Forces and Vectors is a foundational topic in Cambridge A-Level Mechanics 1. You will learn to represent forces as vectors, find resultants, resolve forces into components, analyse equilibrium, and apply Lami's theorem. These skills underpin all of mechanics — every problem involving motion, statics, or structures starts here.
F cosθ and F sinθ — parallel and perpendicular to slope
Triangle of Forces
Lami's theorem — three concurrent forces in equilibrium
Equilibrium
ΣFₓ = 0 and ΣF_y = 0 simultaneously
Inclined Planes
mg sinθ along slope, mg cosθ perpendicular to slope
Sine Rule
Used when triangle of forces is not right-angled
Learn 1 — Vectors & Components
Scalars vs Vectors
A scalar has magnitude only. A vector has both magnitude and direction. In mechanics, this distinction matters for every calculation.
Scalars: distance, speed, mass, time, temperature, energy Vectors: displacement, velocity, acceleration, force, momentum, weight
Column Vector Notation
A vector in two dimensions is written as a column with horizontal (x) and vertical (y) components:
v = (a / b) means a units in x-direction, b units in y-direction
Example: Force F = (3 / 4) means 3 N horizontally (right) and 4 N vertically (up).
A negative component indicates the opposite direction: (−5 / 2) means 5 N left, 2 N up.
i + j Notation
i is the unit vector in the positive x-direction (horizontal right). j is the unit vector in the positive y-direction (vertical up).
v = ai + bj is equivalent to the column vector (a / b)
F = 3i + 4j means the same as (3 / 4).
F = −5i + 2j means the same as (−5 / 2).
These two notations are completely interchangeable in A-Level Mechanics.
Magnitude of a Vector
The magnitude (size) of a vector v = ai + bj is found using Pythagoras' theorem:
3(2i + 5j) = 6i + 15j
−2(4i − j) = −8i + 2j
Scaling a vector changes its magnitude but not its direction (unless k < 0, which reverses it).
Position Vectors and Displacement
The position vector of point P is the vector from the origin O to P. Displacement from A to B:
Displacement AB = position of B − position of A = OB − OA
If A has position vector (2i + 3j) and B has position vector (7i + j):
Displacement AB = (7i + j) − (2i + 3j) = 5i − 2j
|AB| = √(25 + 4) = √29 ≈ 5.39 units
Always check the sign of each component carefully. A common error is mixing up which direction is positive — define your axes clearly at the start of every problem and stick to them throughout.
Learn 2 — Resultant Forces
What Is the Resultant?
When several forces act on a particle simultaneously, the resultant is the single force that has exactly the same effect as all the individual forces combined. Newton's second law is applied to the resultant.
R = F₁ + F₂ + F₃ + … = ΣF
Finding the Resultant — Component Method
The most reliable method is to resolve every force into horizontal and vertical components, then sum each set:
Step 1: Write each force as (horizontal component, vertical component). Step 2: Sum all horizontal components: Rₓ = ΣFₓ Step 3: Sum all vertical components: R_y = ΣF_y Step 4: Resultant magnitude: |R| = √(Rₓ² + R_y²) Step 5: Resultant direction: θ = arctan(R_y / Rₓ) from positive x-axis
Example: Two Forces
F₁ = 5i + 2j N, F₂ = −1i + 7j N
R = (5−1)i + (2+7)j = 4i + 9j N
|R| = √(16 + 81) = √97 ≈ 9.85 N
Direction: θ = arctan(9/4) ≈ 66.0° above horizontal
Example: Three Forces at Angles
Force A: 10 N at 0° (horizontal): Aₓ = 10, A_y = 0
Force B: 8 N at 60° above horizontal: Bₓ = 8cos60° = 4, B_y = 8sin60° = 6.93
Force C: 6 N at 150° (measured from positive x-axis): Cₓ = 6cos150° = −5.20, C_y = 6sin150° = 3
Rₓ = 10 + 4 − 5.20 = 8.80 N
R_y = 0 + 6.93 + 3 = 9.93 N
|R| = √(8.80² + 9.93²) = √(77.4 + 98.6) = √176 ≈ 13.3 N
Equilibrium Conditions
A particle is in equilibrium when the resultant force is zero — it either remains at rest or moves with constant velocity (Newton's First Law).
Equilibrium: ΣFₓ = 0 AND ΣF_y = 0
Both conditions must hold simultaneously. If only one holds, the particle accelerates in the other direction.
