Car banking an angle

  • Thread starter jgens
  • Start date
  • #1
jgens
Gold Member
1,583
50

Homework Statement



A curve of radius 45 m is banked for a design speed of 90 km/hr. If the coefficient of static friction is .3 what is the maximum speed that the car can go around the curve safely. What is the minimum speed?



Homework Equations



n/a

The Attempt at a Solution



θ = tan^-1((25 m/s)^2/(45m * 9.8 m/s^2) = 54.8 degrees

Fn(sinθ) + Ff = (m)(v^2)/r

Fn(cosθ) = mg

Fn= (mg)/(cosθ)

mg(tanθ) + μ(Fn) = (m)(v^2)/r

mg(tanθ) + μ(mg/cosθ) = (m)(v^2)/r

g(tanθ) + μ(g/cosθ) = (v^2)/r

45m((9.8m/s^s(tan(54.8)) + .3(9.8m/s^s/cos(54.8)) = (v^2)

v = 29.23 m/s or 105.25 km/hr.

The key that I'm checking this with says that the above answer is not correct. I haven't tried to find the minimum speed yet because my maximum is incorrect. Could someone please let me know where I'm making a mistake or if I'm going about this the wrong way? Thanks.
 

Answers and Replies

  • #2
Shooting Star
Homework Helper
1,977
4
Draw a freebody diagram. The forces acting on the car on the inclined road at a point are the force of friction inward, the normal reaction and the weight. The sum of the horizontal component of these is equal to the centripetal force.

I’m quite sure that you have done just these, but if you plug in numbers right at the beginning, it makes things look messy and hard to check. Use symbols.
 
  • #3
jgens
Gold Member
1,583
50
I tried this problem again with out putting numbers in at the beginning.

(m)(v^2)/r = Fn(sinθ) + Ff

(m)(v^2)/r = mg(tanθ) + Ff

(m)(v^2)/r = (m)(v1^2)/r + Ff

(m)(v^2)/r = (m)(v1^2)/r + μ(m)(v1^2)/(r(sinθ))

v^2 = (v1^2) + μ(v1^2)/(sinθ)

v^2 = (25 m/s)^2 + 0.3(25 m/s)^2/(sin(54.8))

v = 29.23 m/s or 105.23 km


No matter how I go about this I always get the same answer. Could someone point out what I'm doing wrong or at least steer me in the right direction? Thanks.
 
  • #4
jgens
Gold Member
1,583
50
My freebody diagram looks like that. But aren't the two angles equal? (alpha and beta)
 
  • #5
1,753
1
My freebody diagram looks like that. But aren't the two angles equal? (alpha and beta)
I get 36.286 m/s. What is the actual answer?

My final equation was ...

[tex]v=\sqrt{\frac{-Rg(\mu_s\cos \beta+\sin \alpha)}{\mu_s\sin \beta-\cos \alpha}}}[/tex]

And yes, alpha and beta are the same b/c I used trig identities and I think you did too ...
 
Last edited:
  • #6
jgens
Gold Member
1,583
50
That is the correct answer. I found out how you got that formula and I've also solved for the lowest speed which is about 18.6 m/s or 66.96 km/hr. Thank you for all of your help.
 
  • #7
Pyrrhus
Homework Helper
2,179
1
Alternatively, if you would have chosen your reference axis differently, the only force with an angle would have been gravity.
 
  • #8
1,753
1
Alternatively, if you would have chosen your reference axis differently, the only force with an angle would have been gravity.
You forgot about the acceleration towards the center.
 
  • #9
andrevdh
Homework Helper
2,128
116

Homework Statement



....

Fn(sinθ) + Ff = (m)(v^2)/r

Fn(cosθ) = mg

....
Thanks.

both these equations are wrong they should read

Fn(sinθ) + Ff(cosθ) = (m)(v^2)/r

and

Fn(cosθ) = mg + Ff(sinθ)
 

Related Threads on Car banking an angle

Replies
6
Views
3K
  • Last Post
Replies
6
Views
3K
Replies
3
Views
10K
Replies
1
Views
4K
  • Last Post
Replies
4
Views
1K
Replies
1
Views
7K
  • Last Post
Replies
5
Views
6K
Replies
15
Views
2K
Replies
15
Views
615
Replies
3
Views
7K
Top