How do I draw a FBD for centripetal force

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Biochemgirl2002
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The question
Sometimes road surfaces have banked curves, use a FBD to explain how it helps cars make turns more safely My trial:

I drew this FBD (attached below)

Image1512568486.090307.jpg


And I think it helps make driving more safe because the Fn acting on the car is pointed more towards the middle of the curve, which acts as a centripetal force. Which like in space, where centripetal force acts as your gravity, the centripetal force will cause the car to stick to the road and not slip as much Does this sound correct? How can I elaborate more?
 

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You're on the right track. What other forces might be involved?

rhiana said:
the Fn acting on the car is pointed more towards the middle of the curve, which acts as a centripetal force.
That's good.

In order to turn, you need some centripetal force. Without banking, what force are you relying on?
 
rhiana said:
The question
Sometimes road surfaces have banked curves, use a FBD to explain how it helps cars make turns more safelyMy trial:

I drew this FBD (attached below)

View attachment 216225

And I think it helps make driving more safe because the Fn acting on the car is pointed more towards the middle of the curve, which acts as a centripetal force. Which like in space, where centripetal force acts as your gravity, the centripetal force will cause the car to stick to the road and not slip as muchDoes this sound correct? How can I elaborate more?
Friction is not always reliable at circular turn if high speed and sharp turns are involved.To aviod dependence on friction, the roads are banked at the turn so that the outer part of the road is somewhat lifted up as compared to inner part.The surface of the road makes an angle θ with the horizontal throughout the turn.The normal force Fn makes an angle θ with vertical. At correct speed,the horizontal component of Fn is sufficient to produce the centripetal acceleration
Fnsinθ=mv^2/r
Fncosθ=mg
From these two
tanθ=v^2/rg
The angle θ depends on v as well as r.roads are banked for average expected speed.if the speed of particular vehicle is little more than that the correct speed, the self adjustable static friction operates between the tyres and road and the vehicle doest not skid or slip.
 
Doc Al said:
You're on the right track. What other forces might be involved?That's good.

In order to turn, you need some centripetal force. Without banking, what force are you relying on?

To turn, would you just be relying on the friction of the tires? If there was no banking
 
Abhishek kumar said:
Friction is not always reliable at circular turn if high speed and sharp turns are involved.To aviod dependence on friction, the roads are banked at the turn so that the outer part of the road is somewhat lifted up as compared to inner part.The surface of the road makes an angle θ with the horizontal throughout the turn.The normal force Fn makes an angle θ with vertical. At correct speed,the horizontal component of Fn is sufficient to produce the centripetal acceleration
Fnsinθ=mv^2/r
Fncosθ=mg
From these two
tanθ=v^2/rg
The angle θ depends on v as well as r.roads are banked for average expected speed.if the speed of particular vehicle is little more than that the correct speed, the self adjustable static friction operates between the tyres and road and the vehicle doest not skid or slip.

Okay, so when you go faster, the more grip (or static friction) you have on the road?
 
rhiana said:
To turn, would you just be relying on the friction of the tires? If there was no banking
Exactly.

rhiana said:
Okay, so when you go faster, the more grip (or static friction) you have on the road?
Yes. Static friction will increase as needed to prevent slipping -- up to its maximum for the given surfaces.
 
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rhiana said:
Okay, so when you go faster, the more grip (or static friction) you have on the road?
Upto a certain maximum speed.