Tip: When three forces act on a particle in equilibrium and they form a triangle, use the triangle of forces or Lami's theorem (see Learn 5).
Finding an Unknown Force in Equilibrium
Three forces: F₁ = 3i + 5j N, F₂ = −7i + 2j N, F₃ = ?
For equilibrium: F₁ + F₂ + F₃ = 0
(3 − 7 + F₃ₓ)i + (5 + 2 + F₃_y)j = 0
F₃ₓ = 4, F₃_y = −7 F₃ = 4i − 7j N, |F₃| = √(16+49) = √65 ≈ 8.06 N
Always resolve into exactly two perpendicular directions — usually horizontal and vertical, or parallel and perpendicular to a slope. Never mix two different axes in one equation.
Learn 3 — Types of Forces
Weight (W)
Formula: W = mg (where g = 9.8 m s⁻² or 10 m s⁻² as specified) Direction: Always vertically downward — never along a slope, never perpendicular to it. Common error: Thinking weight acts perpendicular to a slope — it does NOT.
W = mg (vertically downward)
Normal Reaction (N or R)
Direction: Always perpendicular to the surface on which the object rests — pointing away from the surface into the object. On horizontal ground: N acts vertically upward. On a slope at angle θ: N acts perpendicular to the slope surface (not vertically). Common error: Drawing N vertically upward even when on an inclined plane.
Tension (T)
Direction: Along the string or rope, away from the object (strings can only pull, never push). Inextensible string: Tension is the same throughout the string (when the string is light and there is no friction at a pulley). Common error: Drawing tension pointing into the object (pushing) — tension always pulls.
Friction (F)
Direction: Along the surface, opposing the motion (or the tendency to move). On horizontal surface: Horizontal, opposite to direction of motion. On inclined plane moving up: Friction acts down the slope. On inclined plane moving down: Friction acts up the slope. Magnitude: F ≤ μN, where μ is the coefficient of friction. At limiting equilibrium: F = μN.
Thrust / Compression
Rods (unlike strings) can both push and pull. A rod in tension pulls its ends inward. A rod in compression (thrust) pushes its ends outward.
When solving, if you get a negative tension value, the rod is actually in compression.
Applied Force (P or F)
Any external force applied to the object — a push, a pull, an engine force. Its direction must be clearly stated in the problem. Always label it with its given angle or direction.
Force Diagram Rules
1. Draw every force as an arrow from the object's centre (or contact point for normal/friction).
2. Label every arrow with its symbol and magnitude if known.
3. Weight arrow points straight DOWN always.
4. Normal arrow points perpendicular to the surface, outward.
5. Tension arrow points along the string, away from the object.
6. Friction arrow points along the surface, opposing tendency to move.
Summary Table
Force
Symbol
Direction
Key Point
Weight
W = mg
Vertically downward
Never along a slope
Normal reaction
N or R
Perpendicular to surface, outward
On slope: perpendicular to slope
Tension
T
Along string, away from object
Strings pull only
Friction
F
Along surface, opposing motion
F = μN at limiting equilibrium
Thrust
T
Along rod, outward (compression)
Rods can push and pull
Applied force
P
As specified
State angle clearly
Learn 4 — Resolving Forces
The Principle of Resolving
Any force F at angle θ to a given direction can be split into two perpendicular components. Resolving allows complex problems to be reduced to simple algebra.
Horizontal component = F cosθ | Vertical component = F sinθ
where θ is measured from the horizontal. If measured from the vertical, swap sin and cos.
Resolving on a Horizontal Surface
A force P applied at angle α above horizontal:
Horizontal component (along surface): P cosα
Vertical component (away from surface): P sinα
For a 20 N force at 30° above horizontal:
Horizontal = 20 cos30° = 20 × (√3/2) ≈ 17.3 N
Vertical = 20 sin30° = 20 × 0.5 = 10 N
Resolving on an Inclined Plane (Angle θ)
For an object on a slope of angle θ, it is most efficient to resolve parallel and perpendicular to the slope:
Derivation: Weight W = mg acts vertically. The slope makes angle θ with horizontal.
The angle between W and the perpendicular to the slope is also θ (alternate angles).
So: component along slope = mg sinθ, component perpendicular = mg cosθ.
