1.5.1 Effects of Forces – Core (Cambridge IGCSE Physics 0625)
Learning Objective
Determine the resultant of two or more forces acting along the same straight line and explain how the resultant relates to the state of motion of an object.
Key Concepts (Core)
- Force – a push or pull that can change the state of motion or the shape/size of an object. Measured in newtons (N).
- Resultant force – a single force (vector) that has the same effect as all the individual forces acting together.
- Collinear (or coplanar) forces – forces whose lines of action lie on the same straight line.
- Sign convention – choose a convenient positive direction (e.g., rightward or upward). Forces acting in that direction are taken as positive (+); forces acting opposite are negative (–).
- Newton’s 1st law (equilibrium) – if the resultant force is zero, the object remains at rest or continues to move at constant speed.
- Newton’s 2nd law – if the resultant force is non‑zero, the object accelerates: \[ \mathbf{F}_{\text{resultant}} = m\mathbf{a} \] where \(m\) is the mass (kg) and \(\mathbf{a}\) the acceleration (m s⁻²).
Method for Finding the Resultant of Collinear Forces
- Choose a positive direction (e.g., rightward or upward).
- Assign a sign to each force:
- Positive (+) if the force acts in the chosen direction.
- Negative (–) if the force acts opposite to the chosen direction.
- Write the magnitude of each force with its sign, using vector arrows (→ for positive, ← for negative) if desired.
- Add the signed values:
\[
F_{\text{resultant}} = \sum_{i=1}^{n} F_i
\]
- Interpret the sign of the resultant:
- \(F_{\text{resultant}} > 0\) – resultant acts in the chosen positive direction.
- \(F_{\text{resultant}} < 0\) – resultant acts opposite to the chosen direction.
- \(F_{\text{resultant}} = 0\) – forces are balanced; the object is in equilibrium (no change in motion).
- Link to Newton’s laws:
- If \(F_{\text{resultant}} = 0\) → Newton’s 1st law → no acceleration.
- If \(F_{\text{resultant}} eq 0\) → Newton’s 2nd law → the object accelerates in the direction of the resultant, with magnitude \(a = F_{\text{resultant}}/m\).
Worked Example – Horizontal Forces
Three forces act on a sled along a horizontal line. Choose “to the right” as the positive direction.
| Force |
Magnitude (N) |
Direction |
Signed value (N) |
| \(F_1\) |
30 |
Right (→) |
+30 |
| \(F_2\) |
45 |
Left (←) |
–45 |
| \(F_3\) |
20 |
Right (→) |
+20 |
Resultant calculation:
\[
F_{\text{resultant}} = (+30) + (–45) + (+20) = +5\ \text{N}
\]
Interpretation:
- The resultant is \(5\ \text{N}\) to the right (→).
- According to Newton’s 2nd law, the sled will accelerate rightward (\(a = 5/m\) m s⁻²), where \(m\) is the sled’s mass.
Worked Example – Vertical Forces
A person lifts a box upward. Two forces act on the box:
- Upward pull of the hand: \(F_{\text{hand}} = 120\ \text{N}\) (upward = positive).
- Weight of the box: \(W = mg = 80\ \text{N}\) acting downward (negative).
Resultant:
\[
F_{\text{resultant}} = +120\ \text{N} \;-\; 80\ \text{N} = +40\ \text{N}
\]
Because the resultant is non‑zero, the box accelerates upward (\(a = 40/m\) m s⁻²). If the box were already moving upward, it would speed up.
Load‑Extension (Force‑Extension) Graphs – Core Requirement
A load‑extension graph shows how a material (e.g., a spring) deforms under an applied force.
| Feature on the graph |
Physical meaning (syllabus language) |
| Straight‑line region (initial part) |
Force is proportional to extension – the material behaves **elastically** (Hooke’s law). Slope = spring constant \(k\) (N m⁻¹). |
| Limit of proportionality |
The point where the straight line ends. Beyond this the material no longer obeys Hooke’s law; deformation becomes **non‑elastic** (plastic). |
| Maximum extension (breaking point) |
The force at which the material ruptures or yields permanently. The graph typically ends abruptly. |
How to locate the features on a real graph
- Identify the initial straight‑line portion – draw a best‑fit straight line through the low‑force data points.
- The **limit of proportionality** is where the data start to deviate noticeably from that line.
- Continue following the curve until it either levels off (plastic deformation) or ends sharply – that point is the **breaking point**.
Example
- From a graph, the straight‑line region ends at a force of \(30\ \text{N}\) and an extension of \(0.06\ \text{m}\).
- Spring constant: \(k = \dfrac{F}{x} = \dfrac{30\ \text{N}}{0.06\ \text{m}} = 500\ \text{N m}^{-1}\).
Common Mistakes & How to Avoid Them
- Missing sign assignment – always write each force with a sign before adding.
- Adding magnitudes only – this ignores direction and gives an incorrect resultant.
- Inconsistent positive direction – decide on one direction at the start of the problem and stick to it.
- Assuming motion automatically – a non‑zero resultant produces acceleration only if the object has mass and is free to move (Newton’s 2nd law).
- Misreading graphs – locate the limit of proportionality and breaking point by looking for deviation from the straight line and the end of the curve.
Practice Questions (Core)
- Two forces act on a crate along a straight line: \(F_A = 12\ \text{N}\) to the left and \(F_B = 18\ \text{N}\) to the right. Find the resultant force and its direction.
- A tugboat pulls a barge forward with \(F_1 = 250\ \text{N}\). A river current pushes the barge backward with \(F_2 = 80\ \text{N}\). Determine the net force and state what will happen to the barge’s motion.
- Four forces act on a sled: \(+15\ \text{N}\), \(-10\ \text{N}\), \(+5\ \text{N}\) and \(-8\ \text{N}\). What is the resultant?
- Explain why the resultant is zero when \(F_1 = 50\ \text{N}\) to the right and \(F_2 = 50\ \text{N}\) to the left. Relate your answer to Newton’s 1st law.
- Sketch a simple load‑extension graph for a spring that obeys Hooke’s law up to \(30\ \text{N}\) where the extension is \(0.06\ \text{m}\). State the spring constant.
Extension (Supplementary) Topics (Higher‑Tier)
- Resultant of forces acting at right angles – vector addition using the Pythagorean theorem.
- Newton’s 2nd law in quantitative problems: \(F = ma\).
- Elastic vs. plastic deformation; energy stored in a stretched spring (\(E = \tfrac12 kx^2\)).
- Centripetal force as the resultant of radial forces in uniform circular motion.
Summary
To find the resultant of forces acting along the same straight line:
- Choose a positive direction.
- Assign a sign (+ or –) to each force according to that direction.
- Write the signed magnitudes and add them.
- Interpret the sign of the sum:
- Positive → resultant in the chosen direction.
- Negative → resultant opposite to the chosen direction.
- Zero → forces are balanced; the object is in equilibrium (Newton’s 1st law).
- If the resultant is non‑zero, use Newton’s 2nd law (\(F = ma\)) to predict the object’s acceleration.
Further Reading
For students who wish to explore the supplementary material, consult the Cambridge IGCSE Physics textbook sections on:
- Vector addition of perpendicular forces.
- Newton’s second law and its applications.
- Elastic behaviour of springs, Hooke’s law, and elastic potential energy.
- Centripetal force in uniform circular motion.