Physics Help Forum Finding the coefficient of friction

 Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum

 Dec 8th 2015, 08:28 AM #1 Junior Member   Join Date: Dec 2015 Posts: 2 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=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.
 Dec 8th 2015, 11:46 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,333 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.
 Dec 9th 2015, 02:26 AM #3 Junior Member   Join Date: Dec 2015 Posts: 2 Why wouldn't it be mv^2/r=45 N- umg?
 Dec 9th 2015, 04:30 AM #4 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,333 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.

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