Question about Centripetal Force

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Red_CCF
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Hi

For a car making a turn, the friction between the tires causes the centripetal force, but how come in FBD I always see the centripetal force being drawn acting on the center of mass and not the point of contact between the wheels and the ground? Similarly, if I was dragging a pick of block, in the books (and in class) the FBD of the friction force was drawn to be acting on the center of mass and not the point of contact.

Thanks.
 
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In animated diagrams, or in the case of some racing games or simulators, you can see force diagrams at the contact patches. There are inwards centripetal forces at the contact patches. There is an outwards reactive force to the net centripetal acceleration at the center of mass, and this results in an outwards roll torque.
 
rcgldr said:
In animated diagrams, or in the case of some racing games or simulators, you can see force diagrams at the contact patches. There are inwards centripetal forces at the contact patches. There is an outwards reactive force to the net centripetal acceleration at the center of mass, and this results in an outwards roll torque.

Thanks for the response.

I didn't really understand where the outward reactive force applied to the CoM comes from and what is counteracting this roll torque?
 
Red_CCF said:
I didn't really understand where the outward reactive force applied to the CoM comes from
The reactive force is not applied to the CoM, it's the reaction of the mass of the car in response to centripetal acceleration, and it can be considered to be acting at the CoM of the car. The Newton third law pair of forces here are the net centripetal force generated by the pavement onto the tires, with an equal and opposing outwards reactive force generated from the car's mass and applied through the tires to the pavement.

and what is counteracting this roll torque?
An opposing roll torque created by the imbalance of the vertical forces between the inside and outside tires and the pavement. The upwards force from the pavement on the outside tire is greater than the upwards force on the inside tire, and this produces the counteracting torque.
 
rcgldr said:
The reactive force is not applied to the CoM, it's the reaction of the mass of the car in response to centripetal acceleration, and it can be considered to be acting at the CoM of the car. The Newton third law pair of forces here are the net centripetal force generated by the pavement onto the tires, with an equal and opposing outwards reactive force generated from the car's mass and applied through the tires to the pavement.

An opposing roll torque created by the imbalance of the vertical forces between the inside and outside tires and the pavement. The upwards force from the pavement on the outside tire is greater than the upwards force on the inside tire, and this produces the counteracting torque.

So if I draw a FBD of a car going around a circle in an inertial frame, I should have gravity acting down at the CG, a centripetal force at the tires pointing towards the inside of the circle, and two differing forces acting at the inside and outside tires with the outside being bigger? How would the FBD look if it was drawn from an non-inertial frame?

Thank you very much
 
FordRanger4x4 said:
frictional force will be on the tires facing opposite where the car wants to go. so if a car is turning left, the frictional force will be pointing left on the FBD. the CF is summed and shown on the FBD to be at the CM

Hi

What does CF stand for?

Thanks
 
Red_CCF said:
Hi

What does CF stand for?

Thanks

centripetal force, sorry. hope this helps