If you wish to move a 300lb barrell three feet vertically than you know that the vertical component of your incline plane is equal to 3 ft. Next you need to solve what angle the incline plane must be at in order to reduce the amount of force neecded to...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

If you wish to move a 300lb barrell three feet vertically than you know that the vertical component of your incline plane is equal to 3 ft. Next you need to solve what angle the incline plane must be at in order to reduce the amount of force neecded to lift the barrell from 300 lbs to 100lbs.

The formula for an inclined plane is defined as follows:

`MA=L/h=F_w/F_i=R/E`

L= length of the plane

h= height of the plane = 3 ft

R = weight of the barrell = 300 lb

E = force required to raise the barrell

Therefore:

`L=(Rh)/E=((300)(3))/100=9`

Therefore, the inclined plane must be 9 ft in length.

The angle that it is raised at to be 3 ft high is defined as:

`sintheta=h/L=3/9-gttheta=19.5`

**Therefore, a 9ft plank raised at an angle of 19.5 deg will move a 300 lb barrell vertically 3 ft if the force of 100 lbs is applied on the barrell.**

**Further Reading**