Suppose a nonrotating reference frame $S'$ translates relative to an inertial frame $S$ with origin position $\mathbf R(t)$. If a particle has position $\mathbf r'$ in $S'$, then

$$\mathbf r=\mathbf R+\mathbf r'.$$

Differentiating twice gives

$$\boxed{\mathbf a=\mathbf A+\mathbf a'},$$

where

$$\mathbf A=\ddot{\mathbf R}$$

is the acceleration of the moving frame's origin relative to the inertial frame.

Newton's second law is simplest in the inertial frame:

$$m\mathbf a=\mathbf F_{\rm real}.$$

Substituting the acceleration relation gives

$$m(\mathbf A+\mathbf a')=\mathbf F_{\rm real},$$

so an observer using the accelerating frame can write

$$\boxed{m\mathbf a'=\mathbf F_{\rm real}-m\mathbf A}.$$

The extra term

$$\boxed{\mathbf F_{\rm inertial}=-m\mathbf A}$$

is an inertial force or fictitious force. It is not a new physical interaction with another object. It appears because the coordinate frame itself accelerates.

Direction of the inertial force

The inertial force points opposite the frame acceleration.

If a car accelerates forward, an observer inside the car can describe loose objects as experiencing a backward inertial force. In the ground frame, no mysterious backward interaction is needed: objects simply tend to preserve their previous velocity while the car accelerates beneath them.

Both descriptions predict the same relative motion when used consistently.

Effective gravity

Near Earth's surface, let the real gravitational acceleration be $\mathbf g$. In a frame accelerating with $\mathbf A$, the equation of motion can be written

$$m\mathbf a'=m\mathbf g-m\mathbf A+\mathbf F_{\rm other}.$$

Define the effective gravitational acceleration

$$\boxed{\mathbf g_{\rm eff}=\mathbf g-\mathbf A}.$$

Then

$$m\mathbf a'=m\mathbf g_{\rm eff}+\mathbf F_{\rm other}.$$

This lets many equilibrium problems inside an accelerating vehicle be treated as ordinary statics in a tilted or modified effective gravity field.

Example: pendulum in an accelerating vehicle

A vehicle accelerates horizontally to the right with magnitude $A$. A pendulum hanging inside eventually rests at a constant angle relative to the vehicle.

In the vehicle frame, gravity acts downward with magnitude $mg$ and the inertial force acts leftward with magnitude $mA$. The string aligns opposite the vector sum of those two body forces.

If $\theta$ is the angle the string makes from the vertical toward the left, then

$$\tan\theta=\frac{A}{g}.$$

Thus

$$\boxed{\theta=\tan^{-1}\left(\frac{A}{g}\right)}.$$

For

$$A=3.0,\mathrm{m/s^2},$$

$$\theta\approx17.0^\circ.$$

The pendulum therefore points opposite the effective gravity direction inside the accelerating vehicle.

Elevator and apparent weight

For an elevator accelerating upward with acceleration $A$, the inertial force in the elevator frame points downward. A person at rest relative to the elevator satisfies

$$N-mg-mA=0,$$

so

$$\boxed{N=m(g+A)}.$$

The same result follows from ordinary Newton's second law in the inertial ground frame. The accelerating-frame description packages the effect into an effective downward gravity of magnitude $g+A$.

Inertial versus non-inertial frames

If $\mathbf A=0$, the frame moves at constant velocity and the inertial-force term disappears. Nonrotating frames moving at constant relative velocity are therefore equivalent inertial frames in Newtonian mechanics.

If $\mathbf A\ne0$, Newton's second law does not retain the simple form $m\mathbf a'=\mathbf F_{\rm real}$ unless the inertial force is included.

Rotating frames require additional inertial-force terms because their axes change direction as well as their origin potentially accelerating. The translating accelerating frame is the simplest example of how coordinate acceleration changes the form of Newton's equations.