Inclined Planes

• An inclined plane is a type of simple machine

• An inclined plane is a large and flat object that is tilted so that one end is higher than the other

What is an Inclined Plane?

Real World Applications of Inclined Planes

Hadley Canal in Massachusetts used Inclined Plane engineering around 1800’s to raise and lower boats over Great Falls

• The greater the angle of the inclined surface, the faster an object will slide down the incline

• There are always at least 2 forces acting on an object on an inclined plane Fgrav and Fnorm

Background Information

The normal force is always perpendicular to the inclined surface

The gravitational force (WEIGHT) is always in downward direction

Solving for the Forces1. Break down Fgrav into its x and y components

• F║ = mgsinΘ and F┴ = mgcosΘ

HINTS…1. F┴ is always equal and opposite Fnorm

2. When there is NO FRICTION, F║ is the net force

2. Deduce the net force and solve for other unknowns including acceleration and µ

What is the net force for the

Inclined Plane diagram to the

right?

Example to Do Together!

Solve for all unknowns listed. The crate has a mass of 100 kg and the coefficient of friction between the crate and the incline is 0.3

F║ = Ffrict = a =

F┴ = Fnet =

1. Break down Fgrav into its components

F║ = mgsinΘF║ = (100)(9.8)sin30°F║ = 490 N

F┴ = mgcosΘF┴ = (100)(9.8)cos30°F┴ = 849 N

Fgrav = mgFgrav = (100)(9.8)Fgrav = 980 N

2. This example has friction, therefore let’s solve for Ffrict next

Ffrict = µ•Fnorm

Ffrict = 0.3•849

Ffrict = 255 N

F ║ = 490 N

F ┴ = 849 N

Remember,

Fnorm is

ALWAYS equal and

opposite F┴

849

980

3. Now, from our results we can deduce Fnet

Fnet = F║ - Ffrict

Fnet = 490 – 255

Fnet = 235 NF ║ = 490 N

F ┴ = 849 N

849

255

980Remember, when there is

NO FRICTION Fnet = F║

4. Lastly, we need to solve for the acceleration

Fnet = ma

a = Fnet ÷ m

a = 235 ÷ 100

a = 2.35 m/s2

F ║ = 490 N

F ┴ = 849 N

849

255

980Fnet = 235 N

Answers

1. F║ = mgsinΘ

F║ = (100)(9.8)sin30°

F║ = 490 N

2. F┴ = mgcosΘ

F┴ = (100)(9.8)cos30°

F┴ = 849 N

3. F┴ = Fnorm therefore Fnorm = 849 N

Answers Continued

4. Ffrict = μFnorm

Ffrict = .3(849) = 255 N

5. Fnet = F║ - Ffrict

Fnet = 490 – 255 = 235 N

6. F = ma so a = F/m

a = 235/100 = 2.35 m/s2