Banked Curve perfectly banked

AI Thread Summary
The discussion revolves around calculating the coefficient of static friction required for a car to avoid skidding on a perfectly banked curve with a radius of 88 meters while traveling at 95 km/h. The term "perfectly banked" refers to the angle of the curve designed to allow vehicles to navigate it without relying on friction at a specific speed, in this case, 75 km/h. The user attempts to derive equations based on forces acting on the car but struggles with understanding the implications of the banking angle and the coefficient of friction. The final answer for the coefficient of static friction is identified as 0.22, and the radius is confirmed to be in meters. The discussion emphasizes the importance of unit conversion and the relationship between speed, radius, and friction in banking curves.
starji
Messages
16
Reaction score
0
Banked Curve "perfectly banked"

Homework Statement



if a curve with a radius of 88. is perfectly banked for a car traveling 75km/h, what must be the coefficient of static friction for a car not to skid when traveling 95km/h?

Homework Equations


ƩF=ma and a=v^2/r and Ff=μFn

The Attempt at a Solution


below is my attempt but i also want to know what they mean when they say "perfectly banked" is that a specific angle or a specific coefficient of friction? because i can't seem to get anywhere.

a=752/88 Fnx=Fnsinθ, Fny=Fncosθ, Ffx=Ffcosθ, Ffy=Ffsinθ.
in the x-direction:
ƩF=ma
Fnx+Ffx=ma
Fnsinθ+μFncosθ=m(752/88)
(88Fn(sinθ+μcosθ))/752)=m

in the y-direction
ƩF=ma
Fny-Ffy-Fg=ma (no acc in y-dir)
Fncosθ-μFnsinθ=mg
(Fn(cosθ-μsinθ))/g=m

set the 2 equations equal to each other:
(88Fn(sinθ+μcosθ))/(752)=(Fn(cosθ-μsinθ))/(9.8) (Fn divides out)
(88)(9.8)sinθ+(88)(9.8)μcos=752(cosθ)-752(μsinθ)
common factor, take sin and cos out of brackets on both sides, divide:
tanθ=(752-(88)(9.8)μ)/(752μ+(88)(9.8)

and that's as far as i got help me please!
 
Last edited:
Physics news on Phys.org


oh and the answer is 0.22
 


Radius of 88 what? Meters?

I think by perfectly banked they mean no friction.

Remember to convert units as well.

Start by finding what angle its banked at.
 


yes it is 88m
 


and thank you
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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