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.

A second-class lever is used to increase force on an object in the same direction as the force is applied. This lever requires a smaller force to lift a larger load but the force must be applied over a greater distance. The fulcrum is placed at one end of the lever and mechanical advantage increases as the object being lifted is moved closer to the fulcrum or the length of the lever is increased. An example of a second-class lever is a wheelbarrow.

An inclined plane is a simple machine that reduces the force needed to raise an object to a certain height. Work equals force x distance and, by increasing the distance that the object travels, an inclined plane reduces the force necessary to raise it to a particular height. In this case, the mechanical advantage is to make the task easier. An example of an inclined plane is a ramp.

According to Boyle's Law, pressure and volume are inversely proportional:

\( \frac{P_1}{P_2} \) = \( \frac{V_2}{V_1} \)

In this problem, V₂ = 40 ft.³, V₁ = 60 ft.³ and P₁ = 12.0 psi. Solving for P₂:

P₂ = \( \frac{P_1}{\frac{V_2}{V_1}} \) = \( \frac{12.0 psi}{\frac{40 ft.^3}{60 ft.^3}} \) = 18 psi

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

d_b = \( \frac{R_ad_a}{R_b} \) = \( \frac{30 lbs. \times 3 ft.}{50 lbs.} \) = \( \frac{90 ft⋅lb}{50 lbs.} \) = 1.8 ft.