Example: 5 kg object on 25° slope (g = 9.8 m s⁻²):
Along slope: 5 × 9.8 × sin25° ≈ 20.7 N (down the slope)
Perpendicular: 5 × 9.8 × cos25° ≈ 44.4 N (into the slope)
Multiple Forces on a Slope
When there are several forces, resolve every force along and perpendicular to the slope separately, then apply equilibrium or Newton's second law in each direction.
Object on 30° slope with tension T up the slope and friction F down the slope (moving up):
Along slope (taking up as positive): T − mg sin30° − F = ma
Perpendicular to slope: N − mg cos30° = 0 → N = mg cos30°
Then F = μN = μmg cos30° can be substituted.
Resolving an Arbitrary Force on a Slope
A force P at angle α to the slope (measured from the slope direction):
Component along slope: P cosα
Component perpendicular to slope: P sinα
A force P at angle α to the vertical (not the slope):
Draw a diagram and use the geometry of the angles — do NOT assume sin/cos without careful sketch.
Resolving with Three or More Forces
Step 1: Choose two convenient perpendicular directions (usually horizontal/vertical or along/perpendicular to slope).
Step 2: Resolve every force into both directions.
Step 3: For equilibrium: sum in each direction = 0.
Step 4: Solve the two simultaneous equations for unknowns (e.g. T and N, or T₁ and T₂).
Critical: On a slope, mg sinθ is the component along the slope and mg cosθ is perpendicular. Many students swap these. The component that tends to slide the object is along the slope — and it uses sinθ. If θ = 0° (flat), sinθ = 0 (no sliding tendency), which makes sense as a check.
Learn 5 — Triangle of Forces & Lami's Theorem
Triangle of Forces
When exactly three forces act on a particle in equilibrium, they can be drawn as three sides of a closed triangle (head to tail). This is called the triangle of forces.
Method:
1. Draw the three force vectors as arrows, placing the tail of each at the head of the previous one.
2. If the particle is in equilibrium, the three arrows form a closed triangle.
3. Use the sine rule or cosine rule on this triangle to find unknown magnitudes or directions.
Lami's Theorem
For three concurrent coplanar forces in equilibrium, Lami's theorem gives a direct relationship between each force and the angle opposite to it:
F₁/sin α₁ = F₂/sin α₂ = F₃/sin α₃
where α₁ is the angle between F₂ and F₃ (i.e. the angle opposite to F₁), and similarly for α₂ and α₃. Note: these are the angles between the force directions, not the angles of the triangle.
Important: Angle Convention
The angles in Lami's theorem are the angles between the force lines (measured as the angle between the directions of the other two forces). Each angle is typically greater than 90°. They must add up to 360°. If you measure interior angles of the triangle of forces instead, use the supplementary angle: if interior angle is β, Lami's angle is 180° − β.
When to Use Lami's Theorem
Use Lami's theorem when:
• Exactly three forces act on a particle in equilibrium
• The angles between the forces are known or can be found
• You need to find the magnitude of one or two forces
Use resolving when:
• More than three forces act, or
• The angles between forces are not directly given but can be resolved
• The forces are parallel or perpendicular to convenient axes
Lami's Theorem Example
Three forces in equilibrium: P, Q, and W = 50 N downward.
Angle between P and Q = 120° (so angle opposite W = 120°).
P makes 90° with W (angle opposite Q = 90°).
Angle opposite P = 360° − 120° − 90° = 150°.
By Lami's: 50/sin120° = P/sin90° = Q/sin150°
50/sin120° = 50/0.866 = 57.74
P = 57.74 × sin90° = 57.74 N ≈ 57.7 N
Q = 57.74 × sin150° = 57.74 × 0.5 = 28.9 N
Sine Rule on Triangle of Forces
If three forces form a triangle with sides F₁, F₂, F₃ opposite to angles A, B, C respectively:
F₁/sinA = F₂/sinB = F₃/sinC
This is simply the sine rule applied to the force triangle. The interior angles of the force triangle are supplementary to Lami's angles: interior angle = 180° − (Lami's angle).
To avoid confusion between Lami's angles and triangle angles: always draw the force triangle explicitly. Label the interior angles of the triangle. The sine rule on the triangle is straightforward. Then convert to Lami's format only if required by the question.
Worked Examples
8 fully worked examples covering all key topics. Study each step carefully before moving on.
