Equations of motion, inertia and force models

Inertia relates net force to acceleration

For a constant-mass body described in an inertial frame, the sum of external forces equals ma. Along the guide, write m a=Fdrive+Fresistance, using signed components. The support and weight balance vertically in this teaching model. They do not disappear physically; their net vertical contribution is zero under the stated constraint.

Mass measures resistance to a change of translational velocity. It is not an extra opposing “inertia force” to add to this external-force diagram. A moving carriage does not need a net forward force to maintain an unchanged velocity in the ideal model. A drive may still need to balance actual resistance.

A resistance expression carries its own scope

Let the supplied linear resistance be Fresistance=−cv, where c is a positive coefficient with units N·s/m and v is signed velocity. The minus sign makes the resistance oppose motion: if v becomes negative, the resistance becomes positive. At zero velocity, this particular model supplies no resistance. It therefore cannot describe static friction by itself.

A linear viscous law can be useful within a range, but it is not universal. Sliding contact, fluid drag and a controlled actuator can have different force laws, including thresholds and nonlinear speed dependence. Treat c as a supplied model parameter here, not an established property of air around a real cart. Changing the law changes the predicted time history.

A force equation needs an initial state

The balance m a=Fdrive−cv tells us acceleration at a given velocity. To predict the entire motion, we also need initial velocity and, for position, initial position. The graph plots force against speed; it is not a clock record. The crossing shows a possible steady velocity, where the driving and opposing forces balance.

Below the crossing, net force is rightward. Above it, net force is leftward even while the carriage still moves right. Under this simple positive-damping model, small speed deviations tend back toward the crossing. That local stability statement does not establish stability of every motor, controller or flexible structure connected to a real system.

Evaluate acceleration at three states

Use m=3 kg, Fdrive=12 N and c=2 N·s/m. At v=2 m/s, resistance is 4 N left; net force 8 N right gives a=8/3≈2.67 m/s². At v=6 m/s, resistance 12 N balances the drive, so acceleration is zero. At v=8 m/s, resistance 16 N exceeds the drive and a=(12−16)/3≈−1.33 m/s².

The equilibrium speed is Fdrive/c=6 m/s. A carriage starting below it does not instantly jump to 6 m/s; its acceleration changes as velocity changes. Using the initial 2.67 m/s² as a constant over a long interval would contradict the same resistance model. Predicting time requires retaining that changing force relationship.

Try the reasoning

1. Why is the resistance at negative velocity positive in this model?

   Show the answer and reasoning

   Fresistance=−cv reverses sign with v, so it continues to oppose the motion.

2. Does zero acceleration at 6 m/s imply zero drive force?

   Show the answer and reasoning

   No. The 12 N drive and 12 N opposing force balance while the carriage moves.

3. Does −1.33 m/s² at 8 m/s mean the carriage already moves left?

   Show the answer and reasoning

   No. Its current velocity remains rightward; the acceleration reduces that velocity.

What to carry forward

Keep external forces, constitutive force laws, initial conditions and the resulting motion in distinct roles.