How Do You Derive the Normal and Static Frictional Forces on an Inclined Plane?

[acspin]

Homework Statement

Derive the equation for the normal force (N) and the static frictional force (F) in terms of the weight (Mg). $$\beta$$ is the angle of inclination.

Homework Equations

http://img62.imageshack.us/img62/5559/normalforcediagram.png

The Attempt at a Solution

- For N I used N = -mg Sin $$\beta$$. I'm also thinking though, that since friction is present my equation for N should be:

N = -mg Sin $$\beta$$ - F

For Static Frictional force I get:

F = mu_s N

I feel confident about the equation for static friction. I'm not sure if I should be using (-mg) or (mg) for the normal force though.

 

[collinsmark]

- For N I used N = -mg Sin $$\beta$$.

Not quite. http://www.websmileys.com/sm/sad/021.gif

Think about it in terms of extremes. For example, consider the case where $\beta$ = 0. In that case, the normal force is certainly not 0!

You might wish to rearrange the force vectors into a right triangle. Ensure N and F intersect at the 90^o corner. (You might want to use a small value of $\beta$ when you do this, to make it obvious which corner corresponds to $\beta$.) After you draw this triangle, the correct trigonometric function to use in your Normal force equation should become obvious.

I'm also thinking though, that since friction is present my equation for N should be:

N = -mg Sin $$\beta$$ - F

... The frictional force is perpendicular to the normal force. So that tells you...

For Static Frictional force I get:

F = mu_s N

That I like.

I feel confident about the equation for static friction. I'm not sure if I should be using (-mg) or (mg) for the normal force though.

The normal force has its very own direction. And sometimes that specific direction becomes important. But for this particular problem, I think you only need to worry about the magnitude. So for this particular problem, I would trash the negative sign in your normal force equation.

 

[acspin]

Wow, I made such stupid mistakes. I'm studying for my final and I'm really dusty on this subject.

So, the Normal force is equal to the perpendicular component of the weight vector, thus:

N = mg cos $$\beta$$

Since objects don't accelerate perpendicular to the incline, then I only need to consider friction for Net Force calculations and not for calculating N, correct?

Thank you so much for your help!

If I were to express the static frictional force in terms of Mg, would it be ok if i did this:

F = mu_s (mg cos $$\beta$$)

 

[collinsmark]

Wow, I made such stupid mistakes. I'm studying for my final and I'm really dusty on this subject.

So, the Normal force is equal to the perpendicular component of the weight vector, thus:

N = mg cos $$\beta$$

The normal force is always perpendicular to the surface (meaning it's also perpendicular to the frictional force too). But yes, you're equation looks good!

And by the way, I wouldn't call them "stupid" mistakes, just mistakes. Everybody makes mistakes. I make mistakes all the time. Everybody does. http://www.websmileys.com/sm/crazy/146.gif

Since objects don't accelerate perpendicular to the incline, then I only need to consider friction for Net Force calculations and not for calculating N, correct?

Yes that right! In a case where an object is sliding down a constant-angle incline, its acceleration is perpendicular to the Normal of the incline, so its acceleration has no affect on the normal force (this, of course, is assuming that the incline itself is not accelerating, which is a whole different story.) So yes. (Of course there might be a component of the gravitational force vector that fits into the equation of the net force, along with the frictional force vector. But not the normal force vector itself.)

Just keep in mind that the magnitude of the frictional force is a function of the normal force, so they are related. But when summing the vectors to determine the net force of an accelerating body on a (non-accelerating, constant-angle) incline, the normal force vector itself is not a part of it.

(Of course this particular problem deals with static friction, implying that nothing is moving at all. But in general, you're right.)

Thank you so much for your help!

If I were to express the static frictional force in terms of Mg, would it be ok if i did this:

F = mu_s (mg cos $$\beta$$)

Looks good to me.