Why is kinetic friction always lower than maximum static friction?

AI Thread Summary
Kinetic friction is always lower than maximum static friction because static friction acts to prevent movement until a certain threshold is reached. When a force is applied to an object at rest, static friction counteracts this force, maintaining the object's position. Once the applied force exceeds the maximum static friction, the object begins to move, and kinetic friction takes over. Kinetic friction is inherently lower because it involves sliding surfaces, which reduces the interlocking of surface irregularities compared to static friction. Understanding these concepts is crucial for analyzing motion and forces in physics.
Peter G.
Messages
439
Reaction score
0
I was reading about those in my textbook and they drew an example of a block resting on a ramp. The diagram showed the force of friction.

How is there friction if the block is not moving?

Thanks,
Peter G.
 
Physics news on Phys.org
Ok, thanks!
 
Well, if any force is applied on the object (even if it's gravity pulling it down a slope) then there's static friction to prevent movement, or attempt to prevent it anyway (so it'll be opposite to where the force wants to cause movement). Once the force is strong enough to achieve movement on the object, static friction becomes kinetic friction, which-- curiously and interestingly enough, is always lower than the max static friction.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top