Resolving Forces & Inclined Planes (Edexcel A Level Maths: Mechanics): Revision Note
Exam code: 9MA0
Resolving forces
If a single force is applied at an angle to the direction of motion it will need to be resolved into components of the force that act perpendicular to each other.
Why resolve a force acting at an angle?
Resolving a force acting at an angle often helps to simplify a problem
The components of a force parallel and perpendicular to the line of motion allows different types of problems to be solved
The parallel component of a force acting directly on a particle will be the component that causes an effect on the particle
The perpendicular component of a force acting directly on a particle will be the component that has no effect on the particle
Finding the horizontal and vertical components of a force can help solve problems acting in the horizontal plane
The two components of the force will have the same combined effect as the original force
How do I resolve a force acting at an angle?
Use trigonometry to resolve a force acting at an angle
Draw a vector triangle of forces by decomposing the force into its horizontal and vertical components
The original force will be the hypotenuse of the triangle and the two components will make up the opposite and adjacent sides
How do I find the resultant force of two forces acting at different angles?
Force is a vector quantity, so finding the resultant of two or more forces is the same as finding the resultant of two or more vectors
It is possible to use geometry to find the resultant force without decomposing the forces into their horizontal and vertical components
Use the triangle law for vector addition to calculate the magnitude and direction of a resultant force
You will need to use trigonometry to find the missing side or angle in the triangle.
Worked Example
A child is pulling a box of mass 2 kg along a smooth horizontal surface using a force of 10 N at an angle of 30o to the horizontal, as shown in the diagram below.
(i) Find the vertical and horizontal components of the 10 N force.
(ii) Explain why the resultant force acting to move the box is 5 N.
(iii) Work out the acceleration of the box.
(iv) Calculate the magnitude of the normal reaction between the box and the floor.
Answer:
Examiner Tips and Tricks
Always draw a diagram and decompose any angled forces into their horizontal and vertical components. Make sure you are confident with basic trigonometry and angle properties.
Inclined planes
How do I use F = ma on an inclined plane?
Problems involving inclined planes will always be acting along the line of greatest slope of the plane
The line of greatest slope is the most direct route with which it is possible to move from the bottom to the top of the plane
A force said to be acting in the same vertical plane as the line of greatest slope will be acting in the same plane as this most direct route
It is usually easier to resolve parallel and perpendicular to the plane
The normal reaction force always acts perpendicular to the plane and will not need to be resolved
Weight (mg) always acts vertically downwards and so it is easier to use when resolved into components perpendicular and parallel to the plane
When an object is moving up or down the inclined plane, it has acceleration only in the parallel direction
The perpendicular forces will be equal in the positive and negative directions
Newton’s Second Law (F = ma) can be used in the parallel direction to determine acceleration or missing forces
On an inclined plane friction will always act in the opposite direction to the direction the object is moving in or on the point of moving in
Worked Example
A box of mass 5 kg is being pushed up a smooth ramp inclined at 10o to the horizontal by a force of 12 N.
(i) Find the magnitude of the reaction force acting on the box.
(ii) Find the acceleration of the box up the slope.
Answer:
Examiner Tips and Tricks
Use a triangle of components on your force diagrams to help you work out the components of the force but be careful not to count the components as two separate forces. Make sure you are confident with angle properties and always check you are using the correct angle between the force and the direction you are resolving in.
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Questions students ask about Resolving Forces & Inclined Planes
Real questions from students revising this topic on Save My Exams
How do you apply F = ma on an inclined plane?
Resolve every force parallel and perpendicular to the slope, then apply F = ma along the slope, since that is the only direction with acceleration.
Steps
- Draw a diagram with all the forces: weight mg vertically down, normal reaction R perpendicular to the plane, any applied force, and friction if the plane is rough
- Resolve the weight into mg sin θ down the slope and mg cos θ into the slope
- Perpendicular: no acceleration, so the forces balance and R = mg cos θ
- Parallel: resultant force = ma
Page example: 5 kg box pushed up a smooth 10° slope by a 12 N force, g = 9.8
- Perpendicular: R = 5g cos 10° = 48.255... = 48.3 N (3 s.f.)
- Parallel (up the slope positive): 12 − 5g sin 10° = 5a
- 3.49123... = 5a, so a = 0.698 m s⁻² (3 s.f.)
Tips: pick a positive direction along the slope and keep to it; "smooth" means no friction.
How do you determine the direction of components when resolving forces on an inclined plane?
Set your two directions as parallel to the slope and perpendicular to it, then each component points the same way as the original force along those axes.
Force by force
- Weight (mg): acts vertically down, so it resolves into mg sin θ down the slope and mg cos θ into the slope
- Normal reaction (R): already perpendicular, pointing away from the surface, so never needs resolving
- Friction: acts along the slope, opposite to the motion or the intended motion
- Applied force: use the angle it makes with the slope and resolve it the same way
Which gets sin and which gets cos?
- The angle between the weight and the perpendicular to the plane is the slope angle θ — the page marks "this angle is also 10°"
- The component next to that angle is mg cos θ (perpendicular); the other is mg sin θ (parallel)
Signs: choose a positive direction each way (the page uses up the slope ↗ and away from the plane ↖); forces with it are +, against it are −, giving 12 − 5g sin 10° = 5a.
How do you find the direction of a resultant force from the horizontal?
Draw the forces as a vector triangle, find the resultant, then use trigonometry to get the angle it makes with the horizontal.
Page method (vector triangle) Forces: 6 N at 20° below the horizontal and 10 N at 40° above it.
- Draw the vectors tip-to-tail; angle properties give the inside angle 120°
- Cosine rule: R² = 6² + 10² − 2(6)(10) cos 120° = 196, so R = 14 N
- Sine rule: sin(θ + 20°)/10 = sin 120°/14, so sin(θ + 20°) = 5√3/14
- θ + 20° = 38.2132..., so θ = 18.2° to the horizontal (3 s.f.)
Alternative (components)
- Horizontal: 6 cos 20° + 10 cos 40° = 13.298...
- Vertical: 10 sin 40° − 6 sin 20° = 4.375...
- R = √(13.298...² + 4.375...²) = 14 N and tan θ = 4.375.../13.298..., giving θ = 18.2°
Tips: sketch it so you can see whether the angle is above or below the horizontal, and keep unrounded values until the end.
