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

d_b = \( \frac{R_ad_a}{R_b} \) = \( \frac{25 lbs. \times 5 ft.}{65 lbs.} \) = \( \frac{125 ft⋅lb}{65 lbs.} \) = 1.92 ft.

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.

A concentrated load acts on a relatively small area of a structure, a static uniformly distributed load doesn't create specific stress points or vary with time, a dynamic load varies with time or affects a structure that experiences a high degree of movement, an impact load is sudden and for a relatively short duration and a non-uniformly distributed load creates different stresses at different locations on a structure.

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 = \( \frac{R_ad_a}{R_b} \) = \( \frac{5 lbs. \times 3 ft.}{5 lbs.} \) = \( \frac{15 ft⋅lb}{5 lbs.} \) = 3 ft.

The mechanical advantage of a gear train is its gear ratio. The gear ratio (V_r) is the product of the gear ratios between the pairs of meshed gears. Let N represent the number of teeth for each gear:

V_r = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) \( \frac{N_3}{N_4} \) ... \( \frac{N_n}{N_{n+1}} \)

In this problem, we have three gears so the equation becomes:

V_r = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) = \( \frac{22}{10} \) \( \frac{10}{8} \) = \( \frac{22}{8} \) = 2.8