Posted by **TJR** on Wednesday, August 15, 2012 at 9:58am.

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. :(

- Physics -
**Elena**, Wednesday, August 15, 2012 at 12:35pm
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

- Physics -
**bobpursley**, Wednesday, August 15, 2012 at 12:43pm
Thanks, Elena, good work. My analysis was too quick.

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