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 = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) = \( \frac{30}{12} \) \( \frac{12}{6} \) = \( \frac{30}{6} \) = 5

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 = \( \frac{R_bd_b}{d_a} \) = \( \frac{25 lbs. \times 7 ft.}{4 ft.} \) = \( \frac{175 ft⋅lb}{4 ft.} \) = 43.75 lbs.

Specific gravity is the ratio of the density of equal volumes of a substance and water and is measured by a hyrdometer.

Mechanical advantage is a measure of the force amplification achieved by using a tool, mechanical device or machine system. Such a device utilizes input force and trades off forces against movement to amplify and/or change its direction.

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.