Hello all, I also have a calculus thread in the appropriate forum- I wanted to have a separate physics thread in order to make things a little tidier around here! A quickie: A 1,200kg car turns a corner at 8ms^-1 and travels along the arc of a circle radius 9m in the process. How large a horiz. force must the road exert on the tyres in order to hold the car in a circular path & what is the min. coefficient of friction such that the car does not slide? Find horizontal force= Tangential force= mv^2/r (1200kg)(8m/s)^2/9 m= 8533.33N To keep from sliding off: the horizontal force and coefficient friction must work together against the centripetal force! Centripetal force= m[PLAIN]http://theory.uwinnipeg.ca/physics/circ/img24.gifr. [Broken] m[PLAIN]http://theory.uwinnipeg.ca/physics/circ/img24.gifr, [Broken] taking 8^2 as http://theory.uwinnipeg.ca/physics/circ/img24.gif 1200*64*9=691,200N If the centripetal force is 691,200N, the opposing force must be able to deal with this: it is (where u=coefficient of friction) uF. u*8533.33N=691,200N u=81 too large. Another go: ug=[mv^2]/r ug=[mv^2]/r, hence u (9.813)=[ (1,200kg)(8m/s^-1)]/9 =869.59 This seems like far too large of a coefficient of friction, so let's go about it by first principles. u=Ff/FN. Ff=ma, a=rw^2 a=(9)(8)^2=576kg/ms^-1 m=1200kg, as above, so 1200kg *576kg/ms^-1= 691,200 Ff=691,200N FN=Mg, 1200*9.813=11775.6 u=Ff/FN, so 691,200/111775.6= 6.183 This is still too large.. shouldn't coefficient of friction be LESS than one, otherwise it would slide right off the road?? The answer is lurking, but it's being a bit stubborn....help very greatly appreciated!