The mechanical advantage of a wheel and axle is the input radius divided by the output radius:

MA = \( \frac{r_i}{r_o} \)

In this case, the input radius (where the effort force is being applied) is 6 and the output radius (where the resistance is being applied) is 3 for a mechanical advantage of \( \frac{6}{3} \) = 2.0

MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{100 lbs.}{2.0} \) = 50 lbs.

Newton's Law of Univeral Gravitation defines the general formula for the attraction of gravity between two objects:  \(\vec{F_{g}} = { Gm_{1}m_{2} \over r^2}\) . In the specific case of an object falling toward Earth, the acceleration due to gravity (g) is approximately 9.8 m/s². 

Mechanical advantage (MA) can be calculated knowing only the distance the effort (blue arrow) moves and the distance the resistance (green box) moves. The equation is:

MA = \( \frac{E_d}{R_d} \)

where E_d is the effort distance and R_d is the resistance distance. For this problem, the equation becomes:

MA = \( \frac{7 ft.}{3.5 ft.} \) = 2

You might be wondering how having an effort distance of 2 times the resistance distance is an advantage. Remember the principle of moments. For a lever in equilibrium the effort torque equals the resistance torque. Because torque is force x distance, if the effort distance is 2 times the resistance distance, the effort force must be \( \frac{1}{2} \) the resistance force. You're trading moving 2 times the distance for only having to use \( \frac{1}{2} \) the force.

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

d_a = \( \frac{R_bd_b}{R_a} \) = \( \frac{70 lbs. \times 5 ft.}{55 lbs.} \) = \( \frac{350 ft⋅lb}{55 lbs.} \) = 6.36 ft.

Mechanics deals with motion and the forces that produce motion.