007 (part 1 of 4) 10.0 points
A curve in a road is banked. There is a car on the curve.
37◦
What is the component of its weight paral- lel to the incline? The acceleration of gravity is 9.8 m/s2 .
Answer in units of N
008 (part 2 of 4) 10.0 points
If the radius of curvature is 84.1 m, what is the ideal speed of the car such that it doesn’t rely on friction to keep from sliding sideways?
Answer in units of m/s
009 (part 3 of 4) 10.0 points
The next curve that the car approaches also has a radius of curvature of 84.1 m but it is banked at an angle of 30◦.
If the car rounds the curve at a speed of 0.518 times this curve’s ideal speed vc , what is the magnitude of the frictional force needed to keep it from sliding sideways?
1.5 Mg
μ = 0.11
The frictional force keeps the car from
1. moving in a circular path.
2. sliding toward the center of curvature.
3. sliding away from the center of curva- ture.
4. The frictional force is zero.
Help me !!!!
Answers & Comments
Verified answer
1) m(9.89)sin37
2) tanθ = horizontal acceleration / vertical acceleration
tanθ = v²/r / g
tanθ = v²/rg
v = √(rgtanθ)
v = √(84.1(9.8)tan37)
v = 24.9 m/s
3) With a bank angle of 30° the design velocity can be found by
tan30 = (v²/r) / g
v = √(grtan30)
v = √(9.8(84.1)tan30)
v = 21.81 m/s
and the centripetal acceleration is
ac = 21.81²/84.1
ac = 5.658 m/s²
the other velocity is
21.81(0.518) = 11.3 m/s
with a centripetal acceleration of
ac2 = 11.3²/84.1
ac2 = 1.518 m/s²
so friction must create an acceleration of
5.658 - 1.518 = 4.14 m/s²
So the friction force is
F = ma
F = m(4.14)/cos30
plug in the mass