How to Approach Forces Questions
Three laws. One equation. Dozens of question types. Most marks are lost not because students don’t know the laws — but because they skip the setup. This guide walks through exactly how to approach any forces question, from drawing the free body diagram to getting the right answer.
Newton’s Laws questions are a fixed part of every physics course at this level. The laws themselves are short — three sentences that most students can recite. The problem is application. A question drops an object on a ramp, adds friction, connects it to another mass with a string, and suddenly the recitation isn’t enough. What trips students is the method — not the physics. Get the method right and the equations follow.
What This Guide Covers
What Each Law Actually Means
Before you can apply the laws correctly, you need to understand what each one is saying — not just its formula. The formula is the last step, not the first.
An Object Stays As It Is Unless a Net Force Acts On It
A stationary object stays stationary. A moving object keeps moving at the same speed and in the same direction. Neither state changes unless an unbalanced (net) force is applied. This sounds obvious but it has a practical implication: if something is moving at constant velocity, net force is zero. Constant velocity is not evidence of a force — it’s evidence of balanced forces. That’s a common exam trap.
Where this matters on exam questions: Any question that describes an object moving at “constant speed” or “constant velocity” — or sitting still — is telling you F_net = 0. You can immediately write ΣF = 0 and use that to find an unknown force. If a question says a car travels at 60 km/h on a flat road, the driving force equals the resistance forces. No net force. No acceleration. First law in action.Net Force Equals Mass Times Acceleration
This is the central working equation of mechanics. The bigger the net force, the greater the acceleration. The heavier the object, the more force you need to produce the same acceleration. Direction matters — the acceleration is in the same direction as the net force.
The critical word is “net”: F = ma applies to the net force — the vector sum of all forces — not to any individual force acting on the object. If a 10 kg box is being pushed with 50 N and friction is 20 N, the net force is 30 N. Acceleration = 30 / 10 = 3 m/s². Substituting just the 50 N push gives the wrong answer. Always find net force first.m = mass in kilograms (kg)
a = acceleration in m/s²
Apply separately to each axis (x and y) when forces act in multiple directions.
Every Force Has an Equal and Opposite Reaction Force on a Different Object
Forces always come in pairs. When object A exerts a force on object B, object B exerts an equal force in the opposite direction on object A. The two forces are always the same magnitude, opposite in direction, and act on different objects.
The most common third-law misunderstanding: Students ask “if the forces are equal and opposite, why does anything move?” The answer: the two forces act on different objects. When you kick a ball, the force on the ball and the reaction on your foot are separate interactions on separate objects. The ball accelerates because a net force acts on it. To find whether an object moves, look only at forces acting on that object — not forces it exerts on others.The 7-Step Method for Any Forces Question
Every forces question — regardless of how it’s set up — can be handled with the same sequence. Students who skip steps are the ones who get the wrong answer and can’t find where they went wrong.
Read What the Object Is Doing First
Is it stationary? Moving at constant velocity? Accelerating? Decelerating? This immediately tells you about net force. Stationary or constant velocity → F_net = 0. Accelerating or decelerating → F_net ≠ 0. This step costs nothing and saves every calculation that follows.
Draw a Free Body Diagram
Put a dot or box for the object. Draw every force as an arrow from that object: weight (W = mg) straight down, normal force perpendicular to the surface, applied force in its stated direction, friction opposing motion. Label every arrow. If you can’t draw it, you don’t know which forces are acting — and you can’t write the equations.
Choose Your Coordinate Axes
For horizontal/vertical problems: x = horizontal, y = vertical. For inclined planes: tilt your axes so x is along the slope and y is perpendicular to it. Tilting the axes for inclines eliminates the need to resolve the normal force and massively simplifies the working. This is a technique choice, not a rule — but it’s almost always the better choice on a ramp.
Resolve Any Forces Not Aligned With Your Axes
Any force at an angle needs to be split into its x and y components using trigonometry. For a force F at angle θ to the horizontal: F_x = F cos θ, F_y = F sin θ. On a standard inclined plane with tilted axes, it’s the weight (mg) that gets resolved: mg sin θ along the slope, mg cos θ perpendicular. Everything else stays on one axis.
Write F = ma for Each Axis Separately
Sum all forces in the x-direction: ΣF_x = ma_x. Sum all forces in the y-direction: ΣF_y = ma_y. If there’s no acceleration in the y-direction (most standard problems), ΣF_y = 0. This gives you the normal force. Then ΣF_x = ma gives you the acceleration or the unknown force. Two equations, two unknowns if needed.
Solve the Equations
Identify what’s unknown. Substitute what you know. Isolate the unknown. For simultaneous equations (two connected masses), use substitution or elimination. Show your working at every step — examiners award method marks even if the final number is wrong.
