Calculate Force on Bike at 8.3 m/s: umg + mv^2/R

[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.

 

[dextercioby]

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

Daniel.

 

[Naeem]

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

 

[dextercioby]

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

Daniel.

 

[Naeem]

Ok, I agree 3 forces,

but the final eqn,

would it be something like this:

mg + mv^2/R +Us.mg

 

[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.

 

[dextercioby]

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

Daniel.

 

[Felix83]

ok you know that centripital force is given by F=mv^2/r right? since the bike is traveling 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

 

[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