Coefficient of static friction

[huybinhs]

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

I still have my last try. Please help!

Thanks!

 

[rl.bhat]

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.

 

[huybinhs]

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 explanation, I got as follow:

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

I don't 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]

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 … F_total = ma, in the horizontal direction, and F_total = 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? ​

 

[huybinhs]

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 … F_total = ma, in the horizontal direction, and F_total = 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 don't understand "Now take horizontal and vertical components of all this" ?

 

[tiny-tim]

This is hopless.

We'd better do it in stages …

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

 

[huybinhs]

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? ;)

 

[huybinhs]

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).

 

[huybinhs]

then I calculate:

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

What's next, man?

 

[tiny-tim]

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 …

what about the friction force??

Start again.​

 

[huybinhs]

(I've been away for an hour)

No, not correct …

what about the friction force??

Start again.​

The friction force is:

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

What's then?

 

[tiny-tim]

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?

 

[huybinhs]

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]

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.​

 

[huybinhs]

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]

this isn't making any sense

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

 

[huybinhs]

Oh...come on... I still don't get anything from you yet :(

 

[rl.bhat]

From your explanation, I got as follow:

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

I don't 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).

 

[huybinhs]

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?

 

[huybinhs]

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]

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.

 

[huybinhs]

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]

Correct.

 

[huybinhs]

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]

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.

 

[huybinhs]

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]

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?

 

[huybinhs]

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]

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.

 

[huybinhs]

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 :(

 

[huybinhs]

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

You r correct. The correct answer is 0.296. Thanks so much for your help! ;)