Front view Top view - 45 my REASONING AND STRATEGY In this system, the force of static friction provides the centripetal force required for the car to move in a circular path. That is why the force of friction is at right angles to the car's direction of motion; it is directed toward the center of the circle. In addition, the friction in this case is static because the car's tires are rolling without slipping-always making static contact with the ground. Finally, if the car moves faster, more centripetal force (Le., more friction) is required. Thus, the greatest speed for the car corresponds to the maximum static friction, /, = p,N. Hence, by setting a N equal to the centripetal force, mid - me /r, we can solve for v. Known Mass of car, m - 1200 kg corner radius, / - 45 m; coefficient of static friction, A, - 0.82. Unknown Maximum speed without skidding, e - ? SOLUTION 1. Sum the x components of force to relate the force of EF. - 1. - ma, static friction to the centripetal acceleration of the car: 2. The car moves in a circular path, with the center of the circle in the AN - map - m x direction, and hence it follows that a, - a- 3/7. Make this substitution, along with f, - p N for the force of static friction: 3. Next, set the sum of they components of force equal EF - N - W - ma - 0 to zero, since a - 0: 4. Solve for the normal force: N - W - IS 5. Substitute the result N - my in Step 2 and solve for e. Aing - M- Notice that the mass of the car cancels: 6. Substitute numerical values to determine : - V (0.82) (45 m)(9.81 m/s ) - 19 m/s INSIGHT The maximum speed is less if the radius is smaller (tighter corner) or if p, is smaller (slick road). The mass of the vehicle, however, is irrelevant. For example, the maximum speed is precisely the same for a motorcycle rounding this corner as it is for a large, heavily loaded truck. PRACTICE PROBLEM - PREDICT/CALCULATE Suppose the situation described in this Example takes place on the Moon, where the acceleration due to gravity is less than it is on Earth. (a) If a lunar rover goes around this same corner, is its maximum speed greater than, less than, or the same as the result found in Step 4? Explain. (b) Find the maximum speed for a lunar rover when it rounds a corner with r = 45 m and , = 0.82. (On the Moon, g - 1.62 m/s'.) [Answer: (a) The maximum speed is less, because grav- Ity holds the rover to the road with a smaller force. (b) On the Moon we find e - 7.7 m/s, considerably less than the maximum speed of 19 m/s on Earth.] Some related homework problems: Problem 47, Problem 74 RWP If you try to round a corner too rapidly, you may experience a skid; that is, your car may begin to slide sideways across the road. A common bit of road wisdom is that you should turn in the direction of the skid to regain control-which, to most people, seems counterintuitive. The advice is sound, however. Suppose you are turning to the left and begin to skid to the right. If you turn more sharply to the left to try to correct for the kid, you simply reduce the turning radius of your car, r. The result is that the centripetal acceleration, '/r, becomes larger, and an even larger force is required from the road to make the turn. The tendency to skid is therefore increased. On the other hand, if youEXAMPLE 6-15 ROUNDING A CORNER A 1200-kg car rounds a corner of radius r = 45 m. If the coefficient of static friction between the tires and the road is , = 0.82, what is the greatest speed the car can have in the corner without skidding? PICTURE THE PROBLEM In the first sketch we show a bird's-eye view of the car as it moves along its circular path. The next sketch shows the car moving directly toward the observer. Notice that we have chosen the positive x direction to point toward the center of the circular path, and the positive y axis to point vertically upward. We also indicate the three forces acting on the car: gravity, W - -Wy - -myy; the normal force, N = No; and the force of static friction, I, = p Nt.Question 9 (1 point] A ZOOO-kg car goes around a corner of radius 43.6. The coefficient of static friction between the tires and the road is 0.820. What is the maximum speed for which the car does not skid, in meters per second? Use the answer to the solved Example 6-15