<p>I have a very difficult physics problem for a physics genius. It is about centripetal forces.. circular motion... and I warn you that it is extremely difficult. If you can help me answer this problem, please PM me.</p>
<p>how will we know if u don't post it?</p>
<p>we don't do extra credit or homework for people...</p>
<p>lol jp, I suggest posting the problem on the thread</p>
<p>A car is on a banked curve. The radius of the curvature of the curve is 100 meters, and the angle at which it is banked is 10 degrees. The coefficient of statis friction is 0.10.</p>
<p>What is the range of speeds possible?</p>
<p>oh that i did that problem.</p>
<p>first u have to draw a force diagram which includes normal force, gravity, and friction. THen divide it up all into horizontal and vertical vectors. First do that you know that Fc=the sigma of those forces. Then you use the ac=mv^2/r to figure out velocity. First divide it up into vectors and u r half done.</p>
<p>yea the problem is fairly simple actully</p>
<p>Good problem!</p>
<p>Decompose the graviational force into a force parallel to and perpendicular to the road surface. Due the same to the centrifugal force. The frictional force +/- the gravitational force parallel to the surface equals the centrifugal force parallel to the surface. The frictional force is .10 * (gravitational force perpendicular + the centirfugal force perpendicular). </p>
<p>The tricks being that the frictional force opposes the slipping in direction so that you add it to gravitiy in one direction and subtract it in the other (i.e., +/-). And that the frictional force depends on both gravity and centrifugal force (the perpendicular components of them, that is.</p>
<p>Frank</p>
<p>its a centripetal force not a centrifugal force</p>
<p>i did that. Fc = sigma of all what forces? </p>
<p>it's asking for a range of velocity.. it's not as easy as you make it -- or at least not to me.</p>
<p>(sorry) didnt see the previous posts when i posted that</p>
<p>"Decompose the graviational force into a force parallel to and perpendicular to the road surface."</p>
<p>Wait a minute. the gravitational force is pointing straight down on the Y axis no? Should i be setting the Z axis along the bank of the road?</p>
<p>The way i have it set up, it's impossible to draw a gravitational force pararell and perpen. to the road.</p>
<p>of course the gravitational force isn't pararell or perpen. to the road. the friction and the normal r perpendicular. Just dissolve it into horizontal and vertical components. add all the forces(frictional, normal, and gravitational) and that is Fc=mv^2/r</p>
<p>yes.. i get that.. but there's no where to go from there.</p>
<p>i have Fnz + Ffz = mV^2/ r</p>
<p>which gives me</p>
<p>Fn sin 10 + Ff sin 10 = m V^2 / 100</p>
<p>mg sin 10 + (mew) mg sin 10 = m V^2/100</p>
<p>cancel out mass:</p>
<p>g sin 10 + mew g sin 10 = V^2/100</p>
<p>mew = 0.1</p>
<p>1.7 + .10 (1.7) = V^2 / 100</p>
<p>V = 13.68</p>
<p>yeah.... what is that velocity...?</p>
<p>I'm not sure if this is right but I got from 8.57 m/s to 16.6 m/s
This is how I did it(I don't want to use the carat symbol all the time so V=vsquared and v=v):
centripetal force is equal to mV/2 so:
mV/2=Nsin10+.1Nsin80
and
0=Ncos10-mg-.1Ncos80
add mg to second equation and factor out N so:
mg=N(cos10-.1cos80)
mg/(cos100-.1cos80)=N
substitute N into first equation, cancel out m's, and solve.
for the slowest you change your original equations to
MV/2= Nsin10-.1Nsin80
and
0=Ncos10-mg+.1Ncos80</p>
<p>ok nalcoln i agree with your answer.. thanks !</p>