Car negotiating bend - Forces

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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.

W13638035282205281_1.jpg



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.
 

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  • #2
SammyS
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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.

W13638035282205281_1.jpg



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.

[itex]\displaystyle F_\text{f}=\mu N[/itex]

In your equation #5, shouldn't there be a minus sign in the denominator if [itex]\displaystyle \ \frac{v^2}{R}>g\tan(\theta)\ ?[/itex]
 
  • #3
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Equation #2 is the problem.

[itex]\displaystyle F_\text{f}=\mu N[/itex]

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


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 ?
 
  • #4
SammyS
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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:

FC sin(θ) + N = mg cos(θ) .
 
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  • #5
haruspex
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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 Fc 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 Fsf = Nμs. That's the limiting case, i.e. |Fsf| ≤ Nμs. For the max speed, Fsf = Nμs, assuming Fsf is measured down the plane. But, measuring it the same way, there may also be a minimum speed given by Fsf = -Nμs.
 
  • #6
SammyS
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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
[itex]\displaystyle
F_\text{f}=\mu N[/itex]​

The sum of the forces in the vertical direction is:
[itex]\displaystyle
N\cos(\theta)-F_\text{f}\sin(\theta)-mg=0[/itex]

Therefore,

[itex]\displaystyle N\cos(\theta)-\mu N\sin(\theta)-mg=0[/itex]​
Solve that for N.


The sum of the forces in the vertical direction is:
[itex]\displaystyle
N\sin(\theta)+F_\text{f}\cos(\theta)=ma_c\,, \ [/itex] where aC is the centripetal acceleration.

That becomes [itex]\displaystyle \ \ N\sin(\theta)+\mu N\cos(\theta)=ma_c\ . [/itex]​
I realize that I didn't use the centripetal force, FC explicitly. I find it's clearer for me to do it this way.
 
  • #7
SammyS
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If N=m.g.cosθ then there is no net force perpendicular to the plane. But Fc 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 Fsf = Nμs. That's the limiting case, i.e. |Fsf| ≤ Nμs. For the max speed, Fsf = Nμs, assuming Fsf is measured down the plane. But, measuring it the same way, there may also be a minimum speed given by Fsf = -Nμs.
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 .
 
  • #8
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If N=m.g.cosθ then there is no net force perpendicular to the plane. But Fc 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 Fsf = Nμs. That's the limiting case, i.e. |Fsf| ≤ Nμs. For the max speed, Fsf = Nμs, assuming Fsf is measured down the plane. But, measuring it the same way, there may also be a minimum speed given by Fsf = -Nμs.


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.
 

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