The mechanical advantage (MA) of a wedge is its length divided by its thickness:

MA = \( \frac{l}{t} \) = \( \frac{28 in.}{4 in.} \) = 7

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 = \( \frac{R_bd_b}{d_a} \) = \( \frac{25 lbs. \times 9 ft.}{6 ft.} \) = \( \frac{225 ft⋅lb}{6 ft.} \) = 37.5 lbs.

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.

A wheel and axle uses two different diameter wheels mounted to a connecting axle. Force is applied to the larger wheel and large movements of this wheel result in small movements in the smaller wheel. Because a larger movement distance is being translated to a smaller distance, force is increased with a mechanical advantage equal to the ratio of the diameters of the wheels. An example of a wheel and axle is the steering wheel of a car.

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 = \( \frac{R_bd_b}{R_a} \) = \( \frac{65 lbs. \times 5 ft.}{40 lbs.} \) = \( \frac{325 ft⋅lb}{40 lbs.} \) = 8.13 ft.