8.[lpt] A car is travelling along a curve having radius of curvature,r 62.0 m, b

Question

8.[lpt] A car is travelling along a curve having radius of curvature,r 62.0 m, banked at an angle 0-320, as shown. After a freezing rain, the road is covered with ice whose coefficient of static friction is zero. With what single speed can the car negotiate the banked curve? Answer: 9.[1pt] In reality, the coefficient of static friction of ice is not zero but closer to 0.09. In this case, what is the slowest speed the car can negotiate the banked curve with? Answer: 10.[Ip] What is the fastest speed, in this case? Answer: Submit All Answers Submit All Answers Submit All Answers

Explanation / Answer

Theta = 32 degrees
r = 62 m

vmax = (rg(sin(theta) + Mucos(Theta))/(Cos(theta) - Musin(Theta)))^(1/2)

8)
No Friction
Henc,e
Vmax = (rgtan(theta))^(1/2)
Vmax = (62*9.8*tan(32))^(1/2)
Vmax = 19.49 m/s

9)
vmin = (rg(tan(theta) - Mu)/(1 + muTan(theta)))^(1/2)
vmin = (9.8*62(tan(32) - 0.09)/(1 + 0.09Tan(32)))^(1/2)
vmin = 17.54 m/s

10)
vmax = (rg(sin(theta) + Mucos(Theta))/(Cos(theta) - Musin(Theta)))^(1/2)
vmax = (9.8*62(sin(32)+ 0.09cos(32))/(cos(32) - 0.09*sin(32)))^(1/2)
vmax = 21.45 m/s

Related Questions

Navigate

• Browse (All)
• Browse 8
• Subjects

---

---

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