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Newton's Laws

We are now ready to state Newton's three "Laws" of motion, in Newton's own words:

1.

FIRST LAW: Every body continues in its state of rest, or of uniform motion in a right [straight] line, unless it is compelled to change that state by a force impressed on it.

2.

SECOND LAW: The change in motion [rate of change of momentum with time] is proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.

3.

THIRD LAW: To every action there is always opposed an equal reaction; or, the mutual actions of two bodies are always equal, and directed to contrary parts.

Now, Newton's language was fairly precise, but to our modern ears it sounds a bit stilted and not very concise. We also imagine that, with the benefit of several centuries of practice, we have achieved a clearer understanding of these Laws than Newton himself. Regardless of the validity of this conceit, we like to express the Laws in a more modern form with a little mathematical notation thrown in:

1.

FIRST LAW: A body's velocity $\vec{\bf v}$ [which might be zero] will never change unless and until a force $\vec{\bf F}$ acts on the body.

2.

SECOND LAW: The time rate of chage of the momentum of a body is equal to the force acting on the body. That is,

$${d\vec{\bf p} \over dt} = \vec{\bf F}.$$ (5)

3.

THIRD LAW: Whenever a force $\vec{\bf F}_{BA}$ is applied to A by B, there is an equal and opposite reaction force $\vec{\bf F}_{AB}$ on B due to A. That is,

$$\vec{\bf F}_{AB} = - \vec{\bf F}_{BA},$$ (6)

where the subscript _AB (for instance) indicates the force from A to B.

As long as the mass m is constant⁷ we have

$${d\vec{\bf p} \over dt} \; = \; {d \over dt}(m\vec{\bf v}) \; = \; m \, {d\vec{\bf v} \over dt} \; = \; m \, \vec{\bf a}$$

since the derivative of a constant times a variable is the constant times the derivative of the variable. Then the SECOND LAW takes the more familiar form,

$$\vec{\bf F} \; = \; m \, \vec{\bf a}.$$ (7)

This famous equation is often written in scalar form,

$$\dot{p} \; \equiv \; {dp \over dt} \; = \; F \qquad \hbox{\rm or} \qquad F \; = \; m \, a$$

because $\dot{\vec{\bf p}}$ and $\vec{\bf F}$ are always in the same direction.