How Do I Determine Force of Static Friction on a Rolling Car

In summary, a 1450-kg car towing a trailer of mass 454-kg experiences a force of air resistance of 7471N[backward]. The acceleration of both vehicles is 0.225m/s^2[forward]. To find the force of static friction on the wheels from the ground, one must set up an equation using the sum of forces, including the force of air resistance and the force of static friction. The force applied by the tires onto the ground and the reaction force from the applied force are not necessary to include in the equation. Rather, the force of static friction and the force of air resistance are the only two forces needed to find the net force.
  • #1
Liam C
57
4

Homework Statement


A 1450-kg car is towing a trailer of mass 454-kg. The force of air resistance on both vehicles is 7471N[backward]. If the acceleration of both vehicles is 0.225m/s^2[forward], what is the force of static friction on the wheels from the ground?

Homework Equations


Fnet = ma
Ff,s = UsFN (where Ff,s is the force of static friction, Us is the coefficient of static friction and FN is the force of the normal).

The Attempt at a Solution


I know that since the tires are rolling, the force applied must be backward and the force of friction must be forward. As a result, the final answer should be positive.
Mc = 1450-kg, Mt =454-kg, FairR = 7471N[backward], a = 0.225m/s^2[forward], Δm = Mc + Mt, Ff,s = ?
Because of Newton's third law, I know that the applied force has a reactionary force pushing the car forward. As a result, the net force of the entire system(trailer and car) is:
Fnet = Δm x a
Fnet = (1450 + 454)(0.225)
Fnet = 428.4N [forward]
Given this, I think that I should find another equation for net force and then either solve for Ff,s or Us. I have tried this several times, but I can't seem to get it right. I would really appreciate some help as I have been stuck on this problem for hours.
 
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  • #2
Liam C said:

Homework Statement


A 1450-kg car is towing a trailer of mass 454-kg. The force of air resistance on both vehicles is 7471N[backward]. If the acceleration of both vehicles is 0.225m/s^2[forward], what is the force of static friction on the wheels from the ground?

Homework Equations


Fnet = ma
Ff,s = UsFN (where Ff,s is the force of static friction, Us is the coefficient of static friction and FN is the force of the normal).

The Attempt at a Solution


I know that since the tires are rolling, the force applied must be backward and the force of friction must be forward. As a result, the final answer should be positive.
Mc = 1450-kg, Mt =454-kg, FairR = 7471N[backward], a = 0.225m/s^2[forward], Δm = Mc + Mt, Ff,s = ?
Because of Newton's third law, I know that the applied force has a reactionary force pushing the car forward. As a result, the net force of the entire system(trailer and car) is:
Fnet = Δm x a
Fnet = (1450 + 454)(0.225)
Fnet = 428.4N [forward]
Given this, I think that I should find another equation for net force and then either solve for Ff,s or Us. I have tried this several times, but I can't seem to get it right. I would really appreciate some help as I have been stuck on this problem for hours.
I think the first thing you need to do is draw a diagram showing all of the forces and their directions. That should make it a lot easier to write your second equation for the net force.
 
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  • #3
tnich said:
I think the first thing you need to do is draw a diagram showing all of the forces and their directions. That should make it a lot easier to write your second equation for the net force.
Yes, I have already done that, and I have attached it.
Based on this I think the sum of the forces equation looks like this:
Fnet = Fr + Ff,s - FairR - FA
Am I on the right tracK? I am having trouble extracting an answer from this.
 

Attachments

  • freeBodyDiagram.png
    freeBodyDiagram.png
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  • #4
I should mention that that diagram is as if the two vehicles were a single object.
 
  • #5
Liam C said:
I should mention that that diagram is as if the two vehicles were a single object.
I am not sure what all of your forces are, and I don't see ##F_{net} = ma## in the diagram at all. You already know that the force of air resistance is backward. That looks correct in your diagram. You also have the force of static friction in the diagram and you have assigned a direction. What ##F_a## and ##F_r## represent is not clear. I think the three forces, ##F_{net}##, ##F_{airR}## and ##F_{f,s}## are the only ones you need for your ##F_{net}## equation.
 
  • #6
tnich said:
I am not sure what all of your forces are, and I don't see ##F_{net} = ma## in the diagram at all. You already know that the force of air resistance is backward. That looks correct in your diagram. You also have the force of static friction in the diagram and you have assigned a direction. What ##F_a## and ##F_r## represent is not clear. I think the three forces, ##F_{net}##, ##F_{airR}## and ##F_{f,s}## are the only ones you need for your ##F_{net}## equation.
Thank you for your help, I really appreciate it.
FA was supposed to be the applied force from the tire onto the ground and Fr was supposed to be the reaction force from FA. I'm going to try to write an equation with just those three forces now.
 
  • #7
Liam C said:
Thank you for your help, I really appreciate it.
FA was supposed to be the applied force from the tire onto the ground and Fr was supposed to be the reaction force from FA. I'm going to try to write an equation with just those three forces now.
When you make a free-body diagram like that, you need to be careful to just draw in the forces acting on the object, and no forces on other objects. So in your problem, you don't need to draw the force of the wheels on the road. It would just cancel out the force of the road on the wheels. And in this case, the (forward) force of the road on the wheels is just ##F_{f,s}##, the force of static friction on the wheels. So ##F_{f,s}## and ##F_r## mean the same thing in your diagram.
 
