Question

A car whose speed is 90.0 km/h (25 m/s) rounds a curve 180 m in radius that is properly banked for speed of 45 km/h (12.5 m/s). Find the minimum coefficient of friction between tires and road that will permit the car to make a turn. What will happen to the car in this case?


My teacher said we'll use 'tan' but i'm confused. :(

Answer 1

The projections of the forces acting ob the car moving at the velocity v=12.5 m/s are:
x: ma=N•sin α,
y: 0=N•cos α – mg.

The normal force has a horizontal component , pointing toward the center of the curve. Because the ramp is to be designed so that the force of static friction is zero, only the component causes the centripetal acceleration:
m•v²/R=N•sin α= m•g•sin α/cos α =m•g•tan α,
tan α= v²/g•R
α =arctan(v²/g•R)= arctan(12.5²/9.8•180) =5.06º
The equations of the motion of the car moving with velocity V=25 m/s are:
x: ma= N•sin α +F(fr) •cos α,
y: 0=N•cos α – mg- F(fr) •sin α.
Since F(fr) =μ•N, we obtain
ma= N•sin α + μ•N •cos α,
mg =N•cos α – μ•N •sin α.

ma/mg =N(sin α + μ •cos α)/N (cos α – μ •sin α)

a=g(sin α + μ •cos α)/ (cos α – μ •sin α)

But
a=V²/R,
therefore,
V²/R =g(sin α + μ •cos α)/ (cos α – μ •sin α),
V²•cos α -R•g•sin α =μ• (V²• sin α +R•g• cos α),
μ=( V²•cos α -R•g•sin α)/ (V²• sin α +R•g• cos α)=
=(625•0.996-180•9.8•0.088/(625•0.088+180•9.8•0.996)=
= 0.258

Answer 2

Thanks, Elena, good work. My analysis was too quick.