# physics

posted by on .

A banked circular highway curve is designed for traffic moving at 55 km/h. The radius of the curve is 220 m. Traffic is moving along the highway at 38 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)

• physics - ,

0.05

• physics - ,

Use
μmg = mv²/r
Make sure all quantities are in the same units.
Minimum μ comes to 0.05 for 38 km/h, and 0.11 for 55 km/h.
This can be reduced by building super-elevation, or banking at the curve.

• physics - ,

First you need the banking angle, A. The road is designed for V' = 55 km/h, which is 15.3 m/s. At that speed,
M*V'^2/R = M g sin A
sin A = V'^2/(gR) = 0.109
A = 6.23 degrees

For no slipping at that bank angle and a lower speed of V = 38 km/h = 10.56 m/s

M V^2/R = M g sin A - M g cosA*mu

When slipping starts, the friction force at the lower speed will be outward, hence the minus sign.

Cancel out the M's and solve for mu, which will be the minimum static friction coefficient.

0.506 m^2/s = 9.8 sin6.23 - 9.8*0.994*mu

9.74 mu = -0.506 +1.06 = 0.55
mu = 0.057

• physics - ,

Note that they said the curve was already banked for 55 km/h. No friction coefficient is needed at that speed

• physics - ,

How silly of me, I did not notice that the curve is banked.

However, curves need not be banked for the full design speed in case cars drive at a much slower speed than the design speed.

In fact, lateral slopes over 5% will cause a stationary vehicle to slip sideways on iced surfaces.

• physics - ,

Thanks for the help

• physics - ,

I know its been a while, but I'm just wondering how drwls solved the question without resolving friction into its x-component because isn't the centripetal acceleration directed horizontally instead of parallel to the incline and wouldn't the friction force be parallel to the incline. Can someone please explain this to me.

• physics - ,

Also how did you just substitute mg in for the Normal Force. Cause when I solved for the normal force from the vertical components I got:

FNcos A + mu*FNsin A = mg
FN = mg / (cos A + mu*sin A)

And I basically just substituted that into:

FNsin A - mu*FNcos A = m*v^2/R

I solved this and got the same answer as you. So I'm just wondering how did you just substitute mg for FN? Are there some cancellations that I'm unaware of?
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