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Engineering a highway curve. If a car goes through a curve too fast, the car tends to slide out of the curve. For a banked curve with friction, a frictional force acts on a fast car to oppose the tendency to slide out of the curve; the force is directed down the bank (in the direction water would drain). Consider a circular curve of radius R = 180 m and bank angle 8, where the coefficient of static friction between tires and pavement is μs. A car (without negative lift) is driven around the curve as shown in Figure (a). Find an expression for the car speed Vmax that puts the car on the verge of sliding out, in terms of R, 0, and μs. Evaluate Vmax for a bank angle of 0 = 10° and for a) μs = 0.51 (dry pavement) and (b) μs = 0.050 (wet or icy pavement). (Now you can see why accidents occur in highway curves when icy conditions are not obvious to drivers, who tend to drive at normal speeds.) (a) R

Engineering a highway curve. If a car goes through a curve too fast, the car tends to slide out of the curve. For a banked curve with friction, a frictional force acts on a fast car to oppose the tendency to slide out of the curve; the force is directed down the bank (in the direction water would drain). Consider a circular curve of radius R = 180 m and bank angle 8, where the coefficient of static friction between tires and pavement is μs. A car (without negative lift) is driven around the curve as shown in Figure (a). Find an expression for the car speed Vmax that puts the car on the verge of sliding out, in terms of R, 0, and μs. Evaluate Vmax for a bank angle of 0 = 10° and for a) μs = 0.51 (dry pavement) and (b) μs = 0.050 (wet or icy pavement). (Now you can see why accidents occur in highway curves when icy conditions are not obvious to drivers, who tend to drive at normal speeds.) (a) R

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**Engineering a Highway Curve** If a car goes through a curve too fast, the car tends to slide out of the curve. For a banked curve with friction, a frictional force acts on a fast car to oppose the tendency to slide out of the curve; the force is directed down the bank (in the direction water would drain). Consider a circular curve of radius \( R = 180 \, \text{m} \) and bank angle \( \theta \), where the coefficient of static friction between tires and pavement is \( \mu_s \). A car (without negative lift) is driven around the curve as shown in Figure (a). Find an expression for the car speed \( v_{\text{max}} \) that puts the car on the verge of sliding out, in terms of \( R \), \( \theta \), and \( \mu_s \). Evaluate \( v_{\text{max}} \) for a bank angle of \( \theta = 10^\circ \) and for: (a) \( \mu_s = 0.51 \) (dry pavement) and (b) \( \mu_s = 0.050 \) (wet or icy pavement). (Now you can see why accidents occur in highway curves when icy conditions are not obvious to drivers, who tend to drive at normal speeds.) **Diagram Descriptions:** - **Figure (a):** This diagram shows a top view of a circular curve with a car traveling along it. The radius of the curve is labeled \( R \). The velocity vector \( v \) of the car is displayed, indicating the direction of the car's motion along the curve. - **Figure (b):** This diagram provides a side view illustrating the forces acting on the car. The gravitational force \( \mathbf{F}_g \) acts downward, while the normal force \( \mathbf{F}_N \) is perpendicular to the surface of the banked curve. The static frictional force \( \mathbf{F}_r \) acts parallel to the banked surface. The axes are labeled \( x \) and \( y \) to show the components of these forces relative to the bank angle \( \theta \). The acceleration \( \mathbf{a} \) points towards the center of the circular path.
Transcribed Image Text:**Engineering a Highway Curve** If a car goes through a curve too fast, the car tends to slide out of the curve. For a banked curve with friction, a frictional force acts on a fast car to oppose the tendency to slide out of the curve; the force is directed down the bank (in the direction water would drain). Consider a circular curve of radius \( R = 180 \, \text{m} \) and bank angle \( \theta \), where the coefficient of static friction between tires and pavement is \( \mu_s \). A car (without negative lift) is driven around the curve as shown in Figure (a). Find an expression for the car speed \( v_{\text{max}} \) that puts the car on the verge of sliding out, in terms of \( R \), \( \theta \), and \( \mu_s \). Evaluate \( v_{\text{max}} \) for a bank angle of \( \theta = 10^\circ \) and for: (a) \( \mu_s = 0.51 \) (dry pavement) and (b) \( \mu_s = 0.050 \) (wet or icy pavement). (Now you can see why accidents occur in highway curves when icy conditions are not obvious to drivers, who tend to drive at normal speeds.) **Diagram Descriptions:** - **Figure (a):** This diagram shows a top view of a circular curve with a car traveling along it. The radius of the curve is labeled \( R \). The velocity vector \( v \) of the car is displayed, indicating the direction of the car's motion along the curve. - **Figure (b):** This diagram provides a side view illustrating the forces acting on the car. The gravitational force \( \mathbf{F}_g \) acts downward, while the normal force \( \mathbf{F}_N \) is perpendicular to the surface of the banked curve. The static frictional force \( \mathbf{F}_r \) acts parallel to the banked surface. The axes are labeled \( x \) and \( y \) to show the components of these forces relative to the bank angle \( \theta \). The acceleration \( \mathbf{a} \) points towards the center of the circular path.
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