halp!!!!!!physics

E(@M# · 10-22-2007, 11:51 AM

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)...

HALP!!!!!!!

MCS · 10-22-2007, 12:12 PM

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

E(@M# · 10-22-2007, 12:16 PM

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

MCS · 10-22-2007, 02:43 PM

gravity as the downward force - acceleration should be angle of descent subtracting the kinetic friction

Brunotheboxer · 10-22-2007, 03:05 PM

7

Scarecrows25 · 10-22-2007, 03:11 PM

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.

gonzo · 10-22-2007, 04:00 PM

Quote:

Originally Posted by Scarecrows25

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.

..

TurboFan · 10-22-2007, 04:23 PM

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???

TiAg335i · 10-22-2007, 05:00 PM

That looks prety easy. Remember to either make a free body diagram or break the forces into components.

TurboFan · 10-22-2007, 06:37 PM

Quote:

Originally Posted by TiAg335i

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!

!Xoible · 10-22-2007, 06:38 PM

wtf am i doing here?

10-22-2007, 11:51 AM	#1
E(@M# Second Lieutenant 6 Rep 207 Posts Drives: 94' corolla Join Date: Mar 2007 Location: FL iTrader: (0)	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)... HALP!!!!!!!
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10-22-2007, 12:12 PM	#2
MCS e90 newbie 38 Rep 448 Posts Drives: 2012 M3 E93 Join Date: Jul 2005 Location: Northern NJ iTrader: (2) Garage List 2006 330i [0.00]	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 __________________ E93 IL Blue M3 ZPP, ZCW
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10-22-2007, 12:16 PM	#3
E(@M# Second Lieutenant 6 Rep 207 Posts Drives: 94' corolla Join Date: Mar 2007 Location: FL iTrader: (0)	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, 02:43 PM	#4
MCS e90 newbie 38 Rep 448 Posts Drives: 2012 M3 E93 Join Date: Jul 2005 Location: Northern NJ iTrader: (2) Garage List 2006 330i [0.00]	gravity as the downward force - acceleration should be angle of descent subtracting the kinetic friction __________________ E93 IL Blue M3 ZPP, ZCW
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10-22-2007, 03:05 PM	#5
Brunotheboxer Colonel 399 Rep 2,984 Posts Drives: 2014 Shelby GT500. Join Date: Jan 2006 Location: Boston/Cape Cod iTrader: (0)	7 __________________ "...I maybe drunk, but you're ugly, and I can sober up..."
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10-22-2007, 03:11 PM	#6
Scarecrows25 Private First Class 10 Rep 191 Posts Drives: 2007 E90 - Jet Black Join Date: Jul 2007 Location: Dallas, TX iTrader: (1)	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. __________________ Scarecrows25 Jet Black E90 328i / 6MT / Sport / Xenons / Split Fold Rear Seat / OEM Chrome Exhaust Tips
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10-22-2007, 04:23 PM	#8
TurboFan Ski bum 327 Rep 6,198 Posts Drives: sideways Join Date: Aug 2007 Location: Knee deep in the pow iTrader: (8)	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??? __________________ 1999 e46 328i Ti Silver / Black[retired] 2007 e90 335xi Jet Black / Black[retired] 2011 e70 X5 35d Vermillion Red / Cinnamon 2011 e92 M3 LeMans / Fox Red extended
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10-22-2007, 05:00 PM	#9
TiAg335i Trance Life 333 Rep 4,706 Posts Drives: 2007 E92 335i Join Date: Jan 2007 Location: Long Island, NY iTrader: (4)	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, 06:38 PM	#11
!Xoible Banned 875 Rep 46,010 Posts Drives: .... Join Date: Nov 2006 Location: . iTrader: (4) Garage List 2008 M3 [4.00] 2007 335i [9.00] 2008 528i [8.00] 2006 Infiniti - G35 ... [8.00]	wtf am i doing here?
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