A car of mass M = 1500 kg traveling at 50.0 km/hour enters a banked turn covered

Question

A car of mass M = 1500 kg traveling at 50.0 km/hour enters 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 as shown in (Figure 1) . Use g = 9.80 m/s2 throughout this problem.

Part B

Now, suppose that the curve is level (=0) and that the ice has melted, so that there is a coefficient of static friction between the road and the car's tires as shown in (Figure 2) . What is min, 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 50.0 km/hour and that the radius of the curve is 54.1 m .

Express your answer numerically.

Explanation / Answer

Since the radius is part of the formula for centripetal acceleration (ac = v^2/r),

we can solve for r using that formula. You should have learned in your class about dimensional translation/rotation and

m*g = F*N*cos(theta) and m*ac = F*N*sin equals centripetal force.

Since we know mass and g, if we can relate the above expressions, we can find r.

F*N*sin(theta) = m*ac

Since ac = v^2/r

F*N*sin(theta) = m*v^2/r

And we know (from the explanation above) that F*N*cos(theta) = m*g

If we divide the first formula (for F*N*sin(theta)) by the cos(for F*N*cos(theta))

((F*N*sin(theta)) / (F*N*cos(theta)) = (m*v^2/r) / (m*g)

(part B)

In order for the car to not slide around, there needs to be centripetal force.

Since there isn’t a slope, gravity won’t cause the car to slide down towards the center of the circle (F*N*sin(theta)).

So the centripetal force will be coming entirely from the frictional force.

Centripetal force is given by:

Fc = m*ac and frictional force is given by F*fr = F*N.

Centripetal acceleration is given by: ac = v^2/r

From Part A we can plug in values

ac = 13.89^2/54.1

ac = 3.56 m/s^2

Now equate friction with centripetal force:

F*fr = Fc

F*N*u = m*ac

m*g*u = m*ac

g*u = ac

9.8*u = 3.56

u = 0.3638

the minimum value of the coefficient of static friction is

u = 0.3638

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