This is the only problem that I can\'t seem to get the right answer for. Please

Question

This is the only problem that I can't seem to get the right answer for. Please help?
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A curve of radius 20 m is banked so that a 1030 kg car traveling at 40 

km/h can round it even if the road is so icy that the coefficient of static friction is approximately zero.
The acceleration of gravity is 9.81 m/s2 .

a) Find the minimum speed at which a car can travel around this curve without skidding if the coefficient of static friction between the road and the tires is 0.3.Answer in units of m/s.

b) Find the maximum speed under the same conditions.Answer in units of m/s.

Explanation / Answer

The road is designed so that the force pushing the car down the bank (straight towards the center of the 20m curve) is balanced out by the centripetal force when the car is going at the design speed. When the car is going at 40 km/hr its speed is 40 km/hr*3600 s/hr*1000m/km = 11.11 m/s Since the car isn't accelerating up or down the vertical forces must cancel. Therefore the car's weight, W = N cos(theta) This tells us that, at design speed, the normal force is more than the weight. The centripetal force has a component that pushes the car into the road, so the road pushes up with more than the weight even though there is a component of the weight down the bank. If the car weren't moving, the normal force would be less than the weight, so this is novel (unless you've done these before, or worked on problems where people were pushing down on things). Usually the normal force goes to zero as the angle of the hill increases. Here it increases as the angle of the hill increases - again this is due to centripetal force, this force gets really big as the centripetal acceleration increases. The component of the slope pushing the car in is N sin (theta). Since there is no side to side acceleration (right at this point) N sin (theta) = mv^2/R If we divide these two equations with mg = W tan (theta) = v^2/Rg = 11.11^2/(20*9.8) = 0.629 theta = 32.2 a) Wow, that is a steep bank, but 20m radius is a tight loop. Now we add friction and lower the speed. There is probably a way to do it with horizontal and vertical forces and dividing again, but I want to use up the bank and down the bank so we can see how all the components of the forces contribute. Normal force = component of weight into the bank + component of centripetal force into the bank = mg cos(theta) + (mv^2/R)sin(theta) As noted above, we see centripetal force into road growing as angle grows and weight into road shrinking as angle grows Component of weight down the hill = mg sin(theta) Component of centripetal force up the hill (mv^2/R)cos(theta) In this part, friction is up the hill. Force of friction is uN = 0.3(8540.90389+v^2*27.45) when the car is just about to slip down the hill mgsin(32.2) = (mv^2/R)cos(32.2) + 0.3mgcos(32.2) + 0.3(mv^2/R)sin(32.3) divide out the mass bring v^2 to one side v^2(cos(32.2)/R + 0.3sin(32.2)) =gsin(32.2) -0.3g cos(32.2) v^2 = [g sin(32.2)-0.3 g cos (32.2)]/[cos(32.2/R + 0.3 sin(32.2)] v = SQRT([g sin(32.2)-0.3 g cos (32.2)]/[cos(32.2/R + 0.3 sin(32.2)] = 7.37 m/s (b) Friction is a fickle ally and now friction is resisting the swish up the hill. It is the same except it is the falling down the slope force plus the friction force is just balanced by the skidding up the hill force. mgsin(32.2) + 0.3(mgcos(32.2) + (mv^2/R)sin(32.3)) = (mv^2/R)cos(32.2) v = SQRT([g sin(32.2)+ 0.3 g cos (32.2)]/[cos(32.2)/R - 0.3 sin(32.2)/R] = 23.1 m/s In summary, centripetal acceleration force the car into the bank making frictional force greater. Since centripetal force goes as v^2, it is more of a factor at high speeds. This is why minimum speed is closer to design speed than maximum speed is.

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