Example 1 — Find the Resultant of Two Forces
Forces F₁ = 6i + 3j N and F₂ = −2i + 5j N act on a particle. Find the resultant and its magnitude.
Step 1 — Add components: R = (6−2)i + (3+5)j = 4i + 8j N M1
Example 3 — Object on Inclined Plane (No Friction)
A 3 kg block rests on a smooth plane inclined at 20° to the horizontal. A force P acts up the slope. Find P and the normal reaction N. (g = 9.8 m s⁻²)
Step 1 — Forces: Weight W = 3 × 9.8 = 29.4 N downward, P up the slope, N perpendicular to slope. B1
Step 2 — Resolve along slope (↑ slope positive): P − 29.4 sin20° = 0 → P = 29.4 × 0.342 ≈ 10.1 NM1 A1
Step 3 — Resolve perpendicular to slope: N − 29.4 cos20° = 0 → N = 29.4 × 0.940 ≈ 27.6 NM1 A1
Example 4 — Tension in a String Supporting a Weight
A particle of weight 40 N is supported by two strings. String 1 makes 30° with the vertical, String 2 makes 50° with the vertical. Find the tensions T₁ and T₂.
A particle is in equilibrium under three forces: W = 80 N vertically downward, T₁ at 30° to the left of vertical, T₂ horizontally to the right. Apply Lami's theorem.
Step 1 — Find Lami's angles: Angle opposite W (between T₁ and T₂): 90° + 30° = 120°. Angle opposite T₁ (between W and T₂): 90° + 90° = 180°? Recalculate from geometry. M1
Step 2 — Correct angles: T₂ is horizontal (270° from upward vertical). T₁ is at 150° from positive x-axis. W is at 270°. Angle between T₁ and W = 120°. Angle between W and T₂ = 90°. Angle between T₁ and T₂ = 150°. Sum = 360°. M1
8 mistakes that regularly cost marks in Cambridge M1. Read each carefully — recognition is the first step to avoiding them.
Mistake 1 — Normal Reaction on a Slope is NOT Vertical
WRONG: On a 40° slope, N = mg (vertical upward), so N = mg cos40° because "vertical component must balance weight".
CORRECT: N acts perpendicular to the slope surface. By resolving perpendicular to slope: N = mg cos40°. The correct value is the same here but for the right reason. On a slope, N is perpendicular to the slope, NOT vertical.
Mistake 2 — Swapping sin and cos When Resolving on a Slope
WRONG: Component of weight along slope = mg cosθ; component perpendicular = mg sinθ.
CORRECT: Component along slope (causing sliding) = mg sinθ. Component perpendicular to slope = mg cosθ. Memory check: at θ = 0° (flat), sinθ = 0 → no sliding force. This makes physical sense.
Mistake 3 — Drawing Friction in the Wrong Direction
WRONG: Object moving up a slope — friction drawn up the slope "to help".
CORRECT: Friction always opposes the motion or tendency to move. Object moving UP → friction acts DOWN the slope. Object sliding DOWN → friction acts UP the slope. Friction never assists motion.
Mistake 4 — Using Only One Equilibrium Equation
WRONG: Setting ΣFₓ = 0 and declaring the system in equilibrium without checking ΣF_y = 0.
CORRECT: Equilibrium requires BOTH ΣFₓ = 0 AND ΣF_y = 0 to hold simultaneously. One equation gives partial information only — always write both equations, even when one looks trivially satisfied.
Mistake 5 — Tension Drawn Pushing the Object
WRONG: A string attached to an object — tension arrow drawn pointing into the object (as if pushing).
CORRECT: Strings can only pull. Tension acts along the string, away from the object. If a problem gives a negative tension, the member is actually a rod in compression (thrust), not a string.
Mistake 6 — Wrong Angle Convention in Lami's Theorem
WRONG: Using the angles between forces measured as acute angles in the force triangle, instead of the angles between the force directions themselves.
CORRECT: In Lami's theorem, each angle is the angle between the directions of the other two forces (measured going around from one force to the other). These angles typically sum to 360°. Interior angles of the force triangle are supplementary to these: Lami angle = 180° − triangle interior angle.
Mistake 7 — Forgetting Weight Acts Vertically (Not Along the Slope)
WRONG: Drawing the weight arrow pointing down the slope when the object is on an inclined plane.
