Forces On An Inclined Plane

F f

FN

Fg

30°

Think about the forces as she sleds down the hill in the laundry basket (if it is not a non-frictionless surface). Can you label them?

What is the formula used to calculate Fg?

= mg

How many other forces are there? (Don’t make any up!)

Two; the normal force (FN) and the force of friction (Ff).

Fg

30°

Fgll

Fg ┴

The force due to gravity can be broken up into vectors using the tip to tail method. They are Fgll and Fg .┴

The formulas used to calculate the components are:

= Fgcos θ

= Fgsin θ

One way to remember when to use cosine or sine is to look at the formulas on your reference table. Ax = A cos θ (horizontal) and Ay = A sin θ (vertical). Just use the opposite function for that direction when on an incline. Fg ┴= Fgcos θ (vertical) and Fgll = Fgsin θ (horizontal).

• Other formulas you need to know:

–Ff = μFN (μ = coefficient of friction)

–ΣFx = Fnet = ma = Fgll – Ff (ΣFx = sum of the forces in the x- direction)

Example:If a 60 kg person is sledding down a hill on waxed skis (yes, it’s a laundry basket, but just pretend!) at a 30o incline at a constant velocity, calculate Fnet, FN, Fgll, Fg ┴ , Fg, and Ff.

30°

= 588.6N

The easiest thing to find first is the weight (Fg).

Use the weight to find it’s vectors.

= 509.73N

= 294.3N

Fg ┴= Fgcos θ Fgll = Fgsin θ

Fg= mg

Fg ┴= (588.6N)cos(30°)Fg ┴= (588.6N)(0.866)Fg ┴= 509.73N

Fgll = (588.6N)sin(30°)Fgll = (588.6N)(0.500)Fgll = 294.3N

Fg= (60kg)(9.81m/s2)Fg= 588.6N

Fg

Fgll

Fg ┴

What is the normal force equal to? (Remember, it has to be equal and opposite of another force.)

It is equal to Fg┴. The normal force is pushing down with 509.73N of force while the ground is pushing up with an equal amount of force.

Fgll

Fg ┴Fg

FN = 509.73N

30°

To find the force of friction, plug in the normal force and the coefficient of friction that says waxed ski on snow.

F f

Ff = μFN

This would be ________ friction because it is SLIDING.

kinetic

*Since coefficients of friction don’t have units, keep the unit of a Newton.*

Ff = (0.05)(509.73N)Ff = 25.49N

= 25.49N

The last thing to figure out with all of the forces is the net force.

Since she is going down the hill at a constant velocity, what does that mean?

***ZERO acceleration.***

Keep that in mind!

ΣFx = Fnet = ma = Fgll – Ff 0

Fnet = Fgll – Ff

Fnet = 294.3N – 25.49NFnet = 268.81N

30°

Fgll = 294.3N

F f = 25.49N

Subtract the force of friction because it is in the opposite direction of motion.