What Forces Act on a Car Negotiating a Bend at Constant Speed?

[timarli]

Hi,


I am confused a little bit with the forces acting on a vehicle when it's negotiating a bend, at constant speed.

Please click the link to see my sketch.

On the plan view I see that it's moving with constant speed, along a circular route. That means there is unbalanced force causing central acceleration. F=mv2/R.

Then going to the section I showed the forces acting on the vehicle. Gravity, Normal Force and the friction. These should be the forces causing this F (central acceleration).

Equation 1, 2, 3 and 4 are the ones I have but somehow I couldn't get to the 5 which is given as the overall formula for such movements.

Can you please tell me what I am doing wrong? Thanks in advance.

 

[SammyS]

Hi,


I am confused a little bit with the forces acting on a vehicle when it's negotiating a bend, at constant speed.

Please click the link to see my sketch.

On the plan view I see that it's moving with constant speed, along a circular route. That means there is unbalanced force causing central acceleration. F=mv2/R.

Then going to the section I showed the forces acting on the vehicle. Gravity, Normal Force and the friction. These should be the forces causing this F (central acceleration).

Equation 1, 2, 3 and 4 are the ones I have but somehow I couldn't get to the 5 which is given as the overall formula for such movements.

Can you please tell me what I am doing wrong? Thanks in advance.

Equation #2 is the problem.

$\displaystyle F_\text{f}=\mu N$

In your equation #5, shouldn't there be a minus sign in the denominator if $\displaystyle \ \frac{v^2}{R}>g\tan(\theta)\ ?$

 

[timarli]

Equation #2 is the problem.

$\displaystyle F_\text{f}=\mu N$

In your equation #5, shouldn't there be a minus sign in the denominator if $\displaystyle \ \frac{v^2}{R}>g\tan(\theta)\ ?$

Thanks SammyS.

Eq#2: if N=m.g.cos0 is not true then I'm more confused...But equation#5. I hope you are right. That's the main issue that confuses me; every source gives a different formula, some of them take the directions different, some of them find the signs of cos & sin different. I really don't understand how this thing can have so many variations...One other question; instead of breaking down everything along major x and y-axis es, can I use the inclined surface as my x and the perpendicular as y?

Fc.cos0=Ff+m.g.sin0 ?

 

[SammyS]

Thanks SammyS.

Eq#2: if N=m.g.cos0 is not true then I'm more confused...


But equation#5. I hope you are right. That's the main issue that confuses me; every source gives a different formula, some of them take the directions different, some of them find the signs of cos & sin different. I really don't understand how this thing can have so many variations...


One other question; instead of breaking down everything along major x and y-axis es, can I use the inclined surface as my x and the perpendicular as y?

Fc.cos0=Ff+m.g.sin0 ?

Yes, you can do that.

Added in Edit:

The perpendicular component then gives:

F_C sin(θ) + N = mg cos(θ) .

 

[haruspex]

Eq#2: if N=m.g.cos0 is not true then I'm more confused...

If N=m.g.cosθ then there is no net force perpendicular to the plane. But F_c has a component in that direction, so this cannot be true.
Also, it is not true in general that the force of static friction is given by F_sf = Nμ_s. That's the limiting case, i.e. |F_sf| ≤ Nμ_s. For the max speed, F_sf = Nμ_s, assuming F_sf is measured down the plane. But, measuring it the same way, there may also be a minimum speed given by F_sf = -Nμ_s.

 

[SammyS]

Using your sketch showing the vehicle on the inclined roadway, I get the following.

Of course, the magnitude of the force of friction is given by

$\displaystyle F_\text{f}=\mu N$​

The sum of the forces in the vertical direction is:

$\displaystyle N\cos(\theta)-F_\text{f}\sin(\theta)-mg=0$

Therefore,

$\displaystyle N\cos(\theta)-\mu N\sin(\theta)-mg=0$​

Solve that for N.


The sum of the forces in the vertical direction is:

$\displaystyle N\sin(\theta)+F_\text{f}\cos(\theta)=ma_c\,, \$ where a_C is the centripetal acceleration.

That becomes $\displaystyle \ \ N\sin(\theta)+\mu N\cos(\theta)=ma_c\ .$​

I realize that I didn't use the centripetal force, F_C explicitly. I find it's clearer for me to do it this way.

 

[SammyS]

If N=m.g.cosθ then there is no net force perpendicular to the plane. But F_c has a component in that direction, so this cannot be true.
Also, it is not true in general that the force of static friction is given by F_sf = Nμ_s. That's the limiting case, i.e. |F_sf| ≤ Nμ_s. For the max speed, F_sf = Nμ_s, assuming F_sf is measured down the plane. But, measuring it the same way, there may also be a minimum speed given by F_sf = -Nμ_s.

haruspex makes a good point regarding μN being the maximum possible frictional force.

Therefore, your results are for the maximum speed possible, our equivalently they're for the minimum value of μ needed to negotiate the curve under the specified conditions .

 

[timarli]

If N=m.g.cosθ then there is no net force perpendicular to the plane. But F_c has a component in that direction, so this cannot be true.
Also, it is not true in general that the force of static friction is given by F_sf = Nμ_s. That's the limiting case, i.e. |F_sf| ≤ Nμ_s. For the max speed, F_sf = Nμ_s, assuming F_sf is measured down the plane. But, measuring it the same way, there may also be a minimum speed given by F_sf = -Nμ_s.

I couldn't understand the first part. To me, Fc is a resultant force, it's not a force acting on the body so I always thought I can not write it as "m.g.cos0-Fc.sin0 = N". I don't know why because the same thing makes sense to me for the other axis ie. "Fc.cos0=Ff+m.g.sin0". :S

Looks like I am misinterpreting some fundamental rules.


I got it! I understand what you mean now. When I use the secondary axis es (along the slope and perpendicular) and think of the position of the object 2 seconds later I can see the acceleration along both axis es.


Thank you very much SammyS and haruspex! I understand what you are saying about the friction and I will look at the minimum speed version tomorrow - to make sure I understood :) and probably post again if I can't figure that out.

You were really helpful...thanks again.