CORRECT: Weight is mg vertically downward — always, regardless of the orientation of the surface. You then resolve this vertical force into components along and perpendicular to the slope. The weight arrow itself is always vertical.
Mistake 8 — Sign Errors When Adding Vector Components
WRONG: Force F = 5 N at 150° — taking horizontal component as +5cos150° = +4.33 N (forgetting the negative).
CORRECT: cos150° = −√3/2 ≈ −0.866. So horizontal component = 5 × (−0.866) = −4.33 N (pointing left). Always evaluate the trig function including its sign for the given angle, or resolve the acute angle and attach the correct sign separately by inspection of direction.
Key Formulas
Vectors
Formula
Meaning
Notes
|v| = √(a² + b²)
Magnitude of vector v = ai + bj
Pythagoras in 2D
v̂ = v / |v|
Unit vector in direction of v
Magnitude = 1
θ = arctan(b/a)
Angle from positive x-axis
Check quadrant with signs
AB = OB − OA
Displacement from A to B
Always final minus initial
k(ai+bj) = kai+kbj
Scalar multiplication
Multiplies each component
Resultant and Equilibrium
Formula
Meaning
Notes
R = ΣF = ΣFₓi + ΣF_yj
Resultant force
Sum all components
|R| = √(Rₓ² + R_y²)
Magnitude of resultant
After summing components
ΣFₓ = 0 and ΣF_y = 0
Equilibrium conditions
Both must hold
θ = arctan(R_y / Rₓ)
Direction of resultant
Adjust for quadrant
Resolving Forces
Formula
Meaning
Notes
F cosθ
Component in reference direction
θ from that direction
F sinθ
Component perpendicular to reference
θ from reference direction
mg sinθ
Weight component along inclined plane
θ = angle of slope
mg cosθ
Weight component ⊥ inclined plane
θ = angle of slope
N = mg cosθ
Normal reaction on smooth incline
Smooth: no friction
Friction and Weight
Formula
Meaning
Notes
W = mg
Weight
g = 9.8 or 10 m s⁻² as given
F ≤ μN
Friction inequality
Not yet at limit
F = μN
Limiting friction
At point of slipping
μ = tanθ
Coefficient at limiting equilibrium on slope
Elegant result — tanθ = sinθ/cosθ
Lami's Theorem
F₁/sin α₁ = F₂/sin α₂ = F₃/sin α₃
For three concurrent forces in equilibrium. αᵢ is the angle between the other two forces (measured between their directions), NOT the angle in the force triangle. Angles α₁ + α₂ + α₃ = 360°.
Proof Bank
Two proofs required for deep understanding of the topic.
Proof 1 — Equilibrium Condition from Newton's First Law
Newton's First Law states: "An object remains at rest or continues to move with constant velocity unless acted upon by a resultant force."
Let forces F₁, F₂, …, Fₙ act on a particle. Decompose each into horizontal and vertical components:
Fᵢ = (Fᵢₓ)i + (F_{iy})j
The resultant is: R = (ΣFᵢₓ)i + (ΣF_{iy})j
For the particle to remain at rest (or move with constant velocity), by Newton's First Law, R = 0.
A vector is zero if and only if every component is zero:
R = 0 ⟺ ΣFᵢₓ = 0 AND ΣF_{iy} = 0
Therefore: equilibrium requires both ΣFₓ = 0 and ΣF_y = 0 simultaneously. Q.E.D.
Proof 2 — Derive mg sinθ and mg cosθ from Geometry
Consider a particle on a plane inclined at angle θ to the horizontal. Weight W = mg acts vertically downward.
Set up axes: parallel to slope (positive up the slope) and perpendicular to slope (positive away from surface).
The angle between the weight vector and the perpendicular to the slope:
The slope makes angle θ with horizontal. The normal to the slope makes angle θ with the vertical (since slope and vertical are complementary to normal and horizontal respectively — alternate angles in the geometry).
Therefore the angle between W (vertically down) and the inward normal direction is θ.
Resolving W:
Component perpendicular to slope = W cosθ = mg cosθ (since this is the projection onto the normal direction)
Component along slope (down the slope) = W sinθ = mg sinθ (perpendicular component to the above)
Check: θ = 0° (horizontal surface): mg sin0° = 0 (no force along surface ✓), mg cos0° = mg (full weight on surface ✓).
θ = 90° (vertical surface): mg sin90° = mg (weight fully along surface ✓), mg cos90° = 0 (no force into surface ✓).
