AcademyApplying Force Models

Academy

Particle Dynamics with Newton's Second Law

Level 1 - Physics topic page in Applying Force Models.

Principle

Newton's second law turns the external force model for a system into its acceleration.

Notation

\(m\)
mass of the chosen system
\(\mathrm{kg}\)
\(\vec a\)
acceleration of the chosen system
\(\mathrm{m\,s^{-2}}\)
\(\sum\vec F\)
resultant external force
\(\mathrm{N}\)
\(\vec W\)
weight force
\(\mathrm{N}\)
\(g\)
gravitational field strength
\(\mathrm{m\,s^{-2}}\)
\(\theta\)
slope angle
rad or deg

Method

Derivation 1: Choose the system

Newton's second law applies to one chosen system. Only external forces on that system belong in the force sum.

Vector law
\[\sum\vec F_{\mathrm{external}}=m\vec a\]
Internal forces cancel only inside the chosen system.
Resolve by axes
\[\sum F_x=ma_x,\qquad \sum F_y=ma_y\]
Use kinematics after
\[v=v_0+at\]
This step is valid only after the force model gives a constant acceleration.

Derivation 2: Derive the frictionless-slope acceleration

For a block on a frictionless slope, choose one axis parallel to the slope and one perpendicular to it. Gravity is the only force with a component down the slope.

Parallel force balance
\[\sum F_{\parallel}=mg\sin\theta=ma_{\parallel}\]
Slope acceleration
\[a_{\parallel}=g\sin\theta\]
Perpendicular balance
\[\sum F_{\perp}=N-mg\cos\theta=0\]
There is no acceleration through the surface.

The free-body diagram shows the actual forces only: weight downward and the normal force perpendicular to the surface. The component equations above are built from these forces.

mNW
Actual forces on a frictionless slope: weight and the normal reaction.

Rules

These are the compact results from the method above.

Second law
\[\sum \vec F=m\vec a\]
Component law
\[\sum F_x=ma_x,\qquad \sum F_y=ma_y\]
Weight vector
\[\vec W=m\vec g\]
Frictionless slope
\[a=g\sin\theta\]

Examples

Question
A
\[4.0\,\mathrm{kg}\]
cart is pulled forward by
\[18\,\mathrm{N}\]
while a
\[6\,\mathrm{N}\]
resistance acts backward. Find \(a\).
Answer
Choose forward as positive. The net force is
\[\sum F=18-6=12\,\mathrm{N}\]
so
\[a=\frac{12}{4.0}=3.0\,\mathrm{m\,s^{-2}}\]

Checks

  • Only external forces belong in the net-force sum for the chosen system.
  • Acceleration points in the direction of the resultant force.
  • Zero acceleration means zero net force, not zero individual forces.
  • The normal force is found from the perpendicular force equation.