1.5.1 Effects of Forces – Learning Objective
Recall and use the equation F = ma, understand that the net force and the resulting acceleration are always in the same direction, and apply the related concepts required by the Cambridge IGCSE 0625 syllabus.
Key Definitions
- Force ( F ) – a push or pull; a vector quantity measured in newtons (N).
- Mass ( m ) – the amount of matter; a scalar measured in kilograms (kg).
- Acceleration ( a ) – the rate of change of velocity; a vector measured in metres per second squared (m s⁻²).
Newton’s Second Law
The net external force on an object equals the product of its mass and its acceleration:
\(\displaystyle \vec{F}_{\text{net}} = m\vec{a}\)
- The direction of \(\vec a\) is exactly the direction of the net force \(\vec F_{\text{net}}\) (they are parallel).
- For a single force acting on a free body, the force and the acceleration point the same way.
Scalar vs Vector Quantities
| Quantity | Type | Example |
|---|
| Distance | Scalar | 5 m (total path length) |
| Displacement | Vector | 5 m east |
| Speed | Scalar | 10 m s⁻¹ |
| Velocity | Vector | 10 m s⁻¹ north |
| Force | Vector | 20 N down a slope |
| Mass | Scalar | 2 kg |
Resultant of Two or More Forces
1. Collinear (along the same line)
Add algebraically, keeping sign to indicate direction.
Example: 8 N right + 3 N left = 5 N right.
2. Perpendicular (right‑angled) – Parallelogram (or Triangle) Method
Magnitude:
\(R = \sqrt{F{x}^{2}+F{y}^{2}}\)
Direction (measured from the \(x\)-axis):
\(\theta = \tan^{-1}\!\left(\dfrac{F{y}}{F{x}}\right)\)
Newton’s First Law & Equilibrium
- First law (law of inertia): An object remains at rest or moves with constant velocity unless acted on by a non‑zero net external force.
- If \(\vec F_{\text{net}} = 0\) the object is in equilibrium. Example: a book on a table experiences equal and opposite forces – its weight (down) and the normal reaction (up).
Turning Effect – Moments
The turning effect of a force about a pivot is called a moment:
\(M = F \times d\)
- \(F\) – magnitude of the force (N).
- \(d\) – perpendicular distance from the line of action of the force to the pivot (m).
- Moment is a vector; its sense (clockwise or anticlockwise) is indicated by a sign or an arrow.
Hooke’s Law – Load‑Extension Graph for Springs
- Within the linear region the graph of load (force) versus extension is a straight line.
- Gradient = spring constant \(k\) (N m⁻¹).
- Equation: \(\displaystyle F = kx\) where \(x\) is the extension.
- Beyond the limit of proportionality the spring may deform permanently (plastic region) or break.
Friction
- Dry (solid) friction: \(\displaystyle F_{f}= \mu N\)
- \(\mu\) – coefficient of friction (static \(\mu{s}\) or kinetic \(\mu{k}\)).
- \(N\) – normal reaction (N).
- Viscous (fluid) resistance: \(\displaystyle F_{d}= \tfrac12 C\rho A v^{2}\) (drag) – used for objects moving through liquids or gases.
- Typical IGCSE values: \(\mu{s}\approx0.4\!-\!0.6\) for wood on wood; \(\mu{k}\approx0.3\!-\!0.5\).
Centre of Gravity & Stability
- The centre of gravity (CG) is the point at which the weight of an object can be considered to act.
- An object is stable when its CG lies vertically below the base of support; it tips when the line of action of the weight passes outside the base.
- Simple activity: balance a ruler on a fingertip – the balance point is the CG.
Link to Momentum & Impulse (Topic 1.6)
Momentum: \(\displaystyle \vec p = m\vec v\) (vector).
Impulse–momentum theorem: \(\displaystyle \vec J = \Delta\vec p = \vec F_{\text{net}}\Delta t\).
Thus \(F = ma\) is the instantaneous form of the impulse‑momentum relationship.
Derivation of \(F = ma\)
Starting from the definition of acceleration:
\(a = \dfrac{\Delta v}{\Delta t}\)
and momentum \(p = mv\). Newton’s second law states:
\(\displaystyle \vec F_{\text{net}} = \frac{\Delta \vec p}{\Delta t}
= \frac{\Delta (m\vec v)}{\Delta t}
= m\frac{\Delta \vec v}{\Delta t}
= m\vec a\)
For constant mass (the usual IGCSE case) this reduces to the familiar scalar form F = ma.
Direction of Force and Acceleration
- Both \(\vec F_{\text{net}}\) and \(\vec a\) point in the same direction.
- If several forces act, first find the resultant (net) force; the acceleration is parallel to that resultant.
