# Homework Help: Newton's first law

1. Nov 16, 2008

### alpha372

1. The problem statement, all variables and given/known data

What can be said about a car traveling at constant velocity on a street, keeping in mind newton's first law?

2. Relevant equations

3. The attempt at a solution
My first reaction to this question is "But a car is not a particle." Assuming the car is a particle, then it must be safe to say that the car is some how over coming all the forces of friction in order to stay at a constant speed. When I think of something moving at a constant speed without forces acting on it, I think of a ball floating in outer space.

So I feel like I am overlooking something, like, perhaps, that even though the car is traveling at a constant velocity, the sum of the forces acting on it is not zero: The weight of the car is balanced by the normal force, but the friction acting on the tires isn't balanced by any external force; an internal force is counteracting the friction, so it seems to me that it would be more accurate to say that newton's first law doesn't apply to this situation.

2. Nov 16, 2008

### asleight

Your initial reaction was perfect. Newton's First Law applies in all linear situations (if an object isn't acted upon by an unbalanced force, it will maintain its uniform motion)--the sum of all forces acting upon an object equals the mass of the object times its acceleration (or the variable mass of an object times its constant velocity), so you can examine the system using NSL:

$$\sum\vec{F}_x=\vec{F}_{chem}-\vec{F}_f-\vec{F}_{air}=\frac{d}{dt}(m\vec{v})=0\rightarrow\vec{a}_x=0\rightarrow\vec{v}_x=C$$

$$\sum\vec{F}_y=\vec{N}-\vec{F}_g=\frac{d}{dt}(m\vec{v})=0\rightarrow\vec{a}_y=0\rightarrow\vec{v}_y=C=0$$.

Last edited: Nov 16, 2008
3. Nov 16, 2008

### alpha372

Okay, friction is a force acting on the vehicle; what external force is opposing the friction? Friction must be overcome from an external force, not an internal force (I thought) for Newton's first law to be valid ( I believe).

4. Nov 16, 2008

### asleight

Draw a free-body diagram around the car, itself. Only forces within the circle can be considered to analyze Newton's First Law.

5. Nov 16, 2008

### alpha372

Okay, I'll pretend I drew a circle around the car (this doesn't help me determine which forces are relevant and which are not, by the way). I think the forces that are in the circle are: Okay, the whole circle thing is confusing me. I don't know what this is supposed to help me discover. I believe the forces that I am supposed to consider are only the EXTERNAL forces acting on the car; that is... wait a second, the weight of the car is an external force right? I have three forces: Normal force, weight, and friction. I don't have a forth force to balance out the friction.

So I believe there are 3 external forces: weight, normal force, and friction--which has nothing to balance it out.

6. Nov 16, 2008

### asleight

Newton's First Law, "A body continues to maintain its state of rest or of uniform motion unless acted upon by an external unbalanced force." The external force due to friction is balanced by the force due to chemical explosion within the vehicle.

Newton's Second Law, "F = ma: the net force on an object is equal to the mass of the object multiplied by its acceleration." The sum of the forces is zero, the acceleration is zero, the velocity is constant.

Newton's Third Law, "For every action there is an equal and opposite reaction." The car exerts a force on the air in front of it and the air reciprocates this force as drag; the Earth's gravitational field pulls the car against the road, the road exerts an equal and opposite normal force in return.

The circle allows you to examine the forces. Is friction (the contact between the road and car) within the circle? How about weight and the normal forces (same location)? And, the car is inside the system, isn't it? So, its chemical combustion is also allowed to be drawn on the FBD.

7. Nov 16, 2008

### alpha372

No, I'd have to get a second opinion, I'm pretty sure that you aren't supposed to consider internal forces...

my whole point is that because you can't consider the engine as an external force, then newton's first law can't be applied to this situation because it isn't a valid inertial frame of reference.

But I'm probably confusing this with other internal forces, like the internal forces of impact or warping or something...

Maybe you can help clear this up for me if you could give me an example where newton's first law cannot be applied.

8. Nov 16, 2008

### asleight

Okay, then wait for a second opinion. External vs. internal makes no difference. Read Newton's First Law again. The external frictional force is balanced by the external chemical force. The gas is NOT part of the vehicle.

All of Newton's Laws always apply.

9. Nov 16, 2008

### borgwal

Hi Alpha,

Indeed, when asleight "answers" your question it is always wise to get a second opinion. For instance, what he said here:

"Newton's Third Law, "For every action there is an equal and opposite reaction." [...] the Earth's gravitational field pulls the car against the road, the road exerts an equal and opposite normal force in return."

is incorrect: the normal force equals the weight in magnitude NOT because of Newtons' third law, but because of Newton's second law. The forces that form action-reaction pairs are the Earth's gravitational pull on the car and the car's gravitational pull on Earth; as well as the normal force that the road exerts on the car and the force the car exerts on the road.

