Inclined Plane Question

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

justadream

Full Member
10+ Year Member
Joined
Apr 29, 2011
Messages
2,171
Reaction score
862
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?

Members don't see this ad.
 
Yes, you are missing vectors. If it didn't counter exactly the weight, the object would accelerate.
 
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!
 
Members don't see this ad :)
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)
 
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!

 
  • Like
Reactions: 1 user
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. :p
 
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!
 
  • Like
Reactions: 1 user
@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
 
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.
 
Top