Drag is friction that opposes movement through a fluid like liquid or air. The amount of drag depends on the shape and speed of the object with slower objects experiencing less drag than faster objects and more aerodynamic objects experiencing less drag than those with a large leading surface area.

Connected gears of different numbers of teeth are used together to change the rotational speed and torque of the input force. If the smaller gear drives the larger gear, the speed of rotation will be reduced and the torque will increase. If the larger gear drives the smaller gear, the speed of rotation will increase and the torque will be reduced.

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 8 and the output radius (where the resistance is being applied) is 7 for a mechanical advantage of \( \frac{8}{7} \) = 1.14

MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{70 lbs.}{1.14} \) = 61.4 lbs.

W_in = W_out
F_effort x d_effort = F_resistance x d_resistance

In this problem, the effort work is 240 ft⋅lb and the resistance force is 120 lbs. and we need to calculate the resistance distance:

W_in = F_resistance x d_resistance
240 ft⋅lb = 120 lbs. x d_resistance
d_resistance = \( \frac{240ft⋅lb}{120 lbs.} \) = 2 ft.

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 = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) = \( \frac{26}{20} \) \( \frac{20}{6} \) = \( \frac{26}{6} \) = 4.3