# For the N's second law

1. Jan 31, 2010

### jhans9878

So if F=Ma

then if there was no accleration but it was going at constant velocity

there would still be force right? but what? and how

2. Jan 31, 2010

### Redd

If there is a constant velocity, yes, there is no acceleration, and yes, no Force.
Things resist changes in motion, whether that is starting or stopping.
Something moving at a constant velocity will go on forever without another force stopping it. In the real world that doesn't happen because of things like friction.

3. Feb 1, 2010

### wisvuze

a force is only present when an acceleration occurs. If an object is moving at uniform velocity, then there are no forces exerting themselves on that object, you can see from the equation F = Ma. A force might've exerted itself before to get an object to a certain velocity, but if it is moving at a uniform velocity, that force is no longer exerting itself on that object

4. Feb 1, 2010

### Claude Bile

Acceleration is zero only when there is no net force acting on an object.

Claude.

5. Feb 2, 2010

### Redbelly98

Staff Emeritus
Excellent point. As I sit here at the computer, I do not accelerate and yet the Earth exerts a gravitational force on me.

6. Feb 2, 2010

### sophiecentaur

I should hope that you DO accelerate (radially) or you will be following a straight line course out into space.

7. Feb 2, 2010

### Redbelly98

Staff Emeritus
Ha ha, true. But my acceleration is considerably less than g.

8. Feb 2, 2010

### sophiecentaur

Don'tcha just hate a smartypants?

9. Feb 2, 2010

### Forlon

The force is directly proposnal to change in linear momentum so when the momentum is 0 the force is zero.

10. Feb 3, 2010

### sophiecentaur

The Earth is exerting TWO forces on you. Gravity produces your Weight force and the Ground exerts an equal and opposite Reaction force. Net force is Zero so no acceleration. Remove the ground and you accelerate down the lift shaft at g acceleration.

11. Feb 3, 2010

### Redbelly98

Staff Emeritus
If the OP is now learning about Newton's Laws, chances are they have not learned about momentum yet.

Yes, exactly. The Earth is exerting forces on me, but it is zero net force. Thus zero acceleration.

12. Feb 3, 2010

### sophiecentaur

I know you know that I know you know - it needed to be said to avoid confusion.
And doesn't it just?

13. Feb 3, 2010

### thepancakeman

Sorry, I'm feeling dense...

So, F=Ma. Just because F1-F2 = 0 (net force, zero acceleration) how do you have any "F"s to add or subtract, because you don't have any acceleration?

My couch must have force to balance out the force of gravity, but what is the acceleration of my couch? And if it's 0, then wouldn't the force be 0 also??

14. Feb 3, 2010

### sophiecentaur

Mais oui, mon ami.
But, without a force F keeping you from falling, there would be an unbalanced weight force W pulling you down. F - W = 0 (Net)

I suppose it depends your frame of reference. If you're in your chair, the only force you can really feel is the chair pushing up on your bum. Take your chair away and the only clue you would have about any force acting on you would be as you observed the sides of the lift shaft going upwards faster and faster.
Our lives are lived with pairs of balanced forces in almost every experience we have.

15. Feb 3, 2010

### thepancakeman

I'm still confused. If I try to lift a 1000kg weight, it's not going anywhere. I've done no work, and from my limited understanding I have not created/applied any acceleration, but I am still applying force to it, no?

16. Feb 3, 2010

### sophiecentaur

When you are suspending the weight, (it's not accelerating) your force, just like the force which the ground was exerting on it, is balancing its weight force. Balanced forces = no acceleration. Irrespective of what you happen to be doing to the weight, if it is not accelerating, the sum of the forces on it must be zero.
You could also say that you are accelerating it at -g, I suppose, and so the net / resultant acceleration would be g - g = 0 but we don't usually state it in that way.

17. Feb 3, 2010

### thepancakeman

Okay, g-g=0 makes sense to me. BUT, if it's sitting on the ground, it's already at g-g, so my tugging at it does what? The net of the ground and my efforts are still g-g, so if I'm taking away some of a from the ground, then the ground is providing variable acceleration?

Or let's take me out of it. Weight is sitting on the ground. Gravity is pulling down on it. It's not moving, so we have equal forces (gravity and the ground.) If somehow gravity increases to 1.5 g's, the ground will still be supporting the load and we still have equal forces. How does the force of the ground change? Shouldn't that be a constant?

18. Feb 3, 2010

### Staff: Mentor

Don't think of each individual force on the object as providing its own acceleration. It's the net force that produces the acceleration. As long as the net force remains zero, there will be no acceleration. The more you lift, the less force the ground has to exert to support the object.

No, the upward force that the ground exerts is not a constant. Think of it as a 'reactive' force--it's whatever it needs to be to prevent the object from crushing through it (up to the limit of its strength). Say you sat on the object. Now the ground has to push up on the object with a force equal to the combined weight of you and the object. Similarly, if you pull up on the object the ground force on it becomes less.

19. Feb 3, 2010

### thepancakeman

But then we're back to the original issue: if F=ma and there's no a, then how do you have a non-zero F?

