Once in a lifetime opportunity :-)

To come and say hello to me on my very first LoveHoney post.  This chance won't come up again, live for today and do it now :-)

Any questions will be answered honestly and fully (as long as it doesn't require an essay!)

Helloo!!!!

Hi SmoothOne, welcome to the forums. 

Welcome to the forums Smoothone. Hope to see you around.

So Smoothone here's a question for you fella.

 A curve of radius 60 meters is banked for a design speed of 100 km/hr. If the coefficient of static friction is 0.30, at what range of speeds can a car safely make the curve?

SG69

SEXYGET 69 wrote:

Welcome to the forums Smoothone. Hope to see you around.

So Smoothone here's a question for you fella.

 A curve of radius 60 meters is banked for a design speed of 100 km/hr. If the coefficient of static friction is 0.30, at what range of speeds can a car safely make the curve?

SG69

Is that one of your job-related questions? It sounds very specific. No wonder that fella recognised you the other day.

Thanks SexyGet and in answer to your question....without someone in the driving seat the range of speeds will be approx 0km/hr unless someone forgot to put the handbrake on half way up the hill!

TLA wrote:

Is that one of your job-related questions? It sounds very specific. No wonder that fella recognised you the other day.

Ha Ha TLA! Not job related but. . .wierdly enough the same mate sent me that question the other day. I'm working on it! Thought This new bloke might have been a physics buff! My job is developing and testing fire fighting chemicals. Been somewhat bloody hot for me lately!

SG69

Welcome to the forums!

I dont normally post in the Introductions forum but your avatar made me giggle :)

SmoothOne wrote:

Thanks SexyGet and in answer to your question....without someone in the driving seat the range of speeds will be approx 0km/hr unless someone forgot to put the handbrake on half way up the hill!

Ha Ha Nice answer fella! I should've said and what was the name of the driver?

Hope you like it on here!

SG69

Thanks.  Fingers crossed I fit right in.

hello SmoothOne how are you?

Hi Miss Nat....after a few JD's and Coke, I'm good thanks. The world is a good place.

how are you?

hello and welcome

thanks sweetlove.  hope you are well

and great avatar btw

Ha ha, l have a brainy mate who could give you an answer to that SG !!!

TB

tallboy247 wrote:

Ha ha, l have a brainy mate who could give you an answer to that SG !!!

TB

Maybe it's the same bloke who sent me it! He doesn't live on a farm in ellesmere port does he?

SG69

For sexyget

First find centipetal force at 100 km/h:

Fc= mV^2/r

Fc = mV^2/60

V = 27.8 m/s

Fc= m * (27.8)^2 / 60 = 12.88 * m

So now we know the centripetal force is greater than the component of gravity

We need to know the component of centripetal force which will tend to make the car slide up the slope (all this Fu).

Fu = 12.88 * m * cos θ

So if curve is banked, the slope is such that Fu = gravitational force pulling car down the slope.

Fd (gravity component down the slope) = mg * sin θ

mg * sin θ = 12.88 * m * cos θ

9.8 * sin θ = 12.88 cos θ
sin θ / cos θ = tan θ= 12.88/ 9.8 θ = 52.7 degrees

Now the range of speeds:

Note: "m" cancels out.

Maximum speed is where friction force = centripetal force upslope - gravity force downslope

Maximum speed:
Frictional force = 0.30 * normal force
Frictional force = 0.30 * [mg cos θ + (mV^2/60) sin θ)]
Force upslope = (mV^2/60) cos θ - mg sin θ

0.30 * [9.8 * cos 52.7 + (V^2/60) sin 52.7)] = (V^2/60) cos52.7 - 9.8 sin 52.7

0.30 (5.94 + 0.013 * V^2) = 0.011 * V^2 - 7.8

1.78 + .004 * V^2 = .011 * V^2 - 7.8

.007 * V^2 = 15.6

V = 47.2 m/s = 170 km/h (maximum speed)

Now repeat exercise for minimum speed, where
friction force + centripetal force upslope. = gravity force downslope

Anything else?

SmoothOne wrote:

Hi Miss Nat....after a few JD's and Coke, I'm good thanks. The world is a good place.

how are you?

i'm good thank you, slightly tired though

Fr33b1rd wrote:

For sexyget

First find centipetal force at 100 km/h:

Fc= mV^2/r

Fc = mV^2/60

V = 27.8 m/s

Fc= m * (27.8)^2 / 60 = 12.88 * m

So now we know the centripetal force is greater than the component of gravity

We need to know the component of centripetal force which will tend to make the car slide up the slope (all this Fu).

Fu = 12.88 * m * cos θ

So if curve is banked, the slope is such that Fu = gravitational force pulling car down the slope.

Fd (gravity component down the slope) = mg * sin θ

mg * sin θ = 12.88 * m * cos θ

9.8 * sin θ = 12.88 cos θ
sin θ / cos θ = tan θ= 12.88/ 9.8 θ = 52.7 degrees

Now the range of speeds:

Note: "m" cancels out.

Maximum speed is where friction force = centripetal force upslope - gravity force downslope

Maximum speed:
Frictional force = 0.30 * normal force
Frictional force = 0.30 * [mg cos θ + (mV^2/60) sin θ)]
Force upslope = (mV^2/60) cos θ - mg sin θ

0.30 * [9.8 * cos 52.7 + (V^2/60) sin 52.7)] = (V^2/60) cos52.7 - 9.8 sin 52.7

0.30 (5.94 + 0.013 * V^2) = 0.011 * V^2 - 7.8

1.78 + .004 * V^2 = .011 * V^2 - 7.8

.007 * V^2 = 15.6

V = 47.2 m/s = 170 km/h (maximum speed)

Now repeat exercise for minimum speed, where
friction force + centripetal force upslope. = gravity force downslope

Anything else?

Fr33b1rd wrote:

For sexyget

First find centipetal force at 100 km/h:

Fc= mV^2/r

Fc = mV^2/60

V = 27.8 m/s

Fc= m * (27.8)^2 / 60 = 12.88 * m

So now we know the centripetal force is greater than the component of gravity

We need to know the component of centripetal force which will tend to make the car slide up the slope (all this Fu).

Fu = 12.88 * m * cos θ

So if curve is banked, the slope is such that Fu = gravitational force pulling car down the slope.

Fd (gravity component down the slope) = mg * sin θ

mg * sin θ = 12.88 * m * cos θ

9.8 * sin θ = 12.88 cos θ
sin θ / cos θ = tan θ= 12.88/ 9.8 θ = 52.7 degrees

Now the range of speeds:

Note: "m" cancels out.

Maximum speed is where friction force = centripetal force upslope - gravity force downslope

Maximum speed:
Frictional force = 0.30 * normal force
Frictional force = 0.30 * [mg cos θ + (mV^2/60) sin θ)]
Force upslope = (mV^2/60) cos θ - mg sin θ

0.30 * [9.8 * cos 52.7 + (V^2/60) sin 52.7)] = (V^2/60) cos52.7 - 9.8 sin 52.7

0.30 (5.94 + 0.013 * V^2) = 0.011 * V^2 - 7.8

1.78 + .004 * V^2 = .011 * V^2 - 7.8

.007 * V^2 = 15.6

V = 47.2 m/s = 170 km/h (maximum speed)

Now repeat exercise for minimum speed, where
friction force + centripetal force upslope. = gravity force downslope

Anything else?

I looked at that and for a couple of minutes I thought I clicked on a link to a wrong forum... you must be more bored than I am. Now you'll be really popular with all the people here who are still at school or are studying anything maths related.