Circular motion with friction and banking - resultant forces

AI Thread Summary
A car traveling at 76 m/s on a circular track with a radius of 111 m, mass of 2500 kg, and a 21° banking angle experiences various forces. The calculations for the centripetal force (Fc) and frictional force (Fk) yield values of approximately 130090.09 N and 4723.75 N, respectively. The net force required for uniform circular motion is equal to (mv²)/r, which is identified as Fc, emphasizing that Fc represents the net force rather than a single force. The discussion also highlights the confusion surrounding the derivation of certain formulas used to calculate forces, with a suggestion to start from a free body diagram (FBD) to clarify the forces acting on the car. Understanding these concepts is crucial for solving problems related to circular motion with friction and banking.
exi
Messages
85
Reaction score
0

Homework Statement



A car travels 76 m/s around a circular track with a 111 m radius.

The car's mass is 2500 kg, the track is angled at 21°, and the coefficient of friction is 0.18.

What is the magnitude of the resultant force on the car and driver, expressed as kN?

The Attempt at a Solution



Double-checking my work here to see if I'm understanding this correctly.

If:
Fk = µ*Fn
Fn = mg/cosΘ
Fc = mv²/r

Then:
Fc = 130090.09 N
Fk = 4723.75 N

Which leaves me with two questions: Is the above correct, and what am I leaving out of the forces before summing them?

editing this to add: I've gotten a series of formulas that produce the correct answer, but I do not know where they are derived from. If anyone could help explain this, I'd appreciate it.[/color]

1. A = \frac {m(v^2cos\theta - grsin\theta)}{r}

2. B = \frac {mg + Asin\theta}{cos\theta}

3. (\mbox{Answer expressed in kN}) = \frac {Acos\theta + Bsin\theta}{1000}
 
Last edited:
Physics news on Phys.org
You're going around your butt to get to your elbow. ;-)

If an object is moving in uniform circular motion, the NET force on that object must be equal to (mv^2)/r. You have this value recorded as Fc. You should note that Fc is NOT a single force, but rather is the NET force on the object in question.

The three formulas you've listed end up spitting out the exact same value as Fc, although in a very convoluted way.
 
CaptainZappo said:
You're going around your butt to get to your elbow. ;-)

If an object is moving in uniform circular motion, the NET force on that object must be equal to (mv^2)/r. You have this value recorded as Fc. You should note that Fc is NOT a single force, but rather is the NET force on the object in question.

The three formulas you've listed end up spitting out the exact same value as Fc, although in a very convoluted way.

Oh wow, you're right - what I did in the first 30 seconds answered the problem, but I didn't know I actually had the answer. I had to call someone in class with me who was given that formula system by a physics tutor.

That's... definitely something. :shy:
 
hallo exi may you explain the set of fomulas listed below the senteces in red...@all...i have a problem in calculatiion of forces in a free body diagram...any body who can help?
 
We have to start from a FBD.

Try to find the forces which act on the car.
 
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