Boy, my head is starting to hurt from what should be a simple concept. Sorry if I'm being dense.

Last edited by a moderator: Feb 3, 2010
20. Feb 3, 2010

### Staff: Mentor

The F in F=ma refers to the net force on the object. If the acceleration is zero, then the net force is zero.

21. Feb 3, 2010

### sophiecentaur

Try it this way.
If you take a bullworker and pull it a bit, the spring will stretch a bit until the forces are balanced. Exert yourself a bit more and the spring stretches a bit more - until the forces are, again, balanced.
Think of the ground (and the object, of course) as having a finite amount of springiness. Increase the weight of your object and the ground will give a tiny bit, until the restoring force is just enough to balance the additional weight. So the ground will provide just enough force - it doesn't have to 'calculate' it; it doesn't need a servo or 'intelligent control' it just moves until the forces are balanced.
If the weight were on a chair and the strength of the chair were exceeded, it would break - the weight would no longer have a zero net force on it and it would fall. If you partly supported the weight with your hands (the force is shared), the point where the weight broke the chair would be a bit later.

Is your problem with the subjective effect of trying to support something - apparently requiring constant effort? Is that interfering with your understanding? That's because of the way your muscles work; much more complicated than a lump of material.

22. Feb 3, 2010

### thepancakeman

Aha, I think the light bulb just went off with the terrm "reactive." Essentially what you're saying is that a given fixed object (ground, brick wall, etc.) simply applies -a to whatever force impacts it up to the breaking point, right?

Never quite thought of it as a passive or reactive force, which kind of feels like a misnomer, but it makes sense.

Thanks for everyone's help!!

23. Feb 3, 2010

### thepancakeman

And to answer the original question, which I think I've figured out with a lot of thinking, is that force (of a physical object) doesn't exist until it interacts with something.

A bowling ball rolling down the lane doesn't have force until it hits the pins--then the force of the bowling ball is the acceleration of the mass of the pins. Since the ball keeps rolling, not all of the....umm don't know the correct term..."potential force"...is used up. If you want to find the maximum force of the bowling ball, you roll it into a wall, which accelerates it to 0 velocity. Then you have m (mass of the bowling ball) and a (accelerating it--or changing the velocity--to 0 f/s.)

Feel free to correct me if I screwed anything up!

24. Feb 3, 2010

### sophiecentaur

I think you are beginning to get this!
It's kinetic energy that the moving ball has and it loses some each time it hits a pin. As you say, it has no "force". The force exists whilst it is interacting with another object.
To add a bit of confusion, there are actually two quantities associated with moving objects. Each of these quantities is useful in the right application.
1. Kinetic Energy, which is the work that the object could do (like getting it to go so far up a slope) It's given by the formula K.E. = half times mass times velocity squared.
2. Momentum, which is equal to its mass times its velocity. Sometimes the word 'inertia' is used (wrongly, if you want to be accurate). Momentum is a quantity that is always conserved during collisions and that makes it very useful in calculations.

They both relate to your idea of "potential force". When the bowling ball is brought to a halt its momentum is brought to zero by the force from the wall acting for a short time. Or, when you fire a gun, the light bullet goes off fast and the massive gun recoils slowly; momentum in each direction is the same.
There's no end to this sort of stuff.

25. Feb 3, 2010

### gazebo_dude

Force is not a property of an object. The bowling ball does not "have" force. Also the term "potential force" is very misleading. There is no such thing as a store of force inside an object which can be used up. If there were such a thing then how could an object sit on a table for an indefinite amount of time while all of the force is being used up? Force is a property of the interaction between two objects.

There is also no such thing as "the maximum force of the ball." There is a force, which when exerted on the ball for a given amount of time, will bring it to a stop. There is nothing terribly special about this force (from the standpoint of physics - although crash test dummies might care about it), and, by a suitable arrangement, a much larger force could be exerted by the ball on a wall.

When a bowling ball hits a pin the bowling ball exerts a force on the pin. The pin also exerts an equal and opposite force on the ball. Force is mass times acceleration (or rate of change of momentum, technically). Since there is a net force $$\vec{F}$$ on the bowling ball it will accelerate in the direction of the force with acceleration $$\vec{a}=\vec{F}/m_b$$ where $$m_b$$ is the mass of the ball. There is a force $$-\vec{F}$$ on the pin (equal and opposite, remember?) so the pin accelerates with acceleration $$\vec{a}=-\vec{F}/m_p$$ where $$m_p$$ is the mass of the pin. The same story applies for a ball hitting a wall, or two electric charges, or any other collision.

If you want to figure out how something is going to accelerate (or not accelerate, as the case may be), then just find the forces on it due to the interactions it has with all the other objects around, then add them up to find the net force on the object, then divide by the object's mass.

If you know, for instance, that a book on the table is not accelerating then you know that the total force on that book is zero. Since the Earth exerts a downward force of gravity on the book, there must be a compensating force coming from somewhere else. In this case it is the contact force of the table pushing upwards on the book. If that compensating force were not there then the book would accelerate.