Finding the coefficient of friction

Dec 2015
2
0
Ok. This is the verbatim problem: A skateboarder is trying to take a sharp turn (radius of 1.2m). She grabs hold of a tree branch to help make the turn. She pulls with 45 N of force. If she makes the turn going 3.2 m/s, what is the coefficient of friction between the wheels and ground. The mass of the girl and skateboard together is 35 kg.

The first thing I did was write down all my givens and equations. I wrote Ff=uFn, Fc=Ff and Fc= mv^2/r. I plugged in and got Ff=(35(3.2)^2)/1.2 which is equal to 298.666. Therefore uFn= 298.6666 and because Fn =mg, umg=298.666. After plugging mass and gravity into that and solving, I got u=.8707. The problem is that others got another answer by putting that Fc=Fa-Ff and going from there. I'm not sure which answer is correct. I think mine may be wrong because I never included the 45 N force.
 

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
Yes, you need to include the 45N force - it assists in making the turn, so that the force due to friction is less. What you should have is:

mv^2/R = 45N + u mg

Solve for u.
 

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
Two reasons:

1. Try it, and you'll see that with that negative sign u has to be negative to make the math work.

2. Because that would imply that the greater the friction force (umg) the smaller is the value mv^2/R, meaning the slower the skater has to go to make the turn.

Both the 45N force and the friction force point towards the center of curvature, assisting the skater to make her turn. From F=ma you have F = 45 + umg, and ma = mv^2/R.