Coeffincient of static friction

AI Thread Summary
To determine the minimum coefficient of static friction required for a coffee cup to remain on a car's dash while taking a corner, the formula used is μ = v²/(rg), where v is the velocity, r is the radius of the turn, and g is the acceleration due to gravity. Initial calculations suggested a coefficient of 0.19, but it was pointed out that the arithmetic was incorrect, and the correct value is 0.78. The discussion emphasizes the importance of verifying calculations, especially in online systems, to avoid errors. Additionally, it suggests considering the forces involved, including the centripetal force needed and the normal force exerted by the cup. Understanding these forces is crucial for confirming the minimum coefficient of static friction required to prevent slipping.
pinkyjoshi65
Messages
260
Reaction score
0
Kim has her coffee cup on her car's dash when she takes a corner with radius 4m and 20km/hr. What is the minimum coefficient of static friction which would aloow the coffee cup to stay there without slipping?

this is what i did, but i think it is wrong

MU= friction force/normal force= ma/mg= a/g= v^2/rg
so mu= v^2/rg= 30.80/16*9.8=0.19
 
Physics news on Phys.org
pinkyjoshi65 said:
MU= friction force/normal force= ma/mg= a/g= v^2/rg
so mu= v^2/rg= 30.80/16*9.8=0.19
You equation is correct but your arithmetic is not. Did you square the r by mistake?
 
so mu should be = v^2/rg= 30.80/4*9.8=0.78
that is the minimum coefficient of satic friction which would allow the cup to stay in postition without slipping?
 
Yep. (If you're dealing with one of those online systems, you may want to double check the accuracy of your calculations lest you be off a bit. Redo v^2.)
 
how do you know that it is the minimum mu and tht it will make the cup stay in position?
 
Answer these questions:
(1) How much force is required to centripetally accelerate the cup?
(2) What normal force does the cup exert on the surface of the dash?
(3) What's the maximum force that static friction can supply for a given mu?

That should tell you the minimum mu required.
 
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