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Car starting to move, friction and forces

  • Thread starter Karol
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  • #1
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Homework Statement


A car of weight 17[kN] starts to move. the coefficient of friction is 0.6. the dimensions in the drawing.
The height of the CoM, point C, is 0.5[m]. the front wheels drive the car.
What are the pressing forces on the front and rear wheels and what is the friction force on the front wheels.

Homework Equations


Friction: f=μN
Moments: F1L1=F2L2

The Attempt at a Solution


Rf is the reaction on the front and Rr is on the back. the reactions caused by the weight:
$$R_f=\frac{2}{3}\cdot 17=11.33$$
The friction force on the front: f=0.6⋅11.33=6.8. it should be 4.36.
The friction "pulls the carpet" under the CoM and creates a moment which is balanced by the rear wheels:
$$0.5\cdot 6.8=R_r\cdot 2\rightarrow R_r=1.7$$
This force adds to the reaction to the weight: Rr=(17-11.33)+1.7=7.37
Of course it's not true. it should be Rr=4.47. the pressing force on the front wheels should be Rf=7.27
But the motor reacts to the torque applied to the front wheel and this moment increases the pressure on the rear wheels, but for that i need the radius of the wheel.[/SUB]

[Moderator note: repaired subscripting tags - gneill]
 

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Answers and Replies

  • #2
Svein
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The drawing suggests that 2/3 of the "weight" of the car is on the front wheels. This will give you the maximum friction force from the road on the car.
 
  • #3
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Yes, that's what i have done in my second and third lines in my calculations
 
  • #4
Svein
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Yes, that's what i have done in my second and third lines in my calculations
Sorry, but most of your reasoning is all but unreadable, since it is inside a "SUB" clause.
 
  • #5
gneill
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Homework Statement


A car of weight 17[kN] starts to move. the coefficient of friction is 0.6. the dimensions in the drawing.
The height of the CoM, point C, is 0.5[m]. the front wheels drive the car.
What are the pressing forces on the front and rear wheels and what is the friction force on the front wheels.

Homework Equations


Friction: f=μN
Moments: F1L1=F2L2

The Attempt at a Solution


Rf is the reaction on the front and Rr is on the back. the reactions caused by the weight:
$$R_f=\frac{2}{3}\cdot 17=11.33$$
The friction force on the front: f=0.6⋅11.33=6.8. it should be 4.36.
The friction "pulls the carpet" under the CoM and creates a moment which is balanced by the rear wheels:
$$0.5\cdot 6.8=R_r\cdot 2\rightarrow R_r=1.7$$
This force adds to the reaction to the weight: Rr=(17-11.33)+1.7=7.37
Of course it's not true. it should be Rr=4.47. the pressing force on the front wheels should be Rf=7.27
But the motor reacts to the torque applied to the front wheel and this moment increases the pressure on the rear wheels, but for that i need the radius of the wheel.[/SUB]
Sorry, but most of your reasoning is all but unreadable, since it is inside a "SUB" clause.
@Svein : Subscripting tag balance has been restored.
 
  • #6
jbriggs444
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You can solve angular momentum problems by choosing an arbitrary axis. You are not required to choose an axis about which something is actually rotating. You are not required to choose an axis through which an axle passes. This makes the wheel radius irrelevant.

In the usual case, almost any axis will work. You want to choose one that will simplify the problem by eliminating one or more terms from the resulting equations. Good choices are:

An axis positioned at the center of mass.
An axis positioned at the point of application of a key force.
 
  • #7
gneill
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A car of weight 17[kN] starts to move. the coefficient of friction is 0.6. the dimensions in the drawing.
The height of the CoM, point C, is 0.5[m]. the front wheels drive the car.
What are the pressing forces on the front and rear wheels and what is the friction force on the front wheels.
Is that the exact and entire problem statement?

Are you not told how the car starts to move? Surely an acceleration would make a difference in the moments.
 
  • #8
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Is that the exact and entire problem statement?

Are you not told how the car starts to move? Surely an acceleration would make a difference in the moments.
Yes. I agree. I think what the problem statement is implying is that the acceleration is so high that the front wheels are on the verge of peeling out.

Karol:

If the rear wheels are freely rotating and have negligible mass, then the torque on the rear wheels is (a) negligible or (b) not negligible.
Based on this answer, is the frictional force of the ground on the rear wheels (a) negligible or (b) not negligible.

What direction is the frictional force exerted by the ground on the front wheels? (a) rearward or (b) forward?

Chet
 
  • #9
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I don't think it's a problem of angular momentum since they didn't teach this complex subject and the chapter is about moments and forces.
And also i think i need the moment that is exerted on the front wheels and for that i need the radius, but it's not given.
These are all the details given in the question, no acceleration, so i guess it's on the verge of slipping.
Let's assume the frictional force on the rear wheels=0 and with negligible mass. these assumptions are adequate to the material learned.
I guess the frictional force is forward since the car starts to move forward.
 
  • #10
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I don't think it's a problem of angular momentum since they didn't teach this complex subject and the chapter is about moments and forces.
And also i think i need the moment that is exerted on the front wheels and for that i need the radius, but it's not given.
These are all the details given in the question, no acceleration, so i guess it's on the verge of slipping.
Let's assume the frictional force on the rear wheels=0 and with negligible mass. these assumptions are adequate to the material learned.
I guess the frictional force is forward since the car starts to move forward.
This is all correct, except for your indication that there's no acceleration. If the frictional force on the front wheels is forward, then the car is accelerating forward.

Chet
 
  • #11
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yes, correct, forward, simple
 
  • #12
jbriggs444
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I don't think it's a problem of angular momentum since they didn't teach this complex subject and the chapter is about moments and forces.
And also i think i need the moment that is exerted on the front wheels and for that i need the radius, but it's not given.
If they are teaching you about moments then you definitely do need to consider weight transfer from front to rear.

But the radius of the wheels is still irrelevant. Moments can be computed about an arbitrary chosen axis. However, in order to avoid talking about angular momentum, the chosen axis in this case would need to be at the car's center of mass. (With any other axis choice, the car would have an angular momentum that changes over time and one would be hard pressed not to talk about it).
 
  • #13
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If they are teaching you about moments then you definitely do need to consider weight transfer from front to rear.

But the radius of the wheels is still irrelevant. Moments can be computed about an arbitrary chosen axis. However, in order to avoid talking about angular momentum, the chosen axis in this case would need to be at the car's center of mass. (With any other axis choice, the car would have an angular momentum that changes over time and one would be hard pressed not to talk about it).
I respectfully disagree. One can take moments about any axis, even if the center of mass is accelerating, if one includes the moment of the pseudo force -ma. However, in this problem, it is best to take moments about the center of mass.

Chet
 
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  • #14
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There isn't angular momentum in this problem since the material is for a simpler degree than engineering, it's for practical engineers who study 2 years.
I solved 25 problems in this chapter and none with angular momentum. it's not even mentioned.
The chapter is the second in this book which is Statics, the first chapter is about vector algebra, center of mass and etc.
 
  • #15
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Yes. I agree. So where do you stand now on the solution to this problem? What is your force balance in the vertical direction, and what is your moment balance about the center of mass?

Chet
 
  • #16
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I repeat the solution i posted in #1:
Rf is the reaction on the front and Rr is on the back. the reactions caused by the weight:
$$R_f=\frac{2}{3}\cdot 17=11.33$$
The friction force on the front:
##f=0.6\cdot 11.33=6.8## it should be 4.36.
The friction "pulls the carpet" under the CoM and creates a moment which is balanced by the rear wheels:
$$0.5\cdot 6.8=R_r\cdot 2\rightarrow R_r=1.7$$
This force adds to the reaction to the weight: ##R_r=(17-11.33)+1.7=7.37##
It should be Rr=4.47. the pressing force on the front wheels should be Rf=7.27
I don't feel comfortable at all with the solution but i had to do something.
 
  • #17
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Let NR be the normal force on the rear wheels and let NF be the normal force on the front wheels. Let μNF be the frictional force on the front wheels. With this notation, what is the force balance in the vertical direction and what is the moment balance about the center of mass?

Chet
 
  • #18
jbriggs444
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I respectfully disagree. One can take moments about any axis, even if the center of mass is accelerating, if one includes the moment of the pseudo force -ma. However, in this problem, it is best to take moments about the center of mass.
I agree with your disagreement, but was trying not to include that pseudo-force.
 
  • #19
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In posts #1 and #16, you omitted the moment about the center of mass caused by the frictional force.
 
  • #20
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$$\left\{ \begin{array}{l} NF+NR=17 \\ NF(0.5\cdot\mu+1)=2\cdot NR \end{array} \right.$$
$$\rightarrow NF=10.3, \ NR=6.7, \ \mu NF=6.18$$
It should be: NF=7.27, NR=4.47, μNF=4.36
 
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  • #21
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$$\left\{ \begin{array}{l} NF+NR=17 \\ NF(0.5\cdot\mu+1)=2\cdot NR \end{array} \right.$$
$$\rightarrow NF=10.3 \ NR=6.7 \ \mu NF=6.18$$
It should be: NF=7.27, NR=4.47, μNF=4.36
I get the same result that you got, and I'm pretty confident in what I did. In the results that they got, the sum of the normal forces don't add up to the weight.

Chet
 
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  • #22
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Good i will accept that. but what about the reaction to the moment applied to the wheel? i see cars that at start up their rear goes down, but those were old cars, i don't know if they were front or back wheels driven.
Newton's third law applies to the motor and wheel.
 
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  • #23
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Good i will accept that. but what about the reaction to the moment applied to the wheel? i see cars that at start up their rear goes down, but those were old cars, i don't know if they were front or back wheels driven.
Newton's third law applies to the motor and wheel.
I really don't understand what you are asking here. You have already taken into account all the external forces acting on the combination of car and wheels. Also, your results show that the rear of the car would go down when the car is accelerating. After all, the normal force on the rear wheels is higher when the car is accelerating than when the car is not accelerating, and the normal force on the front wheels is lower when the car is accelerating than when the car is not accelerating.

Chet
 
  • #24
Svein
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The height of the CoM, point C, is 0.5[m]. the front wheels drive the car.
Since the front wheels drive the car, the downward force on the rear wheels will be slightly less when accelerating (consider the moments around CoM).
 
  • #25
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Thanks Chet
 

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