Let three forces F₁, F₂, F₃ act at a point in equilibrium. Represent them as vectors from a common point. Since they are in equilibrium: F₁ + F₂ + F₃ = 0.
Rearranging: F₁ = −(F₂ + F₃). This means F₁ is equal in magnitude and opposite in direction to the resultant of F₂ and F₃.
Arrange the three forces head to tail — they form a closed triangle (triangle of forces) with sides of length |F₁|, |F₂|, |F₃|.
Let the interior angles of this triangle opposite to sides F₁, F₂, F₃ be A, B, C respectively.
By the sine rule: F₁/sinA = F₂/sinB = F₃/sinC
Now, the angle between force directions F₂ and F₃ (Lami's angle α₁) is the supplement of the interior angle A:
α₁ = 180° − A → sinα₁ = sin(180°−A) = sinA
Enter up to 4 forces with magnitude and direction (angle in degrees from positive x-axis). The diagram draws each force arrow and displays the resultant vector and magnitude.
Enter force magnitudes and angles, then click Draw Force Diagram.
8 Cambridge M1 style questions with mark schemes. Attempt before revealing.
Q1 [3 marks]
Forces F₁ = (2p − 1)i + 3j N and F₂ = 5i + (p + 2)j N act on a particle. Given that the resultant is parallel to i (horizontal), find p and the magnitude of the resultant.
Resultant parallel to i → j-component = 0: 3 + (p + 2) = 0 → p + 5 = 0 → p = −5 [M1 A1]
i-component: (2(−5)−1) + 5 = −11 + 5 = −6. |R| = 6 N [A1]
Q2 [4 marks]
A particle of mass 4 kg rests on a smooth plane inclined at 30° to the horizontal, held by a force P acting up the slope. Find P and the normal reaction N. (g = 9.8 m s⁻²)
Weight = 4 × 9.8 = 39.2 N [B1]
Along slope: P − 39.2 sin30° = 0 → P = 39.2 × 0.5 = 19.6 N [M1 A1]
Perpendicular: N = 39.2 cos30° = 39.2 × 0.866 = 33.9 N [A1]
Q3 [5 marks]
Three forces act on a particle in equilibrium: F₁ = 12 N at 0°, F₂ = F N at 120°, F₃ = G N at 240°. Find F and G.
Horizontal: 12 + F cos120° + G cos240° = 0 → 12 − 0.5F − 0.5G = 0 → F + G = 24 [M1 A1]
Vertical: F sin120° + G sin240° = 0 → F(√3/2) − G(√3/2) = 0 → F = G [M1 A1]
Substituting: 2F = 24 → F = G = 12 N [A1]
Q4 [5 marks]
A particle of weight W N is suspended by two strings. String 1 makes 40° with the horizontal and has tension T₁. String 2 is horizontal with tension T₂. Given T₁ = 30 N, find T₂ and W.
A 6 kg particle is in limiting equilibrium on a rough plane inclined at θ where tanθ = 3/4. Find the normal reaction, friction force, and coefficient of friction. (g = 10 m s⁻²)
tanθ = 3/4 → sinθ = 3/5, cosθ = 4/5 [B1]
W = 60 N. Perpendicular: N = 60 × 4/5 = 48 N [M1 A1]
Along slope: F = 60 × 3/5 = 36 N [A1]
μ = F/N = 36/48 = 0.75 (= tanθ ✓) [A1]
Q7 [4 marks]
Use Lami's theorem for a particle in equilibrium under: 50 N vertically downward, T₁ at 60° from vertical (to the left), T₂ perpendicular to T₁. Find T₁ and T₂.
Angles between forces: W and T₁ form 60° at their directions; T₁ ⊥ T₂ means 90° between them.
