A Highway curve that has a radius of curvature of 100 meters is banked at an ang

Question

A Highway curve that has a radius of curvature of 100 meters is banked at an angle of 15

A Highway curve that has a radius of curvature of 100 meters is banked at an angle of 15 degree as shown above. Determine the vehicle speed for which this curve is appropriate if there is no friction between the road and the tires of the vehicle. On a dry day when friction is present, an automobile successfully negotiates the curve at a speed of 25 meters per second. On the diagram below, in which the block represents the automobile, draw and albel all the forces on the automobile.

Explanation / Answer

first case, no friction

we use a standard x, y coordinate system even though the track is banked

the forces on the car are the horizontal and vertical components of the normal force (N) and the weight

in the vertical direction, the vertical component of N balances the weight since the car does not accelerate vertically, so

N cos(theta) = mg (eq. 1) draw a diagram showing the normal force (which is perpendicular to the track) and be sure you see what the vertical component is N cos(theta) where theta is the bank angle)

the horizontal component of the normal force must match the centripetal force, so we have

N sin(theta)=mv^/r (eq.2)

now, divide eq. 2 by e1. 1:

Nsin(theta)/Ncos(theta)=v^2/rg

tan(theta)=v^2/rg

for r=100m and theta =15 deg, we have that v=sqrt[ r g tan(theta)]
v=16.2 m/s

second case, with friction

to keep the car from sliding up the track at 25m/s, there must be friction opposing the motion

there is both a horizontal component of friction and a vertical component of friction; the horizontal component of friction acts toward the center of the curve and has magnitude f cos(theta) where f is the total frictional force, the vertical component acts down with magnitude f sin(theta)

remembering that f= u N where u is the coefficient of friction, our force laws become:

in the x direction: N sin(theta) + f cos(theta) = mv^2/r
Nsin(theta)-+u N cos(theta)=mv^2/r
N[sin(theta)+u cos(theta)]=mv^2/r (eq.3)

in the y direction: N cos(theta)-mg - fsin(theta)=0
N[cos(theta)- u sin(theta)]=mg (eq.4)

divide eq. 3 by e1 4

N[sin (theta) + u cos(theta)]/N[cos(theta)- u sin(theta)]=mv^2/r/(mg)

[sin(theta)+u cos(theta)]/[cos(theta)-u sin(theta)=v^2/rg

knowing theta =15 deg, v=25m/s, r=100 m and g=9.8m/s/s, you can solve for u

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.