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justadream

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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?
 

Cawolf

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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!
 
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type12

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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

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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!

 
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Cawolf

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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!
 
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justadream

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@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

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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.
 
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