Coefficient of Static Friction

AI Thread Summary
The experiment aimed to measure static friction by dragging a block across a surface, but static friction is not applicable once the block is in motion; this scenario involves kinetic friction instead. When graphed, the relationship between mass (or normal force) and applied force should yield a linear relationship for static friction, but the experiment's context indicates that kinetic friction is at play. The confusion arises from the misapplication of static friction principles during the experiment. Therefore, the results may not align with expectations for static friction. Understanding the distinction between static and kinetic friction is crucial for accurate experimental analysis.
SwAnK
Messages
51
Reaction score
0
Hey quick question, I did this experiment where I dragged a block across a flat surface with a scale and found the static friction, ehn weight was added to the block and repeated the process. MY question is when graphed with the mass or Fn on the x-axis, and applied force or Fs on the x axis, should it not give a straight line through 0,0 which would be the coeffecient of the static friction?? I am just wondering to know if I am just doing the graph wrong or got messed up results during the experiment.
thanx again
 
Physics news on Phys.org
If a block is being dragged, static friction is not involved. This involves kinetic friction.
 
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top