Resolving Vectors (AQA AS Physics): Revision Note

Exam code: 7407

Lindsay Gilmour

Written by: Lindsay Gilmour

Reviewed by: Tim

Updated on

Resolving vectors

  • Two vectors can be represented by a single resultant vector

    • Resolving a vector is the opposite of adding vectors

  • A single resultant vector can be resolved

    • This means it can be represented by two vectors, which in combination have the same effect as the original one

  • When a single resultant vector is broken down into its parts, those parts are called components

  • For example, a force vector of magnitude F and an angle of θ to the horizontal is shown below

Force F drawn as an arrow at angle θ to the horizontal.
A single force F acting at an angle θ to the horizontal has both a horizontal and a vertical effect
  • It is possible to resolve this vector into its horizontal and vertical components using trigonometry

Force F split into a horizontal component F x and a vertical component F y.
The resultant force F can be split into its horizontal and vertical components
  • For the horizontal component, Fx = F cos θ

  • For the vertical component, Fy = F sin θ

Forces on an inclined plane

  • Objects on an inclined plane is a common scenario in which vectors need to be resolved

    • An inclined plane, or a slope, is a flat surface tilted at an angle, θ

  • Instead of thinking of the component of the forces as horizontal and vertical, it is easier to think of them as parallel or perpendicular to the slope

  • The weight of the object is vertically downwards and the normal (or reaction) force, R is always vertically up from the object

  • The weight W is a vector and can be split into the following components:

    • W cos (θ) perpendicular to the slope

    • W sin (θ) parallel to the slope

  • If there is no friction, the force W sin (θ) causes the object to move down the slope

  • The object is not moving perpendicular to the slope, therefore, the normal force R = W cos (θ)

Object on a slope at angle θ, with its weight split into mg sin θ parallel to the slope and mg cos θ perpendicular to it, and the normal force R.
The weight vector of an object on an inclined plane can be split into its components parallel and perpendicular to the slope

Worked Example

A helicopter provides a lift of 250 kN when the blades are tilted at 15º from the vertical.

Helicopter with its 250 kN lift force acting at 15° to the vertical.

Calculate the horizontal and vertical components of the lift force.

[2]

Answer:

Step 1: Draw a vector triangle of the resolved forces

Vector triangle resolving the 250 kilonewton lift force. The resultant force is 15 degrees left of vertical, with vertical component labelled “250 cos(15)” upwards and horizontal component “250 sin(15)” to the left.

Step 2: Calculate the vertical component of the lift force

Vertical = 250 × cos(15) = 242 kN [1 mark]

Step 3: Calculate the horizontal component of the lift force

Horizontal = 250 × sin(15) = 64.7 kN [1 mark]

Examiner Tips and Tricks

If you're unsure as to which component of the force is cos θ or sin θ, just remember that the cos θ is always the adjacent side of the right-angled triangle AKA, making a 'cos sandwich'

Right-angled triangle showing that the cos θ component is the side adjacent to the angle θ.

Equilibrium

  • Coplanar forces can be represented by vector triangles

  • Forces are in equilibrium if an object is either:

    • at rest

    • moving at constant velocity

  • In equilibrium, coplanar forces are represented by closed vector triangles

    • The vectors, when joined together, form a closed path

  • The most common forces on objects are:

    • weight

    • normal reaction force

    • tension (from cords and strings)

    • friction

  • The forces on a body in equilibrium are demonstrated below:

Three forces on an object in equilibrium joined head-to-tail to form a closed vector triangle. The following are given in textboxes: A vehicle is at rest on a slope and has three forces acting on it to keep it in equilibrium. Step 1: draw all the forces on the free-body diagram. Step 2: remove the object and put all the forces coming from a single point. Step 3: rearrange the forces into a closed vector triangle. keep the same length and direction.
Three forces on an object in equilibrium form a closed vector triangle

Worked Example

A weight hangs in equilibrium from a cable at point X. The tensions in the cables are T1 and T2 as shown.

Diagram of a weight W hanging from point X by a vertical cable, with two upward-sloping cables meeting at X: tension T2 acts up-left and tension T1 acts up-right.

Which diagram correctly represents the forces acting at point X?

Four options A–D show a triangular vector diagram labelled W, T1 and T2. W points down in every option; A has T2 up-left and T1 up-right, B T2 down-right and T1 up-right, C T2 down-right and T1 down-left, and D T2 up-left and T1 down-left.

[1]

Answer:

Solution is A. T1 and T2 act upwards along the cables and W downwards. The vectors are arranged head-to-tail in a closed triangle; the correctly directed triangle is ticked and the reversed version crossed.

[1 mark]

Examiner Tips and Tricks

The diagrams in exam questions about this topic could ask you to draw to scale, so make sure you have a ruler handy!

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Lindsay Gilmour

Author: Lindsay Gilmour

Expertise: Physics Content Creator

Lindsay graduated with First Class Honours from the University of Greenwich and earned her Science Communication MSc at Imperial College London. Now with many years’ experience as a Head of Physics and Examiner for A Level and IGCSE Physics (and Biology!), her love of communicating, educating and Physics has brought her to Save My Exams where she hopes to help as many students as possible on their next steps.

Tim

Reviewer: Tim

Expertise: Content Creator

Timothy graduated with a first class degree in Mathematics and Physics from the University of Warwick. After working as a postgraduate researcher, Timothy has worked as a content creator for various online revision platforms, creating physics resources for a range of levels and exam boards.