1.2 Resolution of Forces
In everyday situations such as pulling a friend on a palm frond (tarik upih), mopping the floor, or pushing a lawn mower, forces are often applied at an angle rather than purely horizontally or vertically. Resolution of forces allows us to break down a single inclined force into two perpendicular components—usually horizontal and vertical, or parallel and perpendicular to an inclined plane. This helps us analyse the actual motion and the forces opposing or supporting it.
Learning Objectives
- Define the resolution of forces.
- Resolve a single force into two perpendicular components (horizontal and vertical components, or components parallel and perpendicular to an inclined plane).
- Determine the magnitudes of the resolved components using trigonometric relationships (\(F_x = F \cos \theta\) and \(F_y = F \sin \theta\), or components of weight on an inclined plane).
- Solve problems involving resultant force and resolution of forces using Newton's second law of motion (\(F = ma\)).
Resolving a Force into Perpendicular Components
Compare how a force is resolved on a flat horizontal plane and on an inclined slope.
Force at an Angle to the Horizontal
When a force \(F\) acts at an angle \(\theta\) above the horizontal, it can be resolved into a horizontal component \(F_x = F \cos \theta\) and a vertical component \(F_y = F \sin \theta\).
Weight on an Inclined Plane
For an object of weight \(W\) on a slope inclined at angle \(\theta\), the weight is resolved into a component parallel to the slope, \(W_x = W \sin \theta\), and a component perpendicular to the slope, \(W_y = W \cos \theta\).
Short Explanation
What is Resolution of Forces?
Resolution of forces is the process of resolving a single force into two components, usually perpendicular to each other.
Why Resolve Forces?
When a force acts at an angle, it affects two directions at once. Resolving the force lets us analyse the horizontal and vertical effects separately, then combine them to find the resultant force and acceleration.
For a force \(F\) acting at an angle \(\theta\) to the horizontal:
\(F_x = F \cos \theta\) \(F_y = F \sin \theta\)\(F_x\) is the horizontal component and \(F_y\) is the vertical component.
For an object of weight \(W\) on a plane inclined at angle \(\theta\) to the horizontal:
\(W_x = W \sin \theta\) \(W_y = W \cos \theta\)\(W_x\) is the component parallel to the plane and \(W_y\) is the component perpendicular to the plane.
| Quantity | Symbol | Formula | SI unit |
|---|---|---|---|
| Horizontal component | \(F_x\) | \(F \cos \theta\) | N |
| Vertical component | \(F_y\) | \(F \sin \theta\) | N |
| Parallel weight component | \(W_x\) | \(W \sin \theta\) | N |
| Perpendicular weight component | \(W_y\) | \(W \cos \theta\) | N |
Try to Answer First
Answer in your mind, then press “Check Answer”.
What is the definition of resolution of forces?
Check Answer
A pushing force of \(32\text{ N}\) is applied at an angle of \(60^\circ\) to the horizontal when mopping the floor. Calculate the magnitude of the horizontal component.
Check Answer
A boy of weight \(300\text{ N}\) sits on a slide inclined at \(50^\circ\) to the horizontal. What is the component of his weight acting parallel to the slide?
Check Answer
Common Mistakes
- Confusing \(\sin \theta\) and \(\cos \theta\) by memorising that the horizontal component is always cosine and the vertical is always sine. This fails when the angle is given relative to the vertical line or on an inclined plane.
- Forgetting that on an inclined plane, the angle of inclination \(\theta\) makes the parallel weight component \(W \sin \theta\) and the perpendicular weight component \(W \cos \theta\).
- Omitting the direction or signs when adding collinear components, for example adding friction instead of subtracting it from the pulling component.
Concept Misunderstandings
Resolving a force creates two new physical forces acting on the body simultaneously with the original force.
The components are not separate additional forces; they are mathematical equivalents of the original single force acting in two perpendicular directions.
The normal reaction on an inclined plane is always equal to the total weight \(W\).
On an inclined plane with no acceleration perpendicular to the surface, the normal reaction balances only the perpendicular component of weight, \(R = W \cos \theta\), not the entire weight \(W\).
Summary
- Resolution of forces splits a force into two perpendicular components.
- For a force \(F\) at angle \(\theta\) to the horizontal: horizontal component \(F_x = F \cos \theta\), vertical component \(F_y = F \sin \theta\).
- On an inclined plane inclined at angle \(\theta\) to the horizontal: parallel weight component \(W_x = W \sin \theta\), perpendicular weight component \(W_y = W \cos \theta\).
- Resultant force is found by summing collinear components algebraically, taking direction into account.
- Acceleration is calculated using \(a = \dfrac{F}{m}\).
Short Activity
Objective: Test your understanding of resolution of forces and its components.
A. Objective Quiz
1 A man pushes a lawn mower with a force of \(90\text{ N}\) inclined at an angle of \(60^\circ\) to the horizontal. What is the magnitude of the horizontal component of the pushing force?
2 A block of weight \(W\) slides down a smooth plane inclined at an angle \(\theta\) to the horizontal. Which expression represents the component of weight acting parallel to the inclined plane?
B. Fill in the Blanks
3 The process of resolving a force into two components is known as of forces.
4 In a tarik upih game, the component of the pulling force acts to overcome friction and move the boy forward.
C. Matching / Drag and Drop
Drag each term to the correct meaning. If using a phone, tap the answer first, then tap the matching box.
