Conceptually, why is Newton's second law of motion F=ma rather than, say, F=m+a ?

heelallev5

heelallev5

Answered question

2022-08-12

Conceptually, why is Newton's second law of motion F = m a rather than, say, F = m + a ?

Answer & Explanation

Tamia Campbell

Tamia Campbell

Beginner2022-08-13Added 14 answers

Because we can measure it.
Think of it like this:
If you push something, it accelerates. Push twice as hard, and you see a doubled acceleration. Push 3 times as hard and you see a trippled acceleration. Acceleration and force clearly follow each other proportionally. This is something we can measure.
But force and acceleration are not equal. From the above we cannot claim F=a. They are proportional, not equal.
If you push with 3 N, then the acceleration is maybe 1 m / s 2 . The force is 3 times larger. Double the force to 6 N and you will see the acceleration double as well to 2 m / s 2 , since they are proportional. This is still a 3 times larger force. We can therefore set up a formula like this, mathematically showing that the force is always 3 times larger than the acceleration:
F=3a
So, what do we call this number "3"? Mathematically it is a proportionality constant. But what is it physically? In some cases it is 3, in other cases another value. It turns out that light objects have a very small value, while heavier objects have larger values. And mathematically, the larger the value is, the smaller is the acceleration caused by the force. So, it is some kind of a "resistance to accelerate".
Let's give it the name mass and the symbol m:
F=ma
We also soon realize that there can be many forces, all giving their contribution to the acceleration. We can thus generalize the formula to the sum of all forces acting:
F = m a
This version of the formula therefore comes from stepwise experimentation. From looking for patterns and dependencies between parameters, such as force and acceleration. It is F = m a, and not F = m + a or F = m a 2 or F = m 3 / a or anything else. If it was, then we would have seen that during the experimentation steps above.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Force, Motion and Energy

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?