# Static and dynamic(kinetic) coefficients of friction

by thestudent101
Tags: dynamic, friction, kinetic coefficient, physics, static
 P: 18 I have a Mathematics C assignment, with one question being about static and dynamic friction. But I think it fits this forum. Anyway, we need to conduct experiments to show if there is a difference between static and dynamic friction or not with three different surfaces. The weight of the object being tested is 0.0685kg and gravity has been assumed as -9.8m/s^2 I have calculated the coefficient of static friction for the three surfaces. This was done so by placing an object of weight 0.0685kg on a surface and increasing the angle until the object starts to slide. For wood, the average angle was 37.3°, for glass - 18.0° and for metal - 14.7°. For calculating the static coefficient of friction the weight force has been calculated as w=mg w=0.0685*-9.8 w=-0.6713 newtons This has then been subbed into 0=wsinθ-Fr 0=-0.6713*sin(37.3)-Fr Fr= 0.407 newtons 0=-wcosθ+N 0=-(-0.6713*cos(37.3))+N N=0.534 newtons μ=Fr/N μ=0.407/0.534 μ=0.762 (wood) This process has been repeated for glass (μ=0.324) and for metal (μ=0.263) Now I am completely stuck on how to calculate the coefficient of dynamic friction. I was thinking about timing how long the object takes to slide down a certain length from a set angle? But I don't know how to calculate the coefficient from that? Any ideas and help will be appreciated thanks.
 Sci Advisor HW Helper PF Gold P: 6,049 You might note from your correct calculations that the static coefficient of friction is simply u_s = tan theta, which you can deduce using letter variables instead of plugging in numeric results. Does this give you a hint as to how you might calculate the dynamic coefficient of friction without using a clock or meter stick?
 P: 18 The friction and normal force also need to be calculated so that's why I did it that way. But no, unfortunately that didn't give me any hints.
 Sci Advisor HW Helper PF Gold P: 6,049 Static and dynamic(kinetic) coefficients of friction How about if the object slides down the incline at constant velocity (no acceleration)?
 P: 18 i still don't know how to incorporate that into it.
HW Helper
PF Gold
P: 6,049
 Quote by thestudent101 i still don't know how to incorporate that into it.
I am not familiar with what is covered in Maths C. You could certainly do it the way you suggested, that is,
 I was thinking about timing how long the object takes to slide down a certain length from a set angle?
where knowing the time, length, and angle, you can calculate the acceleration, then use newton's second law, F_net =ma, to solve ultimately for the kinetic friction coefficient. If you are familiar with this method, this will work fine, with the usual error in the time measurement.

On the other hand, you seem to be familiar with newton's 1st law, which says the net force is zero when an object is not moving (or just about to move). That's how you calculated the coef of static friction (u_s =tan theta). Newton's first law also says that the net force is 0 if an object is moving at constant velocity. So the same equations apply, only this time you must adjust the angle such that the object is moving at constant speed without accelerating. u_k is always equal to or less than u_s, almost always less. Can you think of a way to find this angle such that u_k = tan theta when the object is moving at constant speed ??
 P: 18 what's u_s and u_k? that's probably a really stupid question.
 P: 643 u_s is coefficient of static friction, while u_k is coefficient of kinetic friction.
HW Helper
PF Gold
P: 6,049
 Quote by thestudent101 what's u_s and u_k? that's probably a really stupid question.
Oh, sorry, u_s is the coefficient of static friction, and u_k is the coefficient of kinetic (dynamic) friction. I should have written it in LateX $\mu_s$ and $\mu_k$
 P: 18 T+Wsin(theta)-Fr+n-Wcos(theta)=ma T+Wsin(theta)-Fr+n-Wcos(theta)=0 Therefore -Wcos(theta) and N are the same. T+Wsin(theta)-Fr=0 I'm soooo confused now.