Check the Answer Makes Physical Sense
Is the direction correct? Is the magnitude plausible? A 1 kg object accelerating at 500 m/s² from a 10 N push should flag as wrong. Check units — force in Newtons, acceleration in m/s², mass in kg. A negative acceleration just means the object is decelerating or accelerating in the opposite direction to your positive axis — not necessarily an error.
Drawing a Correct Free Body Diagram
The free body diagram is not a formality — it’s where marks are won and lost. A correct FBD shows every force acting on the object, with correct direction and a label. Nothing else.
Forces That Appear on Most FBDs
- Weight (W = mg) — always straight down from the object’s centre
- Normal force (N) — perpendicular to the contact surface (not always vertical)
- Applied force (F) — in the stated direction of the push or pull
- Friction (f) — along the surface, opposing the direction of motion or impending motion
- Tension (T) — along the string or rope, toward the attachment point
- Air resistance / drag — opposing the direction of motion
What Does NOT Belong on an FBD
- Forces the object exerts on other things (third-law partners go on the other object’s FBD)
- Velocity or acceleration arrows — these are not forces
- Components of forces (these come later, after you’ve resolved — don’t draw both a force and its components on the same diagram)
- Internal forces between parts of the same object
- Imaginary forces like “the force of motion” — no such thing
On a flat, horizontal surface with no vertical applied forces: N = mg. That’s the easy case. On an inclined plane: N = mg cos θ — less than weight. If there’s an applied force with a vertical component pushing into or away from the surface, N changes. If the object is in a lift (elevator) accelerating upward: N = m(g + a). Accelerating downward: N = m(g − a). The normal force adjusts to whatever is needed to prevent the object moving through the surface — it’s not a fixed value tied to weight.
Applying F = ma Correctly
The equation is three letters. The application is where it gets specific. The most important thing to nail before substituting: what is the net force, and in which direction?
Sum All Forces as Vectors, Then Apply F = ma to the Result
Net force is a vector sum. Forces in the same direction add. Forces in opposite directions subtract. Forces at angles need to be resolved into components first, then summed per axis. A question that gives you multiple forces and asks for acceleration is a net-force calculation before it’s an F = ma calculation.
Example sequence: A 5 kg box is pushed with 40 N to the right. Kinetic friction is 15 N acting to the left. What is the acceleration?Step 1 — Identify direction: right is positive. Step 2 — Net force: 40 − 15 = 25 N (to the right). Step 3 — Apply second law: a = F_net / m = 25 / 5 = 5 m/s² to the right.
Students who jump straight to F = ma using 40 N get 8 m/s² — wrong. The net force step cannot be skipped.
| Scenario | What Net Force Equals | How to Apply F = ma |
|---|---|---|
| Object on flat surface, single horizontal force, no friction | F_net = F_applied | a = F / m directly |
| Object on flat surface with friction | F_net = F_applied − f (where f = μN = μmg) | a = (F_applied − μmg) / m |
| Object in free fall (no air resistance) | F_net = mg downward | a = g = 9.8 m/s² downward (same for all masses) |
| Object moving at constant velocity | F_net = 0 | ΣF = 0; use to find an unknown balancing force |
| Object on inclined plane (no friction) | F_net = mg sin θ (along slope) | a = g sin θ along the slope |
| Object on inclined plane (with friction) | F_net = mg sin θ − μmg cos θ | a = g(sin θ − μ cos θ) |
Inclined Planes and Ramps
Inclined plane questions appear constantly. Stairs, ramps, slopes — they’re everywhere in physics exams. The reason they trip students is gravity. Gravity pulls straight down, but the motion is along the slope. You have to bridge that gap with trigonometry.
Align x With the Slope Direction, y Perpendicular to It
Instead of keeping x horizontal and y vertical, rotate your coordinate system so x points along the slope (positive = up the slope or down, your choice — be consistent) and y points away from the surface. Now the normal force lies entirely on the y-axis and the motion lies entirely on the x-axis. The only force you need to resolve into components is weight.
After tilting axes, weight (mg) resolves into:• Component along the slope (x-axis): mg sin θ — acts down the slope
• Component perpendicular to slope (y-axis): mg cos θ — acts into the surface
Since there’s no acceleration perpendicular to the slope (the object stays on the surface), ΣF_y = 0 → N = mg cos θ.
If friction is present: f = μN = μmg cos θ, acting along the slope opposing motion.
Net force along the slope: F_net = mg sin θ − f (for an object sliding down) → a = g sin θ − μg cos θ = g(sin θ − μ cos θ).
That’s the complete setup for any standard inclined plane problem with friction.
