Finding the coefficient of friction
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=FaFf 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.
