# Finding friction w/ out coefficient?

1. Nov 16, 2015

### Mrchilko

1. The problem statement, all variables and given/known data
Hey new to forum here.... Grade 11 physics student... Love physics and math!!!! But this one question seems easy but I can't seem to get through... Here it is : A 1100kg car accelerates at 3.40 m/s. If the wheels exert a force of 5600N on the road, calculate the force of friction resisting the motion?

2. Relevant equations
F net= m x a ... But I don't which equation to use.. I know friction is normally calculated with F fr = u * Fn

3. The attempt at a solution
I have the answer... But don't know where to start other than maybe finding the net force..

2. Nov 16, 2015

### SteamKing

Staff Emeritus
First, acceleration has units of m/s2, not m/s, which are the units for velocity.

Second, you should draw a free body diagram of the car. Remember, the car is accelerating, so what does that tell you about Fnet?

3. Nov 16, 2015

### Mrchilko

Well if its accelerating doesn't that mean... Its an unbalanced force? And that I can determine the Fnet... And ye .. Hahah i know that acceleration has the squared symbol... But I don't know how to put it on my computer :/

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4. Nov 16, 2015

### SteamKing

Staff Emeritus
When you draft a post, the Green tool bar over the edit box contains several different buttons. The button marked x2 supplies superscripts for your text.

Looking at your free body diagram, you have shown the forces acting on the car bass ackwards. The weight of the car is not important here, only the balance between the tractive force (5600 N) and the opposing friction force, which you are supposed to calculate, given that the car is accelerating in a horizontal, rather than a vertical, direction.

You are still trying to calculate the friction force as if you knew a coefficient of friction (which you don't) by turning the tractive force into a part of the normal force (which it isn't).

5. Nov 16, 2015

### Mrchilko

Ahhhh I dont know.... This question must be sooo easy.... Like were talking newtons second law here..... I don't want to give up on it either. :/ physics may not be my thing... If something this easy doesn't click to my head like some people... But how can mass have no significance in this question if Fnet is = mass * acceleration.?

6. Nov 16, 2015

### haruspex

This problem is horribly worded.
"The force the wheels exert on the road" would include the normal force, but since that's about 10,000N they must mean the horizontal force the wheels exert on the road.
Secondly, they don't mean "the force of friction that opposes motion". The friction between wheels and road is providing the acceleration. The forces opposing motion are rolling resistance in the tyres and air drag.
Of these, rolling resistance does not lead to any horizontal forces, only vertical ones. (It generates a torque opposing the rotation of the wheels.)
So what they should ask for is the drag. Draw your FBD on that basis.

7. Nov 16, 2015

### Mrchilko

Ye.. So should it look like this than ?

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8. Nov 16, 2015

### haruspex

No.
Forget the rolling resistance. As SteamKing noted, no vertical forces are interesting here. You just have a car of known mass and acceleration, a given propulsive horizontal force from the tyres, and an unknown (to be determined) drag force; all horizontal.
From those you can construct an equation $\Sigma F = ma$ for the horizontal direction.

9. Nov 17, 2015

### Mrchilko

Hahah I tried that.... And it didn't equal to the answer given which is ( 1.86*103) ... 10780N* 3.40m/s 2= 36652.. Then what ? ( 10780N comes from the... 9.80 * 1100 kg to equal out to N

10. Nov 17, 2015

### haruspex

You seem to be confused about mass and weight.
mass * acceleration = force
kg * m/s2 = N
N * m/s2 = nothing of any meaning

11. Nov 17, 2015

### Mrchilko

Ahhhhhhhhjh damnit.... I forgot... Mass = the same everywhere no matter what and the universal term for mass is kg.... Haha stupid simple mistake... Thank you, very much :/ just getting into this physics business so..

12. Nov 17, 2015

### Mrchilko

But than.. Once I get 3740... Should I proceed with what ?

13. Nov 17, 2015

### haruspex

In the equation $\Sigma F =ma$, where does the 3740N feature? What other forces should be in that equation?

14. Nov 17, 2015

### Mrchilko

well we haven't integrated the 5600N force ... So ∑F= 1100 ( 3.40) = 3740N.... But than doesn't that mean that we add 5600 N? To the overall..? Cause than we would have 9340N... Sorry in tired... Haha I have pre-calculus this semester too... So I'm not thinking straight... But I will figure it out later... Thanks for evrything