# How to determine component forces when net force is known?

Newton's second law of motion states that $$F = ma$$, where $$F$$ is the net force on some object with mass $$m$$ and acceleration $$a$$. This implies that an object with net force $$F$$ acting on it, may have two individual forces in opposite directions on it, $$F_A$$ and $$F_B$$, respectively, such that their vector sum is $$F$$. My question is: how are the values of $$F_A$$ and $$F_B$$ determined? I know that Newton's second law of motion can only be used to determine the difference between them.

For example, if we apply $$2 N$$ on an object in a direction and $$4 N$$ in the opposite direction, and in another case we apply $$1000 N$$ one a direction and $$1002 N$$ in the opposite direction, the acceleration of the object is identical in both the cases.

How do we distinguish between the two situations? How can we assign one pair of values for the first case and another pair of values for the second? Is this not arbitrary? How are these values assigned?

For example, if you apply 2N on an object in one direction and 4N on the object in the opposite direction, and you then apply 1000N one direction and 1002N the other direction, the acceleration of that object remains the same. but how do you distinguish the two situations?

With context. Usually you will have some idea what forces you're working with. For instance, if you're told that you're lifting a boulder upwards with an acceleration of $$a=2\ \rm m/s^2$$ then you know there are at least 2 forces acting on the rock: gravity ($$mg$$) and the force you apply to lift it $$F_A$$. Therefore, $$ma=F_{\rm A}-mg$$, and you can solve for the force you apply, $$F_{\rm A}$$.

However, if all the information you're given that an object has some acceleration with no context as to its environment, then you can't find the forces acting on it. Of course, such a problem is rather useless, and you will always have the necessary information to figure out which forces may act on the object.

If you're told an object is falling down, then for sure gravity is acting on it. There would also be air resistance (which is often ignored in intro physics classes to simplify the calculations, but you would be explicitly told to ignore it). If you're dealing with a charged particle, then you know there are electric forces.

As you proceed to learn physics, you will learn about different types of forces and how they act on objects. But these forces are certainly not arbitrary.

You do not arbitrarily "assign" values to the external forces. The magnitudes and directions of the external forces depend on the type of forces. Once you establish the external forces, you apply the second law.

Here are some examples. Close to the earth the force of gravity is $$mg$$ where m is mass and g is the acceleration of gravity at the earth's surface. The force from a spring is $$kx$$ where k is the spring constant and x is the spring displacement from the un-stretched position. The force for slidding friction is $$\mu N$$ where $$\mu$$ is the coefficient of sliding friction and N is the normal force.

Newton's second law, $$F = ma$$, refers to the total net force $$F$$ applied to an object in a given moment, and gives us it's acceleration $$a$$ at that moment. $$m$$ is the mass of the object.

The total net force $$F$$ is the sum of all forces acting on the object at a given moment.

Knowing only the net force $$F$$ (or the object's mass and acceleration, as knowing $$ma$$ is the same as knowing $$F$$), Newton's three laws do not tell us how we could obtain each individual force acting on the object.

That information is something we may obtain from analyzing the system to see which forces we expect to be acting on the object. This is called a free body diagram.

For example, for a box laying still on a table, on Planet Earth, we know (not from Newton's laws directly, but from experimental knowledge of gravity) that the Earth exerts an individual force on the box, pointing downwards, called gravity force, with a value of $$F_{gravity} = m. g$$, where $$g = 9,8 [m/s^2]$$. Since the box is still, not accelerating, we know $$a = 0$$ and thus $$F = 0$$. Therefore, there must be another individual force acting upwards on the box, with a magnitude exactly equal to $$F_{gravity}$$. We may call this force, normal force, and further analysis will show us that it's the table that's exerting this force on the box.

In other situations, we may be able to measure some individual forces by doing a free body diagram each other object exerting a force on our main object. If we know their acceleration, we may obtain their net $$F$$

Then, we use Newton's third law, that states that for every individual force $$F_{a}$$ exerted by A on B, there is another individual force $$F_{b}$$ exerted by B on A, of equal magnitude and opposite direction.

This will give us a system of equations of various net forces, and various individual forces. By solving it, we may obtain each value of interest. Of course, we will need as much information on the system as we can get.

You need to know about the individual forces, $$\mathbf F_a$$ and $$\mathbf F_b$$. For example, if the body on which the forces act has a mass of 2.0 kg, w e know that one of the forces on it, $$\mathbf F_a$$, say, is the pull of the Earth (pull of gravity) on it, and is 19.6 N downwards. Another force might be an upward force, $$\mathbf F_b$$, say, from a spring, extended by 5.0 cm, attached to the body. In a separate experiment we might have found that when the spring has this extension it exerts a force of 15.0 N. And so on.

As you say, measuring a body's acceleration and knowing its mass will not tell you the magnitudes or directions) of the individual forces acting on it, only their vector sum. Hope I haven't missed the point of your question.

Newton's second law does not direclty tell F=ma, but it tells that F is proportional to the change in momentum. Also, the concept and the formula here talks about 'net force' which same in both the cases. So, you don't differentiate between the cases as they are pretty similar and provide the same effect. This equation was written for constantly accelerated objects under mechanics. So we don't study the forces but the effect of the forces on the object.

How do we distinguish between the two situations? How can we assign one pair of values for the first case and another pair of values for the second? Is this not arbitrary? How are these values assigned?

A you already know, there is no need to distinguish between the two situations as far as their effect on the object in question is concerned since it is only the net force that matters. Furthermore, without additional information on the nature of the individual forces acting on the object, there is no way to know how the values of $$F_{A}$$ and $$F_{B}$$ were determined.

For example, if the object is an electrically charged object, the sources of the forces would be electric fields where the sources of the fields are other electrical charges. Or the forces could be gravitational, in which case the sources of the forces would be other gravitating bodies. Or they could simply be contact forces pushing and/or pulling on the object.

While the two situations have the same effect on the object regardless of the nature of the forces, the two situations can have different effects on the sources of the individual forces, since those effects are based on combining Newton's third law and Newton's second law to each source.

Per Newton's third law source $$A$$ will experience an equal and opposite force by the object of $$-\vec F_{A}$$ and source $$B$$ will experience an equal and opposite force by the object of $$-\vec F_{B}$$. Then applying Newton's second law to each source the acceleration of each source will be

$$\vec a_{A}=\frac {-\vec F_{A}+\sum_{i=1}^{n}\vec F_{Ai}}{M_{A}}$$

$$\vec a_{B}=\frac {-\vec F_{B}+\sum_{i=1}^{n}\vec F_{Bi}}{M_{B}}$$

Where $$\sum_{i=1}^{n}F_{Ai}$$ is the vector sum of all the other external forces acting on $$A$$ and $$\sum_{i=1}^{n} F_{Bi}$$ is the vector sum of all the other external forces acting on $$B$$.

Hope this helps.