1.5 Forces

Now we know how to describe the motion of an object and changes in the motion (kinematics), but we still don’t know how to describe why the motion of an object changes (dynamics). This is the focus of this section.

A force is some influence on an object that changes the object’s motion. In classical mechanics, we describe forces as vectors and denote a generic force with the symbol $F$ . The exact dynamics of how forces affect motion are described in Newton’s laws of motion, which we will write now. In the following the symbol $F$ stands for the sum of all forces acting on the object, otherwise known as the net force.

Definition 1.8 (Newton’s First Law).

An object moving with constant velocity $v$ , will stay at the same constant velocity unless acted upon by a force. In other words:

v=\text{const.}\iff F=0.

(1.35)

Note that this includes an object at rest, which has velocity $v=0$ .

Mass is defined as an objects resistance to acceleration. In order words, a more massive object will accelerate slower relative to a less massive object when under the influence of identical forces. This is quantified by Newton’s Second Law.

Definition 1.9 (Newton’s Second Law).

The net force on the object is equal to the object’s mass times the object’s acceleration.

F=ma.

(1.36)

Note that force is always parallel to acceleration.

Since acceleration is the second derivative of position, we can write Newton’s Second Law as

F(t)=m\frac{\operatorname{\mathrm{d}}\!^{2}x(t)}{\operatorname{\mathrm{d}}\!t^% {2}},

(1.37)

which is a differential equation for $x(t)$ . This is known as an equation of motion, and all classical mechanics problems boil down to solving the equation of motion to obtain the trajectory of the object.

Example 1.4.

Suppose an object is acted upon by a constant force $F_{0}$ , then we align the $x$ axis with the direction of the force and the equation of motion is

\frac{\operatorname{\mathrm{d}}\!^{2}x(t)}{\operatorname{\mathrm{d}}\!t^{2}}=% \frac{F_{0}}{m}.

(1.38)

This is a very easy differential equation to solve. Integrating twice, we get

	$\displaystyle\frac{\operatorname{\mathrm{d}}\!x(t)}{\operatorname{\mathrm{d}}% \!t}=\int\frac{\operatorname{\mathrm{d}}\!^{2}x(t)}{\operatorname{\mathrm{d}}% \!t^{2}}=v_{0}+\frac{F_{0}}{m}t$		(1.39)
	$\displaystyle x(t)=\int\frac{\operatorname{\mathrm{d}}\!x(t)}{\operatorname{% \mathrm{d}}\!t}=x_{0}+v_{0}t+\frac{F_{0}}{2m}t^{2},$		(1.40)

where $v_{0}=v(0)$ and $x_{0}=x(0)$ as before. Note that comparing to equation 1.17, we can identify $a=\frac{F_{0}}{m}$ i.e. acceleration is constant, which is consistent with what we developed before.

It is important to note that Newton’s Laws of Motion are only valid in inertial reference frames, which are reference frames travelling at a constant velocity $v$ . If we are in a nonintertial reference frame, i.e. one that is accelerating, and we try to apply Newton’s Laws, we will encounter odd things such as phantom forces which have no source. One way to test if we are in an inertial frame or not us by using Newton’s First Law. If an object accelerates while under the influence of no forces, then our reference frame must be noninertial.

Definition 1.10 (Newton’s Third Law).

When two objects interact with each other, the forces on each object due to the other are equal in magnitude and opposite in direction. In other words, if object $A$ exerts a force $F_{A\to B}$ on object $B$ , then object $B$ exerts a force $F_{B\to A}$ on object $A$ and we may write

F_{A\to B}=-F_{B\to A}.

(1.41)

These two forces are then known as a “Newton (III) pair”.

Force is a vector quantity, so in more than dimension it can be decomposed into multiple components which are the forces along each axis. In three dimensions, Newton’s Second Law is

$\displaystyle\boldsymbol{F}(t)$	$\displaystyle=F_{x}(t)\hat{\boldsymbol{\imath}}+F_{y}(t)\hat{\boldsymbol{% \jmath}}+F_{z}(t)\hat{\boldsymbol{k}}$	(1.42)
	$\displaystyle=ma_{x}(t)\hat{\boldsymbol{\imath}}+ma_{y}(t)\hat{\boldsymbol{% \jmath}}+ma_{z}(t)\hat{\boldsymbol{k}}$	(1.43)
	$\displaystyle=m\boldsymbol{a}=m\ddot{\boldsymbol{r}}.$	(1.44)

We can now solve any problem in classical mechanics.