## College Physics (4th Edition)

Let $F_1$ be the force required to lift an object without using an incline. Then $F_1 = mg$. Let $F_2$ be the force required to slide an object up a frictionless incline. $F_2$ is equal in magnitude to the component of the object's weight that is directed down the incline. Then $F_2 = mg~sin~\theta$, where $\theta$ is the angle the incline makes above the horizontal. Note that $sin~\theta = \frac{h}{d}$ where $h$ is the height of the incline and $d$ is the length of the incline. Then $F_2 = \frac{mgh}{d}$ We can find the mechanical advantage: $\frac{F_1}{F_2} = \frac{mg}{(\frac{mgh}{d})} = \frac{d}{h}$ Therefore, the mechanical advantage is the length of the incline divided by the height of the incline.