Units and Symbols
| Quantity | Symbol | SI Unit | Unit Symbol |
|---|
| Force | F | newton | N |
| Mass | m | kilogram | kg |
| Acceleration | a | metre per second squared | m s⁻² |
| Moment | M | newton‑metre | N m |
| Friction force | F_f | newton | N |
Worked Examples
Example 1 – Straight‑line motion
Problem: A 2.0 kg cart is pulled horizontally by a constant force of 10 N. Find the acceleration and its direction.
- Write the formula: \(\displaystyle F = ma\).
- Rearrange: \(\displaystyle a = \frac{F}{m}\).
- Substitute: \(\displaystyle a = \frac{10\ \text{N}}{2.0\ \text{kg}} = 5\ \text{m s}^{-2}\).
- Force is to the right ⇒ acceleration is to the right.
Answer: \(a = 5\ \text{m s}^{-2}\) to the right.
Example 2 – Resultant force and acceleration (perpendicular forces)
Problem: A 3 kg object is acted on by two perpendicular forces: 4 N east and 3 N north. Determine the magnitude and direction of its acceleration.
- Resultant force: \(R = \sqrt{4^{2}+3^{2}} = 5\ \text{N}\).
- Direction: \(\theta = \tan^{-1}(3/4) = 36.9^{\circ}\) north of east.
- Acceleration: \(a = \dfrac{R}{m} = \dfrac{5\ \text{N}}{3\ \text{kg}} = 1.67\ \text{m s}^{-2}\) in the same direction as \(R\).
Answer: \(a = 1.7\ \text{m s}^{-2}\) at \(36.9^{\circ}\) north of east.
Example 3 – Turning effect (moment)
Problem: A force of 12 N is applied 0.25 m from a hinge on a door, acting perpendicular to the door. Find the moment produced.
\(M = Fd = 12\ \text{N} \times 0.25\ \text{m} = 3.0\ \text{N m}\) (anticlockwise).
Common Misconceptions
- “A larger force always gives a larger acceleration, regardless of mass.” – Acceleration depends on both force and mass (\(a = F/m\)).
- “Force and acceleration can be opposite in direction.” – For the *net* force they are parallel; only individual forces may oppose each other.
- “If an object moves, a force must be acting in the direction of motion.” – An object can continue moving with no net force (inertia).
- “Friction always opposes motion, so it must be subtracted from the applied force.” – Only the *net* force (applied − friction) determines the acceleration.
Actionable Review Checklist
- Write down the equation \(F = ma\) and identify which quantities are vectors.
- When several forces act, draw a clear free‑body diagram and find the resultant using:
- Algebraic addition for collinear forces.
- Parallelogram (or triangle) method for perpendicular forces.
- State the direction of the acceleration – it is the same as the direction of the resultant (net) force.
- Check equilibrium: if \(\vec F_{\text{net}} = 0\), the object has no acceleration.
- Calculate moments with \(M = Fd\); remember to note the sense (clockwise/anticlockwise).
- For springs, use \(F = kx\) within the linear region of the load‑extension graph.
- Apply \(\displaystyle Ff = \mu N\) for dry friction, distinguishing static (\(\mus\)) and kinetic (\(\mu_k\)) cases.
- Confirm units: N for force, kg for mass, m s⁻² for acceleration, N m for moments.
Practice Questions
- A 5.0 kg block is pushed across a frictionless surface by a horizontal force of 20 N. Calculate the acceleration.
- A 0.8 kg ball is dropped from rest. Ignoring air resistance, what is the net force acting on the ball just before it hits the ground? (Take \(g = 9.8\ \text{m s}^{-2}\).)
- Two forces act on a 3 kg object: 4 N east and 3 N north. Find the magnitude and direction of the resulting acceleration.
- A 0.5 m long lever is pivoted at one end. A 30 N force is applied 0.40 m from the pivot, acting downwards. What moment does it produce? (State the sense.)
- A spring obeys Hooke’s law with a spring constant \(k = 250\ \text{N m}^{-1}\). How far does it stretch when a 75 N force is applied?
- A wooden block of mass 2.0 kg rests on a horizontal table. The coefficient of static friction between wood and the table is 0.45. What is the minimum horizontal force required to start the block moving?
Suggested Classroom Diagrams
- Free‑body diagram of the 2 kg cart pulled by a 10 N force, with the acceleration vector shown in the same direction.
- Parallelogram method for adding the 4 N east and 3 N north forces.
- Simple lever illustrating moment \(M = Fd\) (including clockwise/anticlockwise sense).
- Load‑extension graph of an elastic spring, clearly marking the proportional region and limit of proportionality.
- Block on a table showing weight, normal reaction, and friction force.
- Ruler balanced on a fingertip to demonstrate the centre of gravity.