Centripetal Force on an incline + Fricktion

[Infamous_01]

Homework Statement

If a curve with a radius of 88m is perfectly banked for a car traveling 75kn/h, what must be the coefficient of static friction for a car not to skid when traveling at 95 km/h

Homework Equations

F=m(aR)

F(Normal) x sin0 = m(v^2/r)

The Attempt at a Solution

I've asked my teacher this problem from the book twice and he skimmed through it both times really quickly and I didn't really understand him (hes a busy guy and has a lot of classes right after mine).

What i did so far was convert the km/h to m/s and draw a free body diagram showing the forces that are acting upon the banked car. Assuming friction was 0 in the first case I tried getting the banked angle but not sure if was correct then in the later part (95km/h) I attempted to find Fr. I know that the forces acting upon pulling the car towards the center and pushing away from the center must be equal or else the car wouldn't stay on the road.

 

[LowlyPion]

Homework Statement

If a curve with a radius of 88m is perfectly banked for a car traveling 75kn/h, what must be the coefficient of static friction for a car not to skid when traveling at 95 km/h

Homework Equations

F=m(aR)

F(Normal) x sin0 = m(v^2/r)

The Attempt at a Solution

I've asked my teacher this problem from the book twice and he skimmed through it both times really quickly and I didn't really understand him (hes a busy guy and has a lot of classes right after mine).

What i did so far was convert the km/h to m/s and draw a free body diagram showing the forces that are acting upon the banked car. Assuming friction was 0 in the first case I tried getting the banked angle but not sure if was correct then in the later part (95km/h) I attempted to find Fr. I know that the forces acting upon pulling the car towards the center and pushing away from the center must be equal or else the car wouldn't stay on the road.

What's missing from your equation is the effect of the angle on your centripetal acceleration. (Also F(normal) should really be m*g.) What the problem is asking is that if the curve requires no u of friction at 75 km/h then how much more additional force will be required in terms friction to stop it from skidding when it goes faster.

Hence once you have found the angle from the first equation then you know at the larger speed how much additional force is required. This needs to be supplied by friction.

Don't forget though that the car is experiencing both the weight component from gravity normal to the surface of the incline AND you also have the normal component from the centripetal acceleration.

 

[Infamous_01]

What's missing from your equation is the effect of the angle on your centripetal acceleration. (Also F(normal) should really be m*g.) What the problem is asking is that if the curve requires no u of friction at 75 km/h then how much more additional force will be required in terms friction to stop it from skidding when it goes faster.

Hence once you have found the angle from the first equation then you know at the larger speed how much additional force is required. This needs to be supplied by friction.

Don't forget though that the car is experiencing both the weight component from gravity normal to the surface of the incline AND you also have the normal component from the centripetal acceleration.

Cant we eliminate mass from the equation? In other words in f(Normal) equals mg can't we just say 'a' or (v^2/r) instead.

How does this sound:

(9.8)(sin0)=(v^/r)

 

[LowlyPion]

Cant we eliminate mass from the equation? In other words in f(Normal) equals mg can't we just say 'a' or (v^2/r) instead.

How does this sound:

(9.8)(sin0)=(v^/r)

Mass will tend to cancel out. But I think until you get the right equation it will be better for you to consider it until you can cancel it out.

For instance your equation still does not treat the mv²/r correctly, because it is directed horizontally and it is balancing the m*g directed along the plane of the incline for your 75km/h case, hence you need that component of the force that lays in the plane of the incline.

 

[hemetite]

coincidentally i am doing this question too...

so i got the angle as...

tan teta = V^2/ R
therefore the angle = 26.69 degrees

for it to be stable again. the

frictional force = centripetal force...

the centripetal force => F(Normal) x sin0 = m(v^2/r)

how to find the fricitonal force equation?