A car of mass M = 1300 k g traveling at 55.0 k m / h o u r enters a banked turn

Question

A car of mass M = 1300kg traveling at 55.0km/hourenters a banked turn covered with ice. The road is banked at an angle ?, and there is no friction between the road and the car's tires. (Figure 1) . Use g = 9.80m/s2 throughout this problem.

Now, suppose that the curve is level (theta = 0) and that the ice has melted, so that there is a coefficient of static friction mu between the road and the car's tires.(Fiflure_2) What is mumin, the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from slipping? Assume that the car's speed is still 55.0km/hour and that the radius of the curve is given by the value you found for r in Part A. Express your answer numerically.

Explanation / Answer

PART-A-

FNcos(theta) = mg...................(1)

Since a(c) = v^2/r..............(2)

FNsin(theta) = mv^2/r.......................(3)

divide equation (1) from equation (2)

so, ((FNsin?) / (FNcos(theta))) = (mv^2/r) / (mg)

so, tan(theta) = (mv2/r) / (mg)

so, tan(theta) = v^2 / (rg)..............(4)

v = 55km/h = 55000m/h = 15.277m/s

from equation (4)

tan(theta) = v^2 / (rg)

tan(20) = (15.277^2) / (9.8r)

0.364 = (23.81/r)

r = 65.41m

PART-B-

Centripetal force is given by: F(c) = ma(c) and frictional force is given by F(fr) = FN?.

Centripetal acceleration is given by: a(c) = v^2/r

From Part A we can plug in values

a(c) = 15.2772/65.4

a(c) = 3.57m/s2

Now equate friction with centripetal force:

F(fr) = F(c)

FN(meu) = ma(c)

mg(meu) = ma(c)

g(meu) = a(c)

9.8(meu) = 3.57

(meu) = 0.364

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