# Forces on a Bike

1. Apr 19, 2005

### Naeem

Q. A biology student rides her bike around a corner of radius 24 meter at a steady speed of 8.3 m/sec. The combined mass of the student and the bike is 87 kg. The coefficent of static friction between the bike and the road is μs = 0.39.

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. Apr 19, 2005

### dextercioby

No.The friction force is centripetal force...It prevents slipping.So what are the forces involved there...?

Daniel.

3. Apr 19, 2005

### Naeem

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

4. Apr 20, 2005

### dextercioby

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

Daniel.

5. Apr 20, 2005

### Naeem

Ok, I agree 3 forces,

but the final eqn,

would it be something like this:

mg + mv^2/R +Us.mg

6. Apr 20, 2005

### OlderDan

Forces are vectors. Any addition of forces in this problem is going to involve vector sums.

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. Apr 20, 2005

### dextercioby

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

Daniel.

8. Apr 20, 2005

### Felix83

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

Last edited: Apr 20, 2005
9. Apr 20, 2005

### shyboy

Felix83, it looks like the question about the force which the road exerts on the bike tire. That means it includes the normal recation (horizontal) and the reaction directed towards the center of the turn. Both forces dived between two tires, and they perpendicular to each other. You need to use the Pythagoras' theorem

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