Normal force: N = mg cos θ
Friction force: f = μN = μmg cos θ
Net force along slope (sliding down, with friction): F_net = mg sin θ − μmg cos θ
Acceleration along slope: a = g(sin θ − μ cos θ)
If object moves up the slope: Both mg sin θ and friction act down the slope → F_net = −mg sin θ − μmg cos θ (decelerating). The magnitude of a = g(sin θ + μ cos θ).
Students constantly mix up sin and cos on inclined planes. Here’s the check: when the angle θ = 0° (flat surface), the component along the slope should be zero and the component perpendicular should equal mg. sin(0°) = 0, cos(0°) = 1. So the along-slope component uses sin θ and the perpendicular component uses cos θ. If you ever doubt which is which, test the formula at θ = 0° — it should give a sensible physical result.
Connected Objects and Pulleys
Two masses connected by a string — either on a surface and over a pulley, or both hanging (Atwood machine) — are a standard question type. The method is the same as always: one FBD per object, F = ma for each, then solve simultaneously.
When both masses hang freely over a frictionless pulley, the heavier side accelerates down and the lighter side accelerates up. Define down as positive for the heavier mass (m₁) and up as positive for the lighter mass (m₂).
For m₁ (heavier, going down): m₁g − T = m₁a
For m₂ (lighter, going up): T − m₂g = m₂a
Add both equations: m₁g − m₂g = (m₁ + m₂)a → a = (m₁ − m₂)g / (m₁ + m₂)
Tension: T = 2m₁m₂g / (m₁ + m₂)
If m₁ = m₂, a = 0 — the system is in equilibrium. That’s a useful sanity check.
Mistakes That Lose Marks
Using Total Weight Instead of Net Force in F = ma
The single most common error. When there are multiple forces acting, substituting just the applied force or just the weight into F = ma gives the wrong acceleration. Always find net force first — sum all forces — then divide by mass.
Calculate Net Force Before Applying F = ma
ΣF = F₁ + F₂ + F₃… (as vectors). Only after you have the net force do you apply F_net = ma. This step is not optional on any multi-force question.
Treating Normal Force as Always Equal to mg
N = mg only on a flat horizontal surface with no vertical applied forces. On a slope: N = mg cos θ. In an accelerating lift: N = m(g ± a). With an upward applied force component: N decreases. With a downward push: N increases. Read the scenario before assuming.
Find Normal Force From the Perpendicular Equation
Write ΣF_y = 0 (or ma if accelerating vertically). Solve for N from that equation. Then use N to find friction: f = μN. This sequence ensures the normal force is always correct for the specific setup.
Applying mg sin θ and mg cos θ to the Wrong Axes
Swapping sin and cos on inclined plane problems is one of the most frequent errors at this level. Using mg cos θ along the slope and mg sin θ perpendicular to it gives wrong values for both normal force and net force.
Test at θ = 0° to Confirm Which Ratio Applies
On a flat surface (θ = 0°): the along-slope component should be zero → use sin θ. The perpendicular component should equal mg → use cos θ. Plug in 0° to verify before using the formula in a question.
Skipping the Free Body Diagram to Save Time
Students who skip the FBD “to save time” regularly miss a force entirely — usually friction, the normal force component, or a tension. The FBD takes 60 seconds. Finding a missing force in the middle of algebra takes much longer and often isn’t found at all.
Draw the FBD Before Writing a Single Equation
The FBD is step two in the method — before axes, before resolving, before equations. It’s the inventory of forces. If it’s not on the diagram, it won’t be in the equations. Draw it every time.
Confusing Third-Law Pairs With Balanced Forces
“Action and reaction cancel out” is a misstatement that causes real problems. Third-law pairs act on different objects — they don’t cancel because they’re not on the same object. Only forces on the same object can cancel each other out. Getting this wrong leads to incorrect FBDs and wrong net force calculations.
Draw a Separate FBD for Each Object
Third-law pairs appear on different diagrams. Force A on object B goes on B’s FBD. The reaction force on A goes on A’s FBD. Never on the same diagram. Each FBD should contain only forces acting on that specific object.
Pre-Submission / Pre-Exam Checklist for Forces Questions
Frequently Asked Questions
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Show every step of your working — not just the final answer. Physics examiners award method marks at every stage: identifying forces, resolving components, writing the equation, substituting values. A wrong final answer with correct method can still earn most of the marks. A correct answer with no working earns only the answer mark.
Use consistent notation throughout. If you define right as positive at the start, keep it positive all the way through. A sign error from switching conventions mid-problem is one of the hardest mistakes to spot when checking your work.
If you’re stuck, go back to the free body diagram. Most errors in forces questions trace back to a force that was missed, drawn in the wrong direction, or applied to the wrong axis. The diagram is the foundation — fix the diagram first before touching the algebra.