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

d_a = \( \frac{R_bd_b}{R_a} \) = \( \frac{45 lbs. \times 1 ft.}{25 lbs.} \) = \( \frac{45 ft⋅lb}{25 lbs.} \) = 1.8 ft.

For any given surface, the coefficient of static friction is higher than the coefficient of kinetic friction. More force is required to initally get an object moving than is required to keep it moving. Additionally, static friction only arises in response to an attempt to move an object (overcome the normal force between it and the surface).

A third-class lever is used to increase distance traveled by an object in the same direction as the force applied. The fulcrum is at one end of the lever, the object at the other, and the force is applied between them. This lever does not impart a mechanical advantage as the effort force must be greater than the load but does impart extra speed to the load. Examples of third-class levers are shovels and tweezers.

Collinear forces act along the same line of action, concurrent forces pass through a common point and coplanar forces act in a common plane.

This problem describes an inclined plane and, for an inclined plane, 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 13 ft. = 270 lbs. x 1 ft.
F_e = \( \frac{270 ft⋅lb}{13 ft.} \) = 20.8 lbs.