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Newton's laws and forces - TCE Physics (Tasmania)

Syllabus dot point

“Apply Newton's three laws of motion to objects modelled as point masses.”

TCEPhysicsUnit 3: Gravity and Electromagnetism9 min read

Quick answer

Newton's laws link force and motion: an object keeps constant velocity unless a net force acts (first law), net force equals mass times acceleration Fnet=maF_\text{net} = ma (second law), and every force has an equal and opposite reaction on a different object (third law). You solve problems by drawing a free body diagram and summing forces.

What this dot point is asking

This dot point asks you to connect the forces on an object to its motion. The object is modelled as a point mass, so you can ignore its size and treat every force as acting at one point.

Newton's first law

An object continues at rest or at constant velocity unless a net external force acts on it. This is the law of inertia: motion does not need a force to keep it going, only a force to change it. A puck sliding on frictionless ice would never stop. In real life friction and air resistance are the net forces that bring things to rest.

Newton's second law

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

Fnet=maF_\text{net} = ma

Force is a vector, so this equation holds separately in each direction. The acceleration is always in the same direction as the net force. A larger mass needs a larger force to reach the same acceleration, which is why mass measures inertia.

Net force

The net force is the single vector sum of all the forces acting on an object. If the net force is zero the object is in equilibrium and has zero acceleration, meaning it is either stationary or moving at constant velocity.

Newton's third law

When object A exerts a force on object B, object B exerts an equal and opposite force on A. The two forces are the same size, opposite in direction, and always act on different objects, so they never cancel each other on the same body. A swimmer pushes water backward and the water pushes the swimmer forward.

Common forces and free body diagrams

To analyse a situation, draw a free body diagram showing only the forces on the one object you care about:

  • Weight W=mgW = mg acts downward from the centre of mass.
  • The normal force NN acts perpendicular to a surface.
  • Friction ff acts along a surface, opposing relative motion.
  • Tension TT acts along a rope or string, pulling away from the object.

When forces act at angles, resolve each into components along convenient axes (often horizontal and vertical, or along and perpendicular to a slope), then apply Fnet=maF_\text{net} = ma to each axis separately.

Friction and motion on a slope

Friction is modelled as f=μNf = \mu N, where μ\mu is the coefficient of friction and NN is the normal force. On a slope of angle θ\theta, resolving the weight gives a component mgsin⁡θmg\sin\theta down the slope and mgcos⁡θmg\cos\theta into the surface (so N=mgcos⁡θN = mg\cos\theta for an object on an incline with no other vertical forces). An object slides only when the driving component mgsin⁡θmg\sin\theta exceeds the maximum friction μmgcos⁡θ\mu mg\cos\theta, which is why the angle at which sliding begins gives μ=tan⁡θ\mu = \tan\theta.

Apparent weight comes from the normal force

What a scale reads is the normal force, not your true weight. In a lift accelerating upward the normal force exceeds mgmg, so you feel heavier; accelerating downward it is less, so you feel lighter. In free fall the normal force is zero, which is the sensation of weightlessness.

Worked example: box on a rough floor

Find the net horizontal force

A 5.0 kg5.0\ \text{kg} box is pulled along a level floor by a horizontal force of 20 N20\ \text{N} against a friction force of 8.0 N8.0\ \text{N}. The vertical forces balance, so consider the horizontal direction:

Fnet=20−8.0=12 N.F_\text{net} = 20 - 8.0 = 12\ \text{N}.

Apply the second law

a=Fnetm=125.0=2.4 m s−2.a = \frac{F_\text{net}}{m} = \frac{12}{5.0} = 2.4\ \text{m s}^{-2}.

The box accelerates at 2.4 m s−22.4\ \text{m s}^{-2} in the direction of the pull.

Pairing the wrong forces as action and reaction

A Newton's third law pair always acts on two different objects. The weight of a book and the normal force from the table are not a third law pair, even though they balance, because both act on the book. The reaction to the book's weight is the gravitational pull the book exerts on Earth.

In the exam, always start with a labelled free body diagram, choose your axes, resolve angled forces, then write Fnet=maF_\text{net} = ma for each direction. Treating an equilibrium problem as Fnet=0F_\text{net} = 0 and an accelerating problem as Fnet=maF_\text{net} = ma keeps your method clear.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice3 marks
A person of mass 90.0 kg90.0\ \text{kg} slides down a rope that would break if a force exceeding 700 N700\ \text{N} is exerted on it. Sketch and label the forces on the person, then calculate the minimum acceleration the person must have while sliding so the rope does not break.
Show worked answer →

Two forces act on the person: weight W=mgW = mg down, and tension TT (the rope force) up. By Newton's second law the net downward force gives the downward acceleration.

Weight: W=mg=(90.0)(9.81)=882.9 NW = mg = (90.0)(9.81) = 882.9\ \text{N} down.

The rope can pull up with at most T=700 NT = 700\ \text{N}. Taking down as positive, the net force is F=W−T=882.9−700=182.9 NF = W - T = 882.9 - 700 = 182.9\ \text{N} down.

Acceleration: a=Fm=182.990.0=2.03 m s−2a = \dfrac{F}{m} = \dfrac{182.9}{90.0} = 2.03\ \text{m s}^{-2} downward.

So the person must accelerate downward at no less than about 2.0 m s−22.0\ \text{m s}^{-2}. Sliding faster (gripping harder) raises the tension above 700 N700\ \text{N} and snaps the rope. Markers want the free body diagram with WW down and TT up, and a=mg−Tma = \dfrac{mg - T}{m}.

Practice5 marks
A plane of mass 900 kg900\ \text{kg} turns in a horizontal circle by banking at 30∘30^\circ. Air exerts a lift force on the wings at 90∘90^\circ to their surface. Justify, using Newton's laws, why the vertical forces must add to zero, then calculate the unbalanced (net) force acting on the plane.
Show worked answer →

Vertical balance (Newton's first and second laws): the plane stays in a horizontal circle, so it has no vertical acceleration. Zero vertical acceleration requires zero net vertical force, so the upward vertical component of lift must exactly balance the downward weight.

Weight: W=mg=(900)(9.81)=8829 NW = mg = (900)(9.81) = 8829\ \text{N}.

Lift LL is perpendicular to the wings, tilted 30∘30^\circ from the vertical. Its vertical component balances weight: Lcos⁡30∘=WL\cos 30^\circ = W, so L=8829cos⁡30∘=1.019×104 NL = \dfrac{8829}{\cos 30^\circ} = 1.019 \times 10^4\ \text{N}.

The unbalanced (horizontal) force is the horizontal component of lift, which provides the centripetal force: Fnet=Lsin⁡30∘=(1.019×104)(0.5)=5.10×103 NF_\text{net} = L\sin 30^\circ = (1.019\times10^4)(0.5) = 5.10 \times 10^3\ \text{N}, directed horizontally toward the centre of the circle. Markers reward stating "no vertical acceleration means zero net vertical force" and resolving lift into components.

Practice2 marks
A good-luck charm of mass 20 g20\ \text{g} hangs from a car's rear-view mirror. As the car accelerates forward, the charm swings back to 38∘38^\circ from the vertical. Calculate the acceleration of the car.
Show worked answer →

The charm is acted on by its weight mgmg (down) and the string tension TT (along the string). The horizontal component of tension provides the forward acceleration; the vertical component balances the weight.

Vertical: Tcos⁡38∘=mgT\cos 38^\circ = mg. Horizontal: Tsin⁡38∘=maT\sin 38^\circ = ma.

Dividing the second by the first eliminates both TT and mm:

tan⁡38∘=ag,a=gtan⁡38∘=(9.81)(0.7813)=7.66 m s−2.\tan 38^\circ = \frac{a}{g}, \quad a = g\tan 38^\circ = (9.81)(0.7813) = 7.66\ \text{m s}^{-2}.

The car accelerates forward at about 7.7 m s−27.7\ \text{m s}^{-2}. The mass cancels, so the 20 g20\ \text{g} value is not needed. Markers want a=gtan⁡θa = g\tan\theta.

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Sources & how we know this

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