A crowded city bus leaves a stop with a jerk. The standing passengers sway backward and grab the handles. A water bottle on the floor rolls towards the back. To the passengers it feels as if a force pushed everything backward. A person waiting at the bus stop sees something different: the bottle hardly moves at all, and the bus slides forward under it.
Two observers, one bottle
Take the person at the bus stop first. The ground is an inertial frame: in it, a body with no net force keeps its velocity. The bottle rests on a smooth floor, so no horizontal force acts on it. By the first law it stays where it was. The bus accelerates forward underneath it. That is all.
Now take a passenger. In the frame of the bus, the bottle starts from rest and accelerates backward. Yet no body pushes it backward. In this frame a body accelerates with no force on it, so Newton's first law fails. A frame in which the laws of motion fail is a non-inertial frame. Every frame that accelerates relative to an inertial frame is non-inertial.
The correction
The relative motion of the last unit gives the repair. Let the frame accelerate at \(\vec a_0\) relative to the ground. A body that accelerates at \(\vec a\) relative to the ground accelerates at \(\vec a_{\text{rel}} = \vec a - \vec a_0\) relative to the frame. Newton's law holds in the ground frame.
\[ \sum \vec F = m\,\vec a \]
Why. the second law holds in the inertial ground frame, with the real forces only.
\[ \vec a = \vec a_0 + \vec a_{\text{rel}} \]
Why. the acceleration relative to the ground is the frame's acceleration plus the acceleration relative to the frame.
\[ \sum \vec F = m\,\vec a_0 + m\,\vec a_{\text{rel}} \]
Why. substitute step 2 into step 1.
\[ \sum \vec F + (-m\,\vec a_0) = m\,\vec a_{\text{rel}} \]
Read it. move \(m\vec a_0\) to the left. The law now has the usual form in the accelerating frame, with one extra term that looks like a force.
The extra term \(-m\vec a_0\) is the pseudo force. Add it to the real forces on every body, and the second law gives the correct acceleration relative to the accelerating frame. It points opposite to the acceleration of the frame and is proportional to the mass of the body.
In a frame that accelerates at \(\vec a_0\) relative to an inertial frame, the pseudo force on a body of mass \(m\) is \(-m\vec a_0\). When it is added to the real forces, Newton's second law gives the acceleration of the body relative to that frame.
Be honest about what the pseudo force is. No body exerts it, so it has no agent. It has no third-law partner. It is the term \(m\vec a_0\) moved across the equation. Use it in the accelerating frame only. In the ground frame, where the laws already hold, adding it counts the same physics twice.
In the bus, the pseudo force on the bottle is \(m a_0\) backward, and no real force opposes it. So the bottle accelerates backward at \(a_0\) relative to the bus. The passengers feel thrown back for the same reason: the bus floor drags their feet forward, and their bodies lag behind. From the road nothing threw them back. From inside the bus the pseudo force is the quickest way to get the right answer.
A student says, "The pseudo force on the bottle is \(ma_0\) backward. By the third law, the bottle pushes something forward with \(ma_0\)." What is wrong?
A pendulum in a car
Hang a small bob by a string from the roof of a car. When the car speeds up at a steady \(a\), the string settles at an angle \(\theta\) to the vertical, leaning backward. Solve it both ways.
From the road, the bob moves with the car, so its acceleration is \(a\) forward. Only two real forces act: the tension \(T\) along the string and the weight \(mg\). The horizontal part of the tension provides the acceleration; the vertical part holds the weight.
\[ T\sin\theta = ma, \qquad T\cos\theta = mg \quad\Rightarrow\quad \tan\theta = \dfrac{a}{g} \]
From the car, the bob is at rest. Three forces act on it: \(T\), \(mg\), and the pseudo force \(ma\) backward. They balance. The same two equations appear, now as a balance of forces instead of force equals mass times acceleration.
The bob is an accelerometer. Measure the angle and you know the acceleration of the car, without looking out of the window. The two descriptions always agree. Choose whichever is easier; for a body at rest in the vehicle, the pseudo force turns a dynamics problem into a balance problem.
A handle strap hangs from a bar in a Metro coach. As the train leaves a station, the strap leans back so that \(\tan\theta = 0.12\). Take \(g = 10\,\text{m/s}^2\). Find the acceleration of the train in m/s².
The car now brakes while moving forward. Which way does the bob lean?
A block held still on a pushed wedge
A block rests on the sloping face of a smooth wedge of angle \(\theta\). Left alone, the block slides down at \(g\sin\theta\). Now push the wedge horizontally, towards the side the block would slide to. For one particular acceleration \(a\) of the wedge, the block does not slide at all. Find that acceleration.
In the frame of the wedge the block is at rest. Three forces act on it. They are the weight \(mg\), the normal force \(N\) perpendicular to the face, and the pseudo force \(ma\), horizontal and opposite to the push. Take the axis along the face. The weight pulls the block down the slope with \(mg\sin\theta\). The pseudo force pushes it up the slope with \(ma\cos\theta\).
\[ ma\cos\theta = mg\sin\theta \quad\Rightarrow\quad a = g\tan\theta \]
From the ground the same result comes from the second law. The block accelerates with the wedge, horizontally. The vertical part of \(N\) holds the weight: \(N\cos\theta = mg\). The horizontal part gives the acceleration: \(N\sin\theta = ma\). Dividing gives \(\tan\theta = a/g\) again, and \(N = mg/\cos\theta\), more than the weight.
The wedge problem has a lesson beyond the numbers. While the block stays on the face, its motion and the wedge's motion are tied. The block cannot move into the wedge or leave its face. So their accelerations perpendicular to the face must be equal. This is the wedge constraint, the partner of the string constraint. A block held still on the face meets the constraint in the simplest way: it has exactly the acceleration of the wedge.
A smooth wedge has a face at \(45°\). A \(2\,\text{kg}\) block sits on the face. Take \(g = 10\,\text{m/s}^2\). With what horizontal acceleration must the wedge be pushed so that the block does not slide, and what is the normal force then?
The block stays at rest on the face if \(a = g\tan 45° = 10\,\text{m/s}^2\).
The normal force is \(N = mg/\cos 45° = 20 \times 1.414 = 28.3\,\text{N}\), larger than the weight \(20\,\text{N}\), because it must also give the block its horizontal acceleration.
The wedge of the figure is pushed with an acceleration larger than \(g\tan\theta\). What does the block do, relative to the wedge?
The pendulum in the car was solved twice, once from the road and once from the car. Why do the two methods always give the same answer?
The pseudo force was built from the ground-frame law. Starting from \(\sum \vec F = m\vec a\) and writing \(\vec a = \vec a_0 + \vec a_{\text{rel}}\) gives \(\sum \vec F - m\vec a_0 = m\vec a_{\text{rel}}\). The two methods are the same equation with the term \(m\vec a_0\) written on different sides. So they cannot disagree, as long as the pseudo force is used only in the accelerating frame.
- Adding a pseudo force while working in the ground frame.Pick one frame. In the ground frame use real forces only. Add \(-m\vec a_0\) only when you describe motion relative to the accelerating frame.
- Pointing the pseudo force along the acceleration of the frame.It points opposite to \(\vec a_0\): backward in a car that speeds up, forward in a car that brakes.
- Looking for the reaction to a pseudo force.A pseudo force has no agent and no third-law partner.