Because this lever is in equilibrium, we know that the effort force at the blue arrow is equal to the resistance weight of the green box. For a lever that's in equilibrium, one method of calculating mechanical advantage (MA) is to divide the length of the effort arm (E_a) by the length of the resistance arm (R_a):

MA = \( \frac{E_a}{R_a} \) = \( \frac{8 ft.}{2 ft.} \) = 4

When a lever is in equilibrium, the torque from the effort and the resistance are equal. The equation for equilibrium is R_ad_a = R_bd_b where a and b are the two points at which effort/resistance is being applied to the lever.

In this problem, R_a and R_b are such that the lever is in equilibrium meaning that some multiple of the weight of the green box is being applied at the blue arrow. For a lever, this multiple is a function of the ratio of the distances of the box and the arrow from the fulcrum. That's why, for a lever in equilibrium, only the distances from the fulcrum are necessary to calculate mechanical advantage.

If the lever were not in equilibrium, you would first have to calculate the forces and distances necessary to put it in equilibrium and then divide E_a by R_a to get the mechanical advantage.

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 = \( \frac{R_bd_b}{d_a} \) = \( \frac{55 lbs. \times 6 ft.}{5 ft.} \) = \( \frac{330 ft⋅lb}{5 ft.} \) = 66 lbs.

Torque measures force applied during rotation: τ = rF.  Torque (τ, the Greek letter tau) = the radius of the lever arm (r) multiplied by the force (F) applied. Radius is measured from the center of rotation or fulcrum to the point at which the perpendicular force is being applied. The resulting unit for torque is newton-meter (N-m) or foot-pound (ft-lb).

This problem describes a second-class lever and, for a second class lever, 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 3 ft. = 270 lbs x 0.5 ft
F_e = 135 ft-lb. / 3 ft 
F_e = 45 lbs

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).