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      10-22-2007, 10:51 AM   #1
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halp!!!!!!physics

A block is at rest on an inclined plane whose elevation can be varied. The coefficient of static friction is s= 0.42, and the coefficient of kinetic friction is k = 0.18. The angle of elevation  is increased slowly from the horizontal. What is the acceleration of the block?

i know f=ma

now the question is what the equality of f would be.....ive tried the various identities and i end up with none of the answers provided....mgsintheta(fgx) or the identity of fgx-ffr(simple algebra doesnt work, so no)...


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      10-22-2007, 11:12 AM   #2
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Wow - its been a long time since my high school AP physics class. I did a project on down hill skiing which used these formulas.

I believe the incline has to first be enough to overcome the statisc friction point. At the point static friction is overcome, the kinetic friction will determine how the acceleration is slowed.

Good luck - let me know the answer. I'm interested
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      10-22-2007, 11:16 AM   #3
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ya...i got that....theta, the point at which it begins to slide, is 22.78 with these coefficients.....now the equality is what i need......im confused as to what the net force equation would be....im assuming that the only one it could be is fgx(component x of gravity)....setting it up as gsintheta=a;which results in a wrong answer
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      10-22-2007, 01:43 PM   #4
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gravity as the downward force - acceleration should be angle of descent subtracting the kinetic friction
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      10-22-2007, 02:05 PM   #5
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      10-22-2007, 02:11 PM   #6
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You need to solve the force vectors that act on the block into components that are either perpendicular or parallel to the surface of the incline plane. You can use the sine of the angle to determine the parallel force, and the cosine to determine perpendicular force.
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      10-22-2007, 03:00 PM   #7
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Quote:
Originally Posted by Scarecrows25 View Post
You need to solve the force vectors that act on the block into components that are either perpendicular or parallel to the surface of the incline plane. You can use the sine of the angle to determine the parallel force, and the cosine to determine perpendicular force.
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      10-22-2007, 03:23 PM   #8
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I think the answer is "C". Oh wait, I was guessing at your grade.

What's with all the homework posts??? Funny to read, but I'm tired of doing people's homework! Can't your parents help???
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      10-22-2007, 04:00 PM   #9
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That looks prety easy. Remember to either make a free body diagram or break the forces into components.
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      10-22-2007, 05:37 PM   #10
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Quote:
Originally Posted by TiAg335i View Post
That looks prety easy. Remember to either make a free body diagram or break the forces into components.
yup! It's all in the fee biddy damn! Draw the fee biddy damn!

I had a Russian prof who couldn't say "free body diagram" to save his life. I'll remember that guy to my dying day! But hey, I can draw the fee biddy damn!
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      10-22-2007, 05:38 PM   #11
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