Angle opposite W = 360° − 60° − 90° (supplement) — use resolving instead for clarity. [M1]
Vertical: T₁ cos60° + T₂ cos90° = 50 → T₁(0.5) = 50 → T₁ = 100 N [A1 — wait, T₁ is at 60° from vertical so vertical component = T₁cos60°]
Horizontal: T₁ sin60° − T₂ = 0 → T₂ = 100 × (√3/2) = 86.6 N [M1 A1]
Q8 [6 marks]
Forces (3i + aj) N, (bi − 5j) N, and (−4i + 2j) N act on a particle in equilibrium. Find a and b. Hence find the magnitude of the force (3i + aj) N.
i-components: 3 + b − 4 = 0 → b − 1 = 0 → b = 1 [M1 A1]
j-components: a − 5 + 2 = 0 → a − 3 = 0 → a = 3 [M1 A1]
First force: 3i + 3j N; magnitude = √(9 + 9) = √18 = 3√2 ≈ 4.24 N [M1 A1]
Past Paper Questions
5 questions drawn from Cambridge A-Level 9709 Mechanics 1 past papers on Forces and Vectors. Solutions revealed on demand.
Past Paper Q1 — 9709/41/M/J/20 Q2 [5 marks]
A particle P of weight 18 N is attached to two strings. One string makes an angle of 35° with the vertical and has tension T₁ N. The other string is horizontal with tension T₂ N. Find T₁ and T₂.
Vertical equilibrium: T₁ cos35° = 18 [M1]
T₁ = 18/cos35° = 18/0.8192 ≈ 21.97 N ≈ 22.0 N [A1]
Horizontal equilibrium: T₂ = T₁ sin35° = 21.97 × 0.5736 ≈ 12.6 N [M1 A1]
Check: T₁ forms the hypotenuse of a right-angled triangle with T₂ and W — consistent ✓ [B1]
Past Paper Q2 — 9709/42/O/N/19 Q3 [6 marks]
A particle of mass 5 kg is on a rough plane inclined at 25° to the horizontal. A horizontal force of 20 N acts on the particle. The particle is in limiting equilibrium about to move up the plane. Find the coefficient of friction. (g = 10 m s⁻²)
Weight = 50 N. Forces along and perpendicular to slope: [B1]
Perpendicular: N = 50cos25° + 20sin25° = 45.32 + 8.45 = 53.77 N [M1 A1]
Along slope (up): 20cos25° − 50sin25° − F = 0 → F = 18.13 − 21.13 = −3.00?
Recalculate: 20cos25° = 18.13 N (up), 50sin25° = 21.13 N (down). Net along slope without friction = 18.13 − 21.13 = −2.99 N (down slope). About to move UP so friction acts DOWN: 20cos25° − 50sin25° − μN = 0 → μ = (20cos25°−50sin25°)/N [M1]
μ = (18.13−21.13)/53.77 → this is negative, meaning particle about to move DOWN not up. F acts up: μ = (50sin25°−20cos25°)/53.77 = 2.99/53.77 ≈ 0.0556 [A1 A1]
Past Paper Q3 — 9709/43/M/J/18 Q1 [4 marks]
Forces of magnitudes 8 N and 5 N act at a point. The angle between their directions is 60°. Find the magnitude of the resultant.
Using the cosine rule on the force triangle (angle between the forces = 60°, so interior angle of parallelogram = 120°): [M1]
|R|² = 8² + 5² + 2(8)(5)cos60° [cosine rule with included angle 60°] [M1]
= 64 + 25 + 80 × 0.5 = 64 + 25 + 40 = 129 [A1]
|R| = √129 ≈ 11.4 N [A1]
Past Paper Q4 — 9709/41/O/N/17 Q2 [5 marks]
A particle of weight W N is in equilibrium under three forces: a horizontal force of 12 N, a vertical force W N downward, and a force of 15 N at angle θ above the horizontal. Find θ and W.
A particle P of mass 3 kg is attached to one end of a light string of length 2 m. The other end is fixed at point O. P rests on a smooth inclined plane making angle 30° with the horizontal. The string makes angle 20° with the plane. Find the tension in the string and the normal reaction of the plane on P. (g = 10 m s⁻²)
Weight = 3 × 10 = 30 N downward. Three forces on P: T (along string, at 20° to plane), N (perpendicular to plane), W = 30 N downward. [B1]
Resolve perpendicular to plane: N + T sin20° − 30cos30° = 0 → N = 30cos30° − T sin20° [M1 A1]
= 30 × 0.866 − T × 0.342 = 25.98 − 0.342T
Resolve along plane (up positive): T cos20° − 30sin30° = 0 [M1]
T cos20° = 15 → T = 15/cos20° = 15/0.940 = 15.96 N ≈ 16.0 N [A1]
N = 25.98 − 0.342 × 15.96 = 25.98 − 5.46 = 20.52 N ≈ 20.5 N [M1 A1]