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

[Liam C]

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.

 

[tnich]

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.

 

[Liam C]

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.

 

[Liam C]

I should mention that that diagram is as if the two vehicles were a single object.

 

[tnich]

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.

 

[Liam C]

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.

 

[tnich]

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.

 

[Liam C]

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.

 

[Liam C]

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!

 

[tnich]

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.

 

[jbriggs444]

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.

 

[Liam C]

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.

 

[tnich]

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 $F_{net} = ma$.

 

[jbriggs444]

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.

 

[tnich]

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.

 

[tnich]

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.