Solved: Forces on a Bike: Calculating Friction and Total Force

[Rockdog]

A biology student rides her bike around a corner of radius 30 meter at a steady speed of 8.1 m/sec. The combined mass of the student and the bike is 89 kg. The coefficent of static friction between the bike and the road is ìs = 0.32.

a) If she is not skidding, what is the magnitude of the force of friction on her bike from the road?
b) What is the minimum value the coefficient of static friction can have before the bike tire will skid?
c) What is the magnitude of the total force between the bike tire and the road?
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I'm stuck on part c, but maybe doing part a and b can help out also.

a) A=v^2/R
so A=8.1^2/30=2.187m/s/s
Then I used F=m*a => 89kg*2.187m/sec/sec => 194.643 Newtons

b) Well, I used Newton's second law for the x axis, and figured out that
-fs= -m(v^2/R)
where fs is static friction force, m is mass, v is velocity, R is radius
and I know that fs(max)=Us*N
where Us=static coefficient of friction
N=normal
so I plug in for fs and solve for Us.
Us=(m*v^2)/(N*R)
since N=m*g
Us=(m*v^2)/(m*g*R) or Us= (v^2)/(g*R)
plug in my numbers, and I get minimum Us to be .223. That is when the bike is on the verge of skidding.

c) Here is where I get stuck. I want the total force between bike tire and the road. I did a force diagram of the bike, with a mg force pointing down, a Normal force pointing up, and a frictional force to the left.
So I tried adding weight of the bike/girl to the frictional force, but that doesn't work. I'm pretty sure I need to add the normal force of the girl into there, but to what other force, if any, I'm stuck on.

 

[Chi Meson]

You are correct so far.

Consider the forces from the ground on the bike:

THe normal force, pointing up, and the static friction, pointing to the left. THat's it. Find the resultant of these two perpendicular components.

 

[M98Ranger]

Great job on parts a and b! To solve part c, we need to consider the forces acting on the bike in the horizontal direction. These forces include the force of friction, the force of normal reaction (from the ground), and the force of inertia (centripetal force).

First, let's find the centripetal force using the formula Fc = mv^2/R. Plugging in the values, we get Fc = 89kg * (8.1m/s)^2 / 30m = 194.43 Newtons.

Next, we can use Newton's second law to find the total force in the horizontal direction. This is given by the formula F = ma, where a is the acceleration in the horizontal direction. We can calculate the acceleration using the centripetal force and the mass of the bike and rider. So, F = (89kg * 2.187m/s^2) = 194.643 Newtons.

Now, we can set up an equation for the forces in the horizontal direction:

F - fs = ma

Where fs is the force of friction, and a is the acceleration we just calculated. We can rearrange this equation to solve for fs:

fs = ma - F

Plugging in the values, we get fs = (89kg * 2.187m/s^2) - 194.43 Newtons = 0 Newtons.

This means that the force of friction is equal to the force of inertia (centripetal force), so there is no net force in the horizontal direction. Therefore, the magnitude of the total force between the bike tire and the road is equal to the centripetal force, which we calculated to be 194.43 Newtons.

Hope this helps! Keep up the good work.