What Coefficient of Friction Is Required for a Car on a Banked Curve?

[hofluff87]

A car moving at 41 km/h negotiates a 140 m-radius banked turn designed for 60 km/h. What coefficient of friction is needed to keep the car on the road?


Ok can some one tell me what i did wrong?

F_c= (m*V²)/r = m µ g

Vmax= square root of ((µ) g r)

60 = square root (µ*9.8*140)

for mu i got 2.62391

i used the 60 km/h because that was the maximum velocity of the curve

how ever when i checked it its wrong ... my book does not say anything about the friction involved with a banked curve. help

 

[LowlyPion]

A car moving at 41 km/h negotiates a 140 m-radius banked turn designed for 60 km/h. What coefficient of friction is needed to keep the car on the road? Ok can some one tell me what i did wrong?

F_c= (m*V²)/r = m µ g

Vmax= square root of ((µ) g r)

60 = square root (µ*9.8*140)

for mu i got 2.62391

i used the 60 km/h because that was the maximum velocity of the curve

how ever when i checked it its wrong ... my book does not say anything about the friction involved with a banked curve. help

Isn't there an angle of bank for the curve?

Edit: I see it's designed for 60 km/h.

As for your speed you will need to convert them to m/s from km/h.

 

[LowlyPion]

A car moving at 41 km/h negotiates a 140 m-radius banked turn designed for 60 km/h. What coefficient of friction is needed to keep the car on the road?


Ok can some one tell me what i did wrong?

F_c= (m*V²)/r = m µ g

This equation is not right because you haven't accounted for the angle of the bank.

 

[hofluff87]

i changed the speed from km/h to m/s 60 km/h = 16.67 m/s

there is no given curve however:

if Fc cos (theta) = Weight sin (theta)

(V²/r) cos (theta) = g sin (theta)

then (V²/ g r ) = sin theta / cos theta = tan theta

theta = tan inverse (V²/ g r)

theta = 11.446

but how does that help me find mu?

btw after adjusting the km/h to meters/ s for mu after plugging it in the equation i got .202 but it was also wrong

 

[LowlyPion]

i changed the speed from km/h to m/s 60 km/h = 16.67 m/s

there is no given curve however:

if Fc cos (theta) = Weight sin (theta)

(V²/r) cos (theta) = g sin (theta)

then (V²/ g r ) = sin theta / cos theta = tan theta

theta = tan inverse (V²/ g r)

theta = 11.446

but how does that help me find mu?

btw after adjusting the km/h to meters/ s for mu after plugging it in the equation i got .202 but it was also wrong

Consider first the 60km/h case.
Draw a force diagram. There is the centripetal force. There is the gravitational force down the incline. Since they say it is designed for 60km/h I think they must mean that there is no slipping for any μ.

Mass drops out. Leaving θ, because you know g and V and R

Now develop your equation for the slower speed. There your MV²/R plus your μ*g component must be sufficient to balance the downward force of the m*g*Sinθ component.