A car of mass M = 1400 kg traveling at 40.0 km/hour enters a...

A car of mass M = 1400 kg traveling at 40.0 km/hour enters a banked turn covered with ice. The road is banked at an angle $ heta$, and there is no friction between the road and the car's tires as shown in (Figure 1). Use $g = 9.80 ext{ m/s}^2$ throughout this problem. Part A What is the radius $r$ of the turn if $ heta = 20.0^circ$ (assuming the car continues in uniform circular motion around the turn)? Express your answer in meters. $r = 34.6 ext{ m}$ Part B Now, suppose that the curve is level ($ heta = 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 as shown in (Figure 2). What is $mu_{min}$, 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 40.0 km/hour and that the radius of the curve is 34.6 m. Express your answer numerically.

Transcript

0:00 Hi there.

00:01 So for this problem, we have a card of mass capital m that is equal to 1 ,400 kilograms.

00:13 And it's traveling with a speed.

00:15 We're going to call this the speed.

00:18 B is that is equal to 40 kilometers per hour.

00:24 And enters a banking term that is covered with ice.

00:29 So the road is banking.

00:31 At an angle theta, and there is no friction between the road and the car's tires, as is shown in this figure.

00:38 Now, for part b of this problem, since part a is already solved, we need to suppose that now the curve is level, so that means that theta is equal to zero for this part of the problem, and that the ice has melted, so that there is a coefficient of static friction between the road and the car tires, as is shown in the other figure that is not in here.

01:16 But what we need to determine is what is the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from sleeping.

01:29 And we need to assume that the card is at a speed that is given 40 kilometers per hour, and that the radius of the curve is also given.

01:40 That is the solution for part a, which is 34 .6 meters.

01:47 Now, in this case, since the card is now at level, is not inclined, then we are going to have two forces.

01:57 In the x component.

02:00 So we are going to have the centripetal force towards the center of the circular path.

02:10 This is the centripetal force.

02:13 And the other force is the force of friction.

02:19 Of course, we will have the normal force and the weight.

02:24 So as you can see, if we use newton, second, law to sum all of the forces in the x component.

02:33 We are you going to have that this is.

02:37 The centripetal force should be equal to the friction force, as you can see from here.

02:45 And if we sum all of the forces in the white component, in this case, we're going to have that the normal force, which is pointing upward, is equal to the weight of the card.

02:58 That we know the weight of the car is defined as the mass times the acceleration due to gravity.

03:04 Now, we know that the friction force is defined as the normal force times the coefficient of kinetic friction, in this case, is the minimum value...

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