Normal force (F_N) represents the force a surface exerts when an object presses against it.

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 = \( \frac{R_ad_a}{R_b} \) = \( \frac{25 lbs. \times 5 ft.}{15 lbs.} \) = \( \frac{125 ft⋅lb}{15 lbs.} \) = 8.33 ft.

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 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.

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 = 70 ft.³, V₁ = 75 ft.³ and P₁ = 15.0 psi. Solving for P₂:

P₂ = \( \frac{P_1}{\frac{V_2}{V_1}} \) = \( \frac{15.0 psi}{\frac{70 ft.^3}{75 ft.^3}} \) = 16.1 psi