# Inclined Plane Question

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##### Full Member
7+ Year Member

Mcat-review says:

"When something is laying still on an inclined plane, the normal force and friction force adds up in a vector fashion to equal the weight."

How is this true? Let's set up a random inclined plane with angle = 30 degrees. Place a block of 10kg on it.

Friction Force = Component of Gravity downwards = mgsin(30) = 50 N

Normal Force = mgcos(30) = 86 N

Total Weight = mg = 10*10 = 100 N

The sum of frictional force and normal force (50 + 86) is not equal to the total weight.

Am I missing something?

#### type12

##### Full Member
7+ Year Member
Yes, you are missing vectors. If it didn't counter exactly the weight, the object would accelerate.

#### Cawolf

##### PGY-2
7+ Year Member
Just like @type12 said, you need to break the weight down into vectors also - with directions based on the plane.

It will be equal in value to the forces you listed before - or else it wouldn't be static!

##### Full Member
7+ Year Member
@type12
@Cawolf

Can you show me at least the first step in how to do that?

#### type12

##### Full Member
7+ Year Member
tilted x-axis: Friction force (u_s * N) = "vertical" component of gravity (mg sin theta)
tilted y-axis: Normal force (N) = "horizontal" component of gravity (mg cos theta)

#### Cawolf

##### PGY-2
7+ Year Member
You have the answer, you just need to express the mass in relation to the angle of the plane.

I am back working in the library with easy whiteboard access so here you go!

1 user

#### type12

##### Full Member
7+ Year Member
You have the answer, you just need to express the mass in relation to the angle of the plane.

I am back working in the library with easy whiteboard access so here you go!

Lame, you cheated by providing an image.

#### Cawolf

##### PGY-2
7+ Year Member
Haha - I had already drawn it when I first saw the post. I like to sketch out things to review for myself and make sure I give an accurate answer.

After that when @justadream wanted further info, it was as easy as snapping a pic on my phone in my photobucket app!

1 user

##### Full Member
7+ Year Member
@Cawolf

So frictional force = mgsin(3o) = 50
Normal Force = mgcos(30) = 86

The weight has x-component of mgsin(30) = 50
The weight has y-component of mgcos(30) = 86

Total weight = sqrt(50^2 + 86^2) = 100

#### Cawolf

##### PGY-2
7+ Year Member
Yes!

It is important to remember that you can't just add two scalar components of a vector - the magnitude is what we commonly refer to. You nailed it.