Engineering a highway curve. If a car goes through a curve...

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 $R=200 \mathrm{m}$ 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 Fig. $6-11 .$ (a) Find an expression for the car speed $v_{\text { max }}$ that puts the car on the verge of sliding out. (b) On the same graph, plot $v_{\text { max }}$ versus angle $\theta$ for the range $0^{\circ}$ to $50^{\circ},$ first for $\mu_{s}=0.60$ (dry pavement) and then for $\mu_{s}=0.050($ wet or icy pavement). In kilometers per hour, evaluate $v_{\text { max }}$ for a bank angle of $\theta=10^{\circ}$ and for $(c) \mu_{s}=0.60$ and $(d) \mu_{s}=$ $0.050 .$ (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.)

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 $R=200 \mathrm{m}$ 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 Fig. $6-11 .$ (a) Find an expression for the car speed $v_{\text { max }}$ that puts the car on the verge of sliding out. (b) On the same graph, plot $v_{\text { max }}$ versus angle $\theta$ for the range $0^{\circ}$ to $50^{\circ},$ first for $\mu_{s}=0.60$ (dry pavement) and then for $\mu_{s}=0.050($ wet or icy pavement). In kilometers per hour, evaluate $v_{\text { max }}$ for a bank angle of $\theta=10^{\circ}$ and for $(c) \mu_{s}=0.60$ and $(d) \mu_{s}=$ $0.050 .$ (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.)

Key Concepts

Equilibrium Analysis in Dynamics

Analyzing equilibrium in dynamics involves breaking down the forces acting on an object into their components and ensuring that the net force in any direction is balanced or accounted for. In banked curve problems, it is essential to resolve the gravitational, normal, and friction forces into components parallel and perpendicular to the road surface to determine conditions for equilibrium at the threshold of skidding.

Centripetal Force and Circular Motion

Centripetal force is the net force required to keep an object moving along a circular path at a constant speed. It is directed inward toward the center of the circle and is determined by the mass of the object, its speed squared, and the radius of the circular path. In vehicle dynamics, ensuring that the combined effects of banking and friction can supply the requisite centripetal force is crucial to avoid skidding outwards.

Banked Curve Dynamics

Banked curves involve inclining the road surface at an angle to help provide a component of the normal force that assists in providing the centripetal force required for an object to negotiate a turn. This design reduces reliance on friction alone and helps prevent vehicles from sliding outward on curves when taking them at higher speeds.

Static Friction

Static friction is the force that resists the initiation of sliding between two surfaces in contact. In the context of a banked curve, static friction acts alongside the normal force to provide additional centripetal force, especially when the vehicle’s speed exceeds the ideal condition for a frictionless banked turn. Its maximum value is proportional to the coefficient of static friction, which varies with surface conditions.

Transcript

00:01 So here for part a, we have a banked curve.

00:06 So we can say to be on the verge of sliding out means that the force of the static friction is acting down the bank.

00:14 So we know that the force of friction static acting down the bank if, again, we want to be on the verge of sliding out.

00:31 We can say that we can first consider the vector sum f of the maximum static frictional force and the normal force.

00:43 We know that the, rather the normal force and the frictional force are perpendicular to one another.

00:51 So we can say force friction static perpendicular to force normal.

00:56 And we can say that they are proportional to one another because it's, again, the frictional force depends on the normal force.

01:06 So we can say that we can say that force f would simply be equal to essentially the vector sum.

01:17 So we can say at an angle, we're going to say at an angle phi measured from the vertical axis.

01:43 Given this, we can say that phi is going to be equal to theta plus theta sub s, where here tangent of theta sub s would equal the coefficient of static friction, and theta would simply be the angle of the bank.

02:07 So bank, we can say bank angle.

02:12 And this was the angle that is stated in the problem.

02:15 So we can essentially know that these vectors, sum of f and the vertically downward pull mg must be equal to the centripetal force.

02:27 So we can say that tangent of phi simply equals to the centripetal force mv squared over r divided by that downward force mg of gravity.

02:44 So it divided by mg.

02:46 Therefore, this is going to equal v squared over rg and essentially v max would be equal to the square root of r g tangent of theta rather of theta not phi plus arc tan of the coefficient of static friction given that again these masses are going to cancel out of course so we can say that's the maximum velocity rather let's write over here the maximum velocity would be equal to rg rg open parentheses tangent of theta plus the coefficient of static friction and then this would be divided by one minus the coefficient of static friction multiplied by tangent of theta and at this point we can substitute rather this would be our answer for part a and then for part b when we want to draw the graph we would essentially graph the maximum velocity against the coefficient against the angle using different values for the coefficient of the static friction.

04:24 So what the graph would look like...

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