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.

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 = \( \frac{R_bd_b}{d_a} \) = \( \frac{45 lbs. \times 6 ft.}{4 ft.} \) = \( \frac{270 ft⋅lb}{4 ft.} \) = 67.5 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 = \( \frac{R_ad_a}{d_b} \) = \( \frac{25 lbs. \times 7 ft.}{6 ft.} \) = \( \frac{175 ft⋅lb}{6 ft.} \) = 29.17 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 d_a, our missing value, and plugging in our variables yields:

d_a = \( \frac{R_bd_b}{R_a} \) = \( \frac{55 lbs. \times 7 ft.}{10 lbs.} \) = \( \frac{385 ft⋅lb}{10 lbs.} \) = 38.5 ft.

A gear train is two or more gears linked together. Gear trains are designed to increase or reduce the speed or torque outpout of a rotating system or change the direction of its output. The first gear in the chain is called the driver and the last gear in the chain the driven gear with the gears between them called idler gears.