# Coefficient of static friction

## Homework Statement

If a curve with a radius of 77.0 m is perfectly banked for a car traveling 65.8 km/hr, what must be the coefficient of static friction for a car not to skid when traveling at 91.2 km/hr?

## Homework Equations

$$\mu$$s = v^2/(g*r)

## The Attempt at a Solution

1 Incorrect. (Try 1) 0.609
2 Incorrect. (Try 2) 0.61
3 Incorrect. (Try 3) 0.616
4 Incorrect. (Try 4) 0.443
5 Incorrect. (Try 5) 0.173
6 Incorrect. (Try 6) 0.0335
7 Incorrect. (Try 7) 0.0336
8 Incorrect. (Try 8) 0.287
9 Incorrect. (Try 9) 0.847

Thanks!

rl.bhat
Homework Helper
Using the formula for angle of banking, tan(theta) = V^2/g*r, find theta. At the perfect banking, the car can round the curve without any help of the frictional force. When you increase speed addition force towards the center is needed to keep the car on the track. this force is obtained by the friction.
The additional force is (m*v2^2/r - m*v1^2/r). And the frictional force is mu*m*g*sin(theta). Now solver for mu.

Using the formula for angle of banking, tan(theta) = V^2/g*r, find theta. At the perfect banking, the car can round the curve without any help of the frictional force. When you increase speed addition force towards the center is needed to keep the car on the track. this force is obtained by the friction.
The additional force is (m*v2^2/r - m*v1^2/r). And the frictional force is mu*m*g*sin(theta). Now solver for mu.

From your explaination, I got as follow:

tan $$\theta$$ = v^2 / (g * r) = 18.28^2 / (9.81 * 77) => $$\theta$$ = 23.86 degree.

I dont understand "The additional force is (m*v2^2/r - m*v1^2/r). And the frictional force is mu*m*g*sin(theta). Now solver for mu." ??? What is the equation here? Thanks!

tiny-tim
Homework Helper
Hi huybinhs!

(have a theta: θ and a degree: º and try using the X2 tag just above the Reply box )
What is the equation here?

It's good ol' Newton's second law … Ftotal = ma, in the horizontal direction, and Ftotal = 0, in the vertical direction.

You know from the given measurements that a = mv2/r, horizontally,

and you know that the only two forces are the normal force, N, and the down-slope friction force of (maximum) µN.

Now take horizontal and vertical components of all this …

what do you get?

Hi huybinhs!

(have a theta: θ and a degree: º and try using the X2 tag just above the Reply box )

It's good ol' Newton's second law … Ftotal = ma, in the horizontal direction, and Ftotal = 0, in the vertical direction.

You know from the given measurements that a = mv2/r, horizontally,

and you know that the only two forces are the normal force, N, and the down-slope friction force of (maximum) µN.

Now take horizontal and vertical components of all this …

what do you get?

N = F = ma; but a = mv^2/r => N=F = (M2 * v2) /r.

And Fs = µs * N.

Is this correct?

I dont understand "Now take horizontal and vertical components of all this" ???

tiny-tim
Homework Helper
This is hopless.

We'd better do it in stages …

what are the three forces on the car, and what are their directions?​

This is hopless.

We'd better do it in stages …

what are the three forces on the car, and what are their directions?​

1. mg-down vertical direction

2. F(N)*cos(theta)- up vertical direction.

3. F(N)*sin(theta)- horizontal direction.

Correct? What then? ;)

With vertical direction, we have:

F(N)*cos(theta) - mg = 0 <=> F(N) = mg/cos(theta) (i)

With horizontal direction, we have:

F(N)*sin(theta) = m (v^2/r) (ii)

From i and ii, we have:

tan(theta) = v^2 (r*g).

then I calculate:

tan(theta) = 18.28^2 / (9.81 * 77) => = 23.86 degree.

What's next, man?

tiny-tim
Homework Helper
1. mg-down vertical direction

2. F(N)*cos(theta)- up vertical direction.

3. F(N)*sin(theta)- horizontal direction.

Correct? What then? ;)

(I've been away for an hour)

No, not correct …

Start again.​

(I've been away for an hour)

No, not correct …

Start again.​

The friction force is:

F(fr) = $$\mu$$* N.

What's then?

tiny-tim
Homework Helper
The friction force is:

F(fr) = $$\mu$$* N.

What's then?

ok, so what are the three forces on the car, and what are their directions?

ok, so what are the three forces on the car, and what are their directions?

1. F(N) = mg - vertical up

2. F(G) = mg - vertical down

3. F(fr) = $$\mu$$s * F(N).

Correct? What's next?

tiny-tim
Homework Helper
1. F(N) = mg - vertical up

2. F(G) = mg - vertical down

3. F(fr) = $$\mu$$s * F(N).

Correct? What's next?

No, not correct …

are we doing the same question?

the track is banked, at an angle θ.

Try again.​

No, not correct …

are we doing the same question?

the track is banked, at an angle θ.

Try again.​

1. F(N) = F(N)* cos (theta)- vertical up

2. F(G) = mg - vertical down

3. F(fr) = $$\mu$$s * F(N) = $$\mu$$s * F(N)* cos (theta) - horizontal right.

????

tiny-tim
Homework Helper
this isn't making any sense

i'm going to bed now :zzz: … i'll look at it again in the morning

Oh.....come on... I still dont get anything from you yet :(

rl.bhat
Homework Helper
From your explaination, I got as follow:

tan $$\theta$$ = v^2 / (g * r) = 18.28^2 / (9.81 * 77) => $$\theta$$ = 23.86 degree.

I dont understand "The additional force is (m*v2^2/r - m*v1^2/r). And the frictional force is mu*m*g*sin(theta). Now solver for mu." ??? What is the equation here? Thanks!
When the angle of banking is $$\theta$$, the car will go round the curve with the given velocity without the help of frictional force. The necessary centripetal force is provided by the component of N. When the velocity of the car increases, centripetal force provided by the component N is not sufficient to keep the car on the road. The additional force is needed to keep the on the road. And this force can be provided by the frictional force between the tire and the road. The frictional force is mu*N. On the flat road, normal force in mg. On the banked road what is the normal force? The additional force is (m*v2^2/r - m*v1^2/r).

Last edited:
When the angle of banking is $$\theta$$, the car will go round the curve with the given velocity without the help of frictional force. The necessary centripetal force is provided by the component of N. When the velocity of the car increases, centripetal force provided by the component N is not sufficient to keep the car on the road. The additional force is needed to keep the on the road. And this force can be provided by the frictional force between the tire and the road. The frictional force is mu*N. On the flat road, normal force in mg. On the banked road what is the normal force? The additional force is (m*v2^2/r - m*v1^2/r).

You mean:

mu*m*g*sin(theta) + (m*v2^2/r - m*v1^2/r) = 0 ????

And solve for mu???

No.
mu*m*g*cos(theta) = (m*v2^2/r - m*v1^2/r)

Ok, I calculated as follow:

tan(theta) = v^2 / (g * r) = 18.28^2 / (9.81 * 77) => theta = 23.9 degree

then:

mu*m*g*cos(theta) = (m*v2^2/r - m*v1^2/r)

=> mu = 0.442

Here is my forth try and it was incorrect:

4 Incorrect. (Try 4) 0.443 ???

Why?

rl.bhat
Homework Helper
No. I will go through it again.
In this case three forces are being considered.
i) mg, vertically down ward through C.G
ii) The normal reaction N, perpendicular to the surface inclined at an angle theta.
iii) force of friction F, parallel to the plane surface towards the center of the curvature.
In the Y direction
mg + F*sin(theta) = N*cos(theta)......(1)
In the x direction
N*sin(theta) + F*cos(theta) = m*v^2/r......(2)
Put F = mu*N .......
Eliminate m and solve for mu for the new velocity.

No. I will go through it again.
In this case three forces are being considered.
i) mg, vertically down ward through C.G
ii) The normal reaction N, perpendicular to the surface inclined at an angle theta.
iii) force of friction F, parallel to the plane surface towards the center of the curvature.
In the Y direction
mg + F*sin(theta) = N*cos(theta)......(1)
In the x direction
N*sin(theta) + F*cos(theta) = m*v^2/r......(2)
Put F = mu*N .......
Eliminate m and solve for mu for the new velocity.

Plug in theta = 23.9 degree. Correct?

rl.bhat
Homework Helper
Correct.

Correct.

mg + mu*N*sin(23.9) = N*cos(23.9) (i)

N*sin(23.9) + mu*N*cos(23.9) = m*v^2/r (ii)

Sorry. Now I have no ideas to solve these equation :(

rl.bhat
Homework Helper
m*v^2/r = N*sin(23.9) + μ*N*cos(23.9) .........(1)
mg = N*cos(23.9) - μ*N*sin(23.9)...........(2)
Divide (1) by (2) and solve for μ.

m*v^2/r = N*sin(23.9) + μ*N*cos(23.9) .........(1)
mg = N*cos(23.9) - μ*N*sin(23.9)...........(2)
Divide (1) by (2) and solve for μ.

I solved for $$\mu$$ = 5.08 ???

rl.bhat
Homework Helper
v^2/rg = [sin(23.9) + μ*cos(23.9)]/[cos(23.9) - μ*sin(23.9)]
using v = 25.33 m/s and r = 77 m, find v^2/rg. Substitute this value in the above equation and find μ. What is the expected answer?

v^2/rg = [sin(23.9) + μ*cos(23.9)]/[cos(23.9) - μ*sin(23.9)]
using v = 25.33 m/s and r = 77 m, find v^2/rg. Substitute this value in the above equation and find μ. What is the expected answer?

I'm not sure whether it's correct or not. The answer is supposed to be in 0.1 - 0.9. And with your equation, the result excesses the range :(

rl.bhat
Homework Helper
I'm not sure whether it's correct or not. The answer is supposed to be in 0.1 - 0.9. And with your equation, the result excesses the range :(
I am getting 0.29
0.85 = [ 0.405 + μ*0.914]/[0.914 - μ*0.405]
Now solve for μ.

I am getting 0.29
0.85 = [ 0.405 + μ*0.914]/[0.914 - μ*0.405]
Now solve for μ.

Therefore, the coefficient of static friction for a car not to skid when traveling at 91.2 km/hr is 0.29 = final answer, right? (because this is my last try :(