Newton's laws and forces - TCE Physics (Tasmania)
“Apply Newton's three laws of motion to objects modelled as point masses.”
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 (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:
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.
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 acts downward from the centre of mass.
- The normal force acts perpendicular to a surface.
- Friction acts along a surface, opposing relative motion.
- Tension 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 to each axis separately.
Friction and motion on a slope
Friction is modelled as , where is the coefficient of friction and is the normal force. On a slope of angle , resolving the weight gives a component down the slope and into the surface (so for an object on an incline with no other vertical forces). An object slides only when the driving component exceeds the maximum friction , which is why the angle at which sliding begins gives .
What a scale reads is the normal force, not your true weight. In a lift accelerating upward the normal force exceeds , 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.
Find the net horizontal force
A box is pulled along a level floor by a horizontal force of against a friction force of . The vertical forces balance, so consider the horizontal direction:
Apply the second law
The box accelerates at in the direction of the pull.
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 for each direction. Treating an equilibrium problem as and an accelerating problem as 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 marksA person of mass slides down a rope that would break if a force exceeding 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 down, and tension (the rope force) up. By Newton's second law the net downward force gives the downward acceleration.
Weight: down.
The rope can pull up with at most . Taking down as positive, the net force is down.
Acceleration: downward.
So the person must accelerate downward at no less than about . Sliding faster (gripping harder) raises the tension above and snaps the rope. Markers want the free body diagram with down and up, and .
Practice5 marksA plane of mass turns in a horizontal circle by banking at . Air exerts a lift force on the wings at 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: .
Lift is perpendicular to the wings, tilted from the vertical. Its vertical component balances weight: , so .
The unbalanced (horizontal) force is the horizontal component of lift, which provides the centripetal force: , 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 marksA good-luck charm of mass hangs from a car's rear-view mirror. As the car accelerates forward, the charm swings back to from the vertical. Calculate the acceleration of the car.Show worked answer →
The charm is acted on by its weight (down) and the string tension (along the string). The horizontal component of tension provides the forward acceleration; the vertical component balances the weight.
Vertical: . Horizontal: .
Dividing the second by the first eliminates both and :
The car accelerates forward at about . The mass cancels, so the value is not needed. Markers want .