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8.9 Resolution of Forces

GCSE Physics · AQA · Chapter 8: Forces in Balance

A free GCSE Physics lesson for AQA (8463) from OBA School: (Higher tier) Splitting a force into perpendicular components, forces on slopes, and objects in equilibrium. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

  • (Higher tier) explain what is meant by resolving a force into two perpendicular components
  • (Higher tier) resolve a force using a scale drawing, including the weight of an object on a slope
  • (Higher tier) state the conditions for an object to be in equilibrium
  • (Higher tier) use components or a closed vector triangle to decide whether an object is in equilibrium

Key words

  • Component: One of two perpendicular forces that together have the same effect as the original force.
  • Resolving a force: Replacing a force by two perpendicular components whose resultant is the original force.
  • Parallel component: The component of a force that acts along a slope.
  • Perpendicular component: The component of a force that acts at 90° to a slope (also called the normal component).
  • Equilibrium: The state of an object with zero resultant force and no overall turning effect.
  • Vector diagram: A scale drawing in which force arrows are joined tip to tail; a closed shape means the resultant is zero.

Resolving a force into two components

Resolving is the opposite of finding a resultant. One force is replaced by two perpendicular components that together have exactly the same effect. Usually we choose horizontal and vertical directions, or directions along and at 90° to a slope.

1. Draw the force to scale as an arrow from a point O.

2. Draw two lines from O in the two chosen perpendicular directions.

3. Draw a rectangle with these lines as two sides and the force arrow as its diagonal.

4. Measure the two sides of the rectangle and convert with the scale: these are the components.

Worked example: the sledge rope

Draw the 50 N pull at 30° to the horizontal (e.g. 1 cm ≡ 5 N, so 10.0 cm long) and complete the rectangle.
Horizontal side ≈ 8.7 cm → 43 N, which pulls the sledge forward.
Vertical side = 5.0 cm → 25 N, which lifts slightly and reduces the push of the sledge on the snow.

Answer: components ≈ 43 N horizontally and 25 N vertically

Forces on a slope

It is harder to cycle uphill because part of your weight acts down the slope. Resolve the weight of an object on a slope into a component parallel to the slope and a component perpendicular to it.

  • The parallel component (≈ 17 N) tries to make the box slide down. If the box stays at rest, friction acts up the slope and is equal to it (17 N).
  • The perpendicular component (≈ 36 N) presses the box onto the slope. The slope pushes back with an equal normal contact force.
  • The steeper the slope, the bigger the parallel component and the smaller the perpendicular component.

Exam tip: In a scale drawing on a slope, the rectangle’s sides must be parallel and perpendicular to the slope, not horizontal and vertical.

Equilibrium

An object at rest (or moving at a constant velocity) is in equilibrium. Two conditions must be met:

  • the resultant force on the object is zero
  • the forces have no overall turning effect (the moments balance).

If all the forces are parallel, the total force in one direction must equal the total force in the opposite direction. If they are not parallel, resolve every force into components along the same two perpendicular directions: the components must balance in each direction. Alternatively, draw the forces tip to tail: for an object in equilibrium they form a closed triangle.

Worked example: tension in the cables

Horizontal: the two tensions are equal and their horizontal components cancel.
Vertical: each tension has a vertical component of half its size (from a scale drawing of a 30° line).
Vertical components must balance the weight: ½T + ½T = 60 N, so T = 60 N.

Answer: each cable has a tension of 60 N

Testing an incline

A trolley is held still on a board inclined at a fixed angle by a newton-meter parallel to the board. The reading F equals the component of the trolley’s weight down the slope (friction in the wheels is small). Weigh the trolley with the newton-meter, then add masses to change W and record F each time. Repeat at a different angle.

Common mistakes

  • Drawing the components so that they do not make a rectangle with the force as the diagonal.
  • On a slope, resolving the weight horizontally and vertically instead of along and perpendicular to the slope.
  • Thinking friction on a resting object is always as big as possible. It only matches the parallel component needed to stop sliding.
  • Checking only one direction for equilibrium. The components must balance in both perpendicular directions.

Revision questions

Try these, then check your answers below each one.

Question 1. (Higher tier) What does resolving a force mean? (Choose from: replacing it by two perpendicular components that together have the same effect, adding it to another force, finding the moment of the force, measuring the force with a newton-meter.)

Answer: replacing it by two perpendicular components that together have the same effect

Working: Resolving splits one force into two perpendicular components whose resultant is the original force.

Question 2. (Higher tier) A box of weight 40 N rests on a slope at 25° to the horizontal. Use a scale drawing to find the component of its weight acting down the slope.

Answer: 17 N

Working: The side along the slope measures about 17 N (40 × sin 25° = 16.9 N). The other side, into the slope, is about 36 N.

Question 3. (Higher tier) True or false? A parachutist falling at a constant velocity is in equilibrium.

Answer: True

Working: True. Constant velocity means zero resultant force (and no turning), which are the conditions for equilibrium.

Related lessons

Continue in the full course

This lesson is section 8.9 (AGCPH0809) of our free GCSE Physics course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Physics course.

 

 

 

 

 

 

 

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