Grade 11 Physical Sciences · Mechanics

Resolve a vector into components

Split a force at an angle into horizontal and vertical parts, for example F cos θ and F sin θ, or mg sin θ along a slope.

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Vectors in 2D grade 11: Resolving a vector into perpendicular and parallel componentsMiss Martins Maths and Science

2) Component method gr 11 | Part 1Kevinmathscience

Grade 11 and 12 Forces: Weight and components of WeightMiss Martins Maths and Science

Grade 11 Newton Laws: Objects on a slopeKevinmathscience

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Question WE12 · Siyavula Grade 11 Physical Sciences, 1.3 Components of vectors, Worked example 12 · 2 marks

WE12 A force of

250

N acts at an angle of

30°

to the positive

x

-axis. Resolve this force into components parallel to the

x

- and

y

-axes.(2)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: resolve a vector into components

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3. Question 2.3.1 · November 2025 · Paper 1 · National (DBE) · 2 marks
A block of mass $5$ kg is at rest on a rough horizontal surface. When a constant force F of magnitude $18$ N acts on the block at an angle $θ$ to the horizontal, the block experiences maximum static friction. See the diagram below.
2.3 The horizontal component of force F is $15$ N.

Calculate:
2.3.1 $θ$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Substitution $F_{x} = F cos θ$
  $15 = 18 cos θ$
  1 mark · Resolve a vector into components
2. ✓
  Answer $θ = 33, 56°$
  1 mark · Resolve a vector into components
4. Question WE13 · Siyavula Grade 11 Physical Sciences, 1.3 Components of vectors, Worked example 13 · 2 marks
WE13 A force of $12, 5$ N acts at an angle of $230°$ to the positive $x$ -axis. Resolve this force into components parallel to the $x$ - and $y$ -axes.(2)
Show the memo (2 ticks)Memo
1. ✓
  y-component $F_{y} = 12, 5 sin (230°) = - 9, 58$ N, so $F_{y}$ is $9, 58$ N in the negative $y$ -direction
  1 mark · Resolve a vector into components
2. ✓
  x-component $F_{x} = 12, 5 cos (230°) = - 8, 03$ N, so $F_{x}$ is $8, 03$ N in the negative $x$ -direction
  1 mark · Resolve a vector into components
5. Question WE12 · Siyavula Grade 11 Physical Sciences, 2.3 Newton's laws, Worked example 12 · 3 marks
WE12 A man is pulling a $20$ kg box with a rope that makes an angle of $60°$ with the horizontal. If he applies a force of magnitude $150$ N and a frictional force of magnitude $15$ N is present, calculate the acceleration of the box.(3)
Show the memo (3 ticks)Memo
1. ✓
  Horizontal component of the applied forceRight is positive. Only forces parallel to the motion count.
  $F_{x} = F_{a ppl i e d} cos (θ) = 150 cos (60°) = 75$ N
  1 mark · Resolve a vector into components
2. ✓
  Fnet = ma substitution $F_{R} = ma$
  $F_{x} + F_{f} = (20) a$
  $(75) + (- 15) = (20) a$
  1 mark · Apply Newton's second law
3. ✓
  Answer with unit and direction $a = \frac{60}{20} = 3$ m·s $^{- 2}$ to the right
  1 mark · Apply Newton's second law
6. Question 2.2.1 · November 2025 · Paper 1 · Provincial · 9 marks
Three forces, $F_{1}$ , $F_{2}$ and $F_{3}$ , NOT DRAWN TO SCALE, are applied to a crate of $25$ kg as shown in the diagram below.
2.2 Calculate the:
2.2.1 Magnitude and direction of the resultant of the three forces(9)
Show the memo (9 ticks)Memo
1. ✓
  x-component of F1 $F_{1 x} = 700 cos 60°$ (or $700 sin 30°$ ) $= 350$ N right
  1 mark · Resolve a vector into components
2. ✓
  x-component of F2 $F_{2 x} = 800 cos 40°$ (or $800 sin 50°$ ) $= 612, 84$ N right
  1 mark · Resolve a vector into components
3. ✓
  Net x-component $F_{3 x} = 650$ N left
  $F_{x} = 350 + 612, 84 - 650 = 312, 84$ N right
  1 mark · Find a resultant in one dimension
4. ✓
  y-component of F1 $F_{1 y} = 700 sin 60°$ (or $700 cos 30°$ ) $= 606, 22$ N upward
  1 mark · Resolve a vector into components
5. ✓
  y-component of F2 $F_{2 y} = 800 sin 40°$ (or $800 cos 50°$ ) $= 514, 23$ N downward