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How do Newton's three laws relate the forces on an object to its motion?

Apply Newton's three laws of motion, drawing free-body diagrams and resolving forces to determine the acceleration of objects and the forces in interacting systems.

Newton's three laws of motion, free-body diagrams, resolving forces on inclines, and using net force to find acceleration, with worked examples.

Generated by Claude Opus 4.78 min answer

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  1. What this dot point is asking
  2. The three laws
  3. Free-body diagrams
  4. Forces on an inclined plane
  5. Connected systems
  6. Equilibrium

What this dot point is asking

You need to state and apply Newton's three laws, draw free-body diagrams, resolve forces (including on inclined planes), and calculate acceleration from the net force on an object or connected system.

The three laws

The second law is the workhorse. "Net force" means the vector sum of all forces acting on the object. If the net force is zero the object is in equilibrium (first law); if not, it accelerates in the direction of the net force.

Free-body diagrams

A free-body diagram shows one object as a point (or simple shape) with every force on it drawn as an arrow from that point. Typical forces are:

  • Weight W=mgW = mg, always vertically down.
  • Normal force NN, perpendicular to the surface.
  • Tension TT, along a rope or cable, pulling away from the object.
  • Friction ff, parallel to the surface, opposing relative motion.
  • Applied force FF, whatever pushes or pulls.

Once drawn, resolve the forces into perpendicular axes (usually horizontal and vertical, or along and across an incline) and apply Fnet=ma\vec{F}_{\text{net}} = m\vec{a} to each axis separately.

Forces on an inclined plane

For an object on a frictionless incline at angle θ\theta, choose axes along and perpendicular to the slope. The weight resolves into:

W=mgsinθ(down the slope)W=mgcosθW_{\parallel} = mg\sin\theta \quad (\text{down the slope}) \qquad W_{\perp} = mg\cos\theta

Perpendicular to the slope there is no acceleration, so N=mgcosθN = mg\cos\theta. Along the slope, mgsinθmg\sin\theta produces an acceleration a=gsinθa = g\sin\theta (frictionless).

Connected systems

For objects joined by a rope (e.g. a trolley pulled by a hanging mass over a pulley), treat the whole system to find the common acceleration, then isolate one object to find the tension. For the whole system:

a=net external forcetotal massa = \frac{\text{net external force}}{\text{total mass}}

Then apply Newton's second law to a single object to get the internal tension. The rope tension is an internal force for the system but an external force for each block.

Equilibrium

When Fnet=0\vec{F}_{\text{net}} = 0 the object is in equilibrium (at rest or constant velocity). Then the components of all forces balance in each direction. This is how you find unknown tensions in a hanging sign or the friction holding a box still on a slope.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 SACE Stage 22 marksExplain how the banked section of a racetrack reduces a car's reliance on friction to move around the curve.
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On a flat curve, friction alone must supply the entire centripetal force directed toward the centre of the curve.

On a banked curve the track is tilted, so the normal force (which acts perpendicular to the road surface) now has a horizontal component pointing toward the centre of the circle. This horizontal component of the normal force provides part, or all, of the required centripetal force.

Because the normal force now contributes to turning the car, less friction is needed to keep the car on the circular path. 1 mark for stating that the normal force gains a horizontal component directed toward the centre, 1 mark for linking this to providing the centripetal force and so reducing the friction needed.

2023 SACE Stage 22 marksA delivery van and a Formula One racing car travel at the same speed. Explain whether the van or the racing car would experience a smaller drag force.
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Drag force depends on the object's speed (equal here), its frontal cross-sectional area, and how streamlined its shape is (its drag coefficient).

The racing car is low, smooth and streamlined with a small frontal area, so air flows over it with little turbulence. The van is tall and box-shaped with a large flat frontal area, producing more turbulent airflow and a larger drag coefficient.

Therefore the racing car experiences the smaller drag force at the same speed. 1 mark for identifying the relevant factors (frontal area and streamlining), 1 mark for the correct conclusion that the racing car has less drag.