b} \) = \( \frac{60 lbs. \times 3 ft.}{50 lbs.} \) = \( \frac{180 ft⋅lb}{50 lbs.} \) = 3.6 ft.

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

Mass is a measure of the amount of matter in an object.  In general, larger objects have larger mass than smaller objects but mass ultimately depends on how compact (dense) a substance is.

The mechanical advantage of a gear train is its gear ratio. The gear ratio (V_r) is the product of the gear ratios between the pairs of meshed gears. Let N represent the number of teeth for each gear:

V_r = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) \( \frac{N_3}{N_4} \) ... \( \frac{N_n}{N_{n+1}} \)

In this problem, we have three gears so the equation becomes:

V_r = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) = \( \frac{24}{14} \) \( \frac{14}{4} \) = \( \frac{24}{4} \) = 6