Question

Help me set this up im lost....The equation v max = sqrt(14.88fr) gives the maximum velocity in miles per hour that a vehicle can safely travel around a curve that has a radius of r feet. If the velocity is greater than v max , the tires will slip. Engineers find that under snowy conditions, v max = 15mi/h for a freeway off-ramp that has a radius of 50 ft. To the nearest tenth, what is the coefficient of friction for the off-ramp in these conditions? (formula)>> Vmax=√14.88fr

Answer 1

I cant continue until i solve this but idk to solve it.

Answer 2

i believe the formula for coeff of friction, normal, and Force are:
F = u n
u = F/n

Answer 3

it gives me Vmax=SQrt14.88fr as formula but ive never seen it before

Answer 4

but as is:

15 = sqrt(14.88 fr) we know r = 50

Answer 5

if we square both sides we get:

225 = 14.88(2500) f

Answer 6

lol, 50 is fine not 2500 .....

Answer 7

ive never seen that setup either, can we assume f is force?

Answer 8

idk it says friction... but its all new to me so not sure

Answer 9

oh, then f = friction coeff ....

Answer 10

its just solving for "f" is all

Answer 11

15 = sqrt(14.88(50)f)

15^2 = 14.88(50) f

15^2/(14.88(50)) = f

Answer 12

so f would =.3?

Answer 13

if thats what the calculator determines it to be, then yes :)

Answer 14

ok thankyou for the help.... i didnt realize how simple the problem actually was lol

Answer 15

good luck ;)