How do forces determine how an object moves?
Apply Newton's three laws of motion to objects modelled as point masses.
Newton's three laws of motion, free body diagrams, resolving forces, and how to find net force and acceleration including weight, friction, tension and the normal force.
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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.
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.
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 practice questions
Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2022 TASC3 marksA person of mass 90.0 kg slides down a rope that would break if a force exceeding 700 N is exerted on it. Sketch and label the forces on the person, then calculate the minimum acceleration the person must have while sliding in order to not break the rope.Show worked answer →
Two forces act on the person: weight W = mg acting down, and tension T (the rope force) acting up. By Newton's second law the net downward force gives the downward acceleration.
Weight: W = mg = 90.0 x 9.81 = 882.9 N (down).
The rope can pull up with at most T = 700 N. Taking down as positive, the net force is F = W - T = 882.9 - 700 = 182.9 N (down).
Acceleration: a = F / m = 182.9 / 90.0 = 2.03 m s-2 downwards.
So the person must accelerate downwards at no less than about 2.0 m s-2. Sliding faster (smaller grip) raises the tension above 700 N and snaps the rope. Markers want the free body diagram with W down and T up, and a = (mg - T)/m.
2022 TASC5 marksA plane of mass 900 kg turns in a horizontal circle by tilting 30 degrees and slightly increasing its speed. Air exerts a force called 'lift' on the wings at 90 degrees to their surface. Justify, in terms of 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. A 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 x 9.81 = 8829 N.
Lift L is perpendicular to the wings, tilted 30 degrees from the vertical. Its vertical component balances weight: L cos30 = W, so L = 8829 / cos30 = 1.019 x 10^4 N.
The unbalanced (horizontal) force is the horizontal component of lift, which provides the centripetal force: F net = L sin30 = 1.019 x 10^4 x 0.5 = 5.10 x 10^3 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.
2019 TASC2 marksA car has a good-luck charm of mass 20 g hanging from the rear-view mirror. As the car accelerates forward, the charm swings back to an angle of 38 degrees to the vertical. Calculate the acceleration of the car.Show worked answer →
The charm is acted on by its weight mg (down) and the string tension T (along the string). The horizontal component of tension provides the forward acceleration; the vertical component balances the weight.
Vertical: T cos38 = mg
Horizontal: T sin38 = ma
Dividing the second equation by the first eliminates both T and m:
tan38 = a / g
a = g tan38 = 9.81 x tan38 = 9.81 x 0.7813 = 7.66 m s-2.
So the car accelerates forward at about 7.7 m s-2. The mass cancels, so the 20 g value is not needed. Markers want a = g tan(theta).