And this statement:

"External vs. internal makes no difference."

is nonsense of course: it makes all the difference.

To answer your question about friction forces: let's consider a 4-wheel drive for simplicity. Then all 4 wheels are made to rotate such that they exert a force on the ground in the *backward* direction: the frictional force of the road on the tires is, therefore, *forward*. That's how a car can move *forward*. When the car is moving at constant velocity, there is only a small friction force in the backward direction from air resistance, and that is compensated for by the small friction forces on the tires.

(You can also think of a car on frictionless ice: it wouldn't be able to accelerate!)

10. Nov 16, 2008

### alpha372

The last pair: "the normal force that the road exerts on the car and the force the car exerts on the road,"

The normal force is equal in magnitude to the force that the car exerts on the road. But, the force that the car exerts on the road is it's weight (mg); so why isn't it correct to say that the action is the weight and the reaction is the normal force?

Or is that the two cannot be action - reaction pairs because the normal force is only a contact force, while the earth's gravitational pull is a long range force?

11. Nov 16, 2008

### alpha372

Okay, "All of Newton's Laws always apply." is an erroneous assumption. Newton's first law is only valid in an inertial frame of reference, which means that the frame of reference cannot be accelerating.

You've convinced me to consider that "The external frictional force is balanced by the external chemical force. The gas is NOT part of the vehicle."

If this is safe to assume, then the forces balance. So thanks.

12. Nov 16, 2008

### alpha372

This makes sense, I'm thankful that you pointed it out.

Just to be clear:
Are you saying that the gas in the car doesn't matter-- which I am starting to consider... let's say the car ran out of gas; the tires moving backward doesn't help much to overcome the friction, the car slows down and eventually stops. Letting off the gas in our own vehicles we find naturally that the car slows down due to friction and eventually comes to a stop. Therefore, it is clear that "the external gas force" that pushes the car forward to overcome friction is needed. So, wouldn't this be the balancing force to the friction?

I'm really trying to determine weather we can assume the car is in equilibrium because it is traveling at a constant rate because the force causing the car to move forward (I presumed the thrust of the engine) is valid/ external. I believe I remember hearing in class a few summers ago that we're only supposed to consider forces acting on the object, not forces the object exerts. As long as the pushing force of the engine is acting on the object, then the car is in equilibrium. It's just confusing because the engine is the object, working on the object, so it doesn't make any sense to me.

13. Nov 17, 2008

### borgwal

An action-reaction pair of forces always involves just two objects, A and B, and the forces of the pair are the one that A exerts on B and the one that B exerts on A.

In the case of weight and normal force, we're talking about Earth exerting a force on the car, and the road exerting a force on the car. The fact that these two forces happen to be balanced only follows from the fact the car is not accelerating in the vertical direction.

14. Nov 17, 2008

### borgwal

The engine plus gasoline plus tires plus dashboard plus windows etc. etc. etc. are all part of "the car". All forces that act internally to the car are parts of action-reaction forces and therefore do not give rise to a net external force. The motion of the car as a whole (more precisely, its center-of-mass motion) is determined *only* by the net external force.

15. Nov 17, 2008

### alpha372

Okay, just for super clarity, Will you answer these three quick questions please? :)

1. Can we assume that the gas, the engine thrust, is an internal force, and does not count?

2. Can we assume that the balance of the friction force is friction it's self and wind resistance?

3. Is it safe to assume that an inertial frame of reference is a frame of reference that is not accelerating?

16. Nov 17, 2008

### alpha372

So you are saying we are dealing with two different forces? One from the earth and one from the road?

17. Nov 17, 2008

### borgwal

1. yes, with a small proviso: the gas that leaves the exhaust pipe would be considered as leaving the car, and thus contributing a tiny force in the forward direction!

2. yes, apart from the tiny effect mentioned above, the only external forces on the car in the horizontal direction are friction/drag forces, exerted by the road and air.

3. in Newtonian physics, yes.

18. Nov 17, 2008

### borgwal

Yes, there are in fact two action-reaction force pairs, one between Earth and car (gravity), the other between road and car (contact).

19. Nov 17, 2008

### alpha372

Yeah!! You just made my day! I finally got the info I was looking for! Thank you! Thank you! Thank you! :D

20. Nov 17, 2008

Sure :-)