- #1

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What is the magnitude of the total force between the bike tire and the road?

Is it F = umg + mv^2 / R [ 'u' here is the static firction 0.39 ]

Is this right.

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- Thread starter Naeem
- Start date

- #1

- 194

- 0

What is the magnitude of the total force between the bike tire and the road?

Is it F = umg + mv^2 / R [ 'u' here is the static firction 0.39 ]

Is this right.

- #2

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Daniel.

- #3

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Force of gravity and Normal force, what else can it be?

- #4

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There must be 3:what about the friction force...?

Daniel.

Daniel.

- #5

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Ok, I agree 3 forces,

but the final eqn,

would it be something like this:

mg + mv^2/R +Us.mg

but the final eqn,

would it be something like this:

mg + mv^2/R +Us.mg

- #6

OlderDan

Science Advisor

Homework Helper

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Centripetal acceleration has to come from somewhere. The equation that tells you how big it must be, does not tell you what provides the force. Two of the forces in your equation are the same force.

- #7

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Just write Newton's second law for radial and vertical direction...

Daniel.

Daniel.

- #8

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ok you know that centripital force is given by F=mv^2/r right? since the bike is travelling in a horizintal circle, the centripital force has to be a lateral force towards the center of the circle. the only lateral force between the tire and the road is friction. in the vertical direction you have weight and normal force but they cancel each other out so they are irrelevant. therefore the total net force between the bike tire and road = Ff (force friction) and that has to equal centripital force, so the force F=Ff=mv^2/r

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