Newton's laws and forces (QCE Physics Unit 2)

What this dot point is asking

QCAA expects you to state and apply Newton's three laws of motion, and to construct free-body diagrams to analyse the forces on a single body. The standard problem types are level-surface motion with friction, motion on an inclined plane, connected bodies (pulleys and trains of carts), and tension in a hanging string or cable.

Newton's three laws

First law (inertia). An object continues in a state of rest or uniform motion in a straight line unless a net external force acts on it. A net force of zero means constant velocity (which includes being at rest).

Second law. The net force on an object equals its mass times acceleration:

Force has SI unit newton (N): $1$ N is the force that accelerates a $1$ kg mass at $1$ m s$^{-2}$. Force is a vector. Add forces vectorially.

Third law. For every action there is an equal and opposite reaction. If body A exerts a force on body B, then body B exerts a force of equal magnitude and opposite direction on body A. The forces act on different bodies and never on the same body, which is why a third-law pair does not "cancel out" the way two opposing forces on one body do.

Free-body diagrams

Isolate one object and draw every external force acting on it as an arrow from the object's centre. Standard forces:

• Weight ($W = mg$). Straight down. Always present unless explicitly in deep space.
• Normal force ($N$). Perpendicular to the contact surface, pointing away from the surface.
• Friction ($f$). Parallel to the surface, opposing relative motion or relative motion tendency. Kinetic: $f_k = \mu_k N$. Static: $f_s \le \mu_s N$.
• Tension ($T$). Along a string or cable, pulling away from the body.
• Applied force. As stated.

Sum forces vectorially in two perpendicular directions and apply $F = ma$ in each.

Inclined planes

On a frictionless ramp at angle $\theta$:

• Weight component along the slope (down the slope): $W_\parallel = mg \sin\theta$.
• Weight component perpendicular to slope: $W_\perp = mg \cos\theta$.
• Normal force balances $W_\perp$: $N = mg \cos\theta$.
• If friction acts up the slope (object sliding down): $F_{\text{net}} = mg \sin\theta - \mu_k mg \cos\theta$.

Choose the $x$-axis along the slope and the $y$-axis perpendicular. This eliminates the need to resolve the normal force or friction.

Connected bodies

For two masses joined by a light, inextensible string over a frictionless pulley, both masses have the same magnitude of acceleration and the tension is the same throughout the string. Write $F = ma$ for each mass, then solve simultaneously.

Examples in context

Example 1. A Gladstone port crane lifts a $4000 \text{ kg}$ container at constant velocity. The free-body diagram shows weight $mg = 39200 \text{ N}$ downward balanced by cable tension $T = 39200 \text{ N}$ upward (Newton's first law). Accelerating the same container upward at $1.2 \text{ m s}^{-2}$ requires $T - mg = ma$, so $T = m(g+a) = 44000 \text{ N}$ - an extra $4800 \text{ N}$ from Newton's second law that is the design basis for the crane's safe working load margin.

Example 2. A Cairns light-rail tram traverses a $5^\circ$ incline approaching a stop. The tram (mass $45000 \text{ kg}$) experiences a component of weight along the incline $mg\sin\theta = 38500 \text{ N}$ plus rolling resistance $1800 \text{ N}$. The traction motor must apply $40300 \text{ N}$ to hold the tram stationary, by Newton's third law equally pushing the rail backward. QCAA Unit 2 EA Paper 2 incline-stem decomposition follows this template.

Try this

Q1. State Newton's third law and give one example. [2 marks]

• Cue. Forces on two interacting bodies are equal in magnitude and opposite in direction; e.g. rocket pushes gas down, gas pushes rocket up.

Q2. A $3.0 \text{ kg}$ block on a horizontal surface ($\mu_k = 0.20$) is pulled with $15 \text{ N}$ horizontal. Calculate the friction force, the net force and the acceleration. [4 marks]

• Cue. $f_k = \mu_k mg = 5.88 \text{ N}$; $F_{net} = 9.12 \text{ N}$; $a = 3.04 \text{ m s}^{-2}$.

Q3. A $1500 \text{ kg}$ car ascends a $6^\circ$ incline at constant velocity. (a) Draw the free-body diagram. (b) Calculate the driving force and the normal reaction. (c) If the car then accelerates at $1.5 \text{ m s}^{-2}$ up the slope, determine the new driving force. [3+3+2 marks; ISMG: Knowledge and conceptual understanding, Analysis and interpretation]

• Cue. (a) weight, normal, drive, opposed by gravity-component (and friction if specified); (b) $F = mg\sin 6 = 1538 \text{ N}$, $N = mg\cos 6 = 14620 \text{ N}$; (c) $F = 1538 + 2250 = 3788 \text{ N}$.