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  • #8
tnich said:
When you make a free-body diagram like that, you need to be careful to just draw in the forces acting on the object, and no forces on other objects. So in your problem, you don't need to draw the force of the wheels on the road. It would just cancel out the force of the road on the wheels. And in this case, the (forward) force of the road on the wheels is just ##F_{f,s}##, the force of static friction on the wheels. So ##F_{f,s}## and ##F_r## mean the same thing in your diagram.
Ohh that makes so much more sense! I will think of that from now on.
 
  • #9
tnich said:
When you make a free-body diagram like that, you need to be careful to just draw in the forces acting on the object, and no forces on other objects. So in your problem, you don't need to draw the force of the wheels on the road. It would just cancel out the force of the road on the wheels. And in this case, the (forward) force of the road on the wheels is just ##F_{f,s}##, the force of static friction on the wheels. So ##F_{f,s}## and ##F_r## mean the same thing in your diagram.
Fnet = Ff,s + FairR
Ff,s = Fnet - FairR
Ff,s = 7899.4
Ff,s = 7.90 x 10 ^ 3 N [forward]
Thanks again!
 
  • #10
Liam C said:
Fnet = Ff,s + FairR
Ff,s = Fnet - FairR
Ff,s = 7899.4
Ff,s = 7.90 x 10 ^ 3 N [forward]
Thanks again!
In your diagram ##F_{airR}## and ##F_{f,s}## point in different directions, but in your equation you give them the same sign. So your answer must be incorrect. Here is the way to do it without errors - all the force have to add to zero, so set 0 equal to their sum. Give all of the forward-point forces a positive sign in the sum, and all of the backward-pointing forces a negative sign.
 
  • #11
tnich said:
In your diagram ##F_{airR}## and ##F_{f,s}## point in different directions, but in your equation you give them the same sign. So your answer must be incorrect. Here is the way to do it without errors - all the force have to add to zero, so set 0 equal to their sum. Give all of the forward-point forces a positive sign in the sum, and all of the backward-pointing forces a negative sign.
For a static equilibrium, all forces must balance. Such is not the case here.

F=ma. All the forces have to add to ma.
 
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  • #12
tnich said:
In your diagram ##F_{airR}## and ##F_{f,s}## point in different directions, but in your equation you give them the same sign. So your answer must be incorrect. Here is the way to do it without errors - all the force have to add to zero, so set 0 equal to their sum. Give all of the forward-point forces a positive sign in the sum, and all of the backward-pointing forces a negative sign.
In my equation they do not have the same sign. In the first line I am adding a positive number (Ff,s) to a negative number (FairR). When I move the FairR to the other side of the equation and subtract, it becomes a double negative, making it a positive. As a result, I add the net force to the force of the air resistance.
 
  • #13
jbriggs444 said:
For a static equilibrium, all forces must balance. Such is not the case here.

F=ma. All the forces have to add to ma.
Right. One of his forces is ##F_{net} = ma##.
 
  • #14
tnich said:
Right. One of his forces is ##F_{net} = ma##.
Normally students in introductory courses are taught to work in inertial frames of reference where "ma" is on one side of the equation and all of the forces ##F_{net}## are on the other. I was hoping to guide @Liam C toward the more standard way of casting the equation.
 
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  • #15
Liam C said:
In my equation they do not have the same sign. In the first line I am adding a positive number (Ff,s) to a negative number (FairR). When I move the FairR to the other side of the equation and subtract, it becomes a double negative, making it a positive. As a result, I add the net force to the force of the air resistance.
OK. I think that works.
 
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  • #16
jbriggs444 said:
Normally students in introductory courses are taught to work in inertial frames of reference where "ma" is on one side of the equation and all of the forces ##F_{net}## are on the other. I was hoping to guide @Liam C toward the more standard way of casting the equation.
You are right. And I think that is what he did anyway.
 
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Related to How Do I Determine Force of Static Friction on a Rolling Car

1. What is static friction?

Static friction is the force that prevents two surfaces from sliding against each other when there is no relative motion between them. It acts in the opposite direction of the applied force and increases as the applied force increases.

2. How do I calculate the force of static friction on a rolling car?

The force of static friction on a rolling car can be calculated by multiplying the coefficient of static friction between the car's tires and the road surface by the weight of the car. This will give you the maximum amount of force that can be applied to the car before it starts to slide.

3. What factors affect the force of static friction on a rolling car?

The force of static friction on a rolling car can be affected by factors such as the weight of the car, the surface roughness of the tires and road, and the coefficient of static friction between the tires and the road surface.

4. How does the coefficient of static friction affect the force on a rolling car?

The coefficient of static friction is a measure of the roughness of the surfaces in contact. A higher coefficient of static friction means that the surfaces have a stronger grip on each other, resulting in a higher maximum force of static friction on the rolling car.

5. Can the force of static friction on a rolling car be greater than the weight of the car?

No, the force of static friction on a rolling car cannot be greater than the weight of the car. This is because the maximum force of static friction is determined by the weight of the car and the coefficient of static friction, and it cannot exceed the weight of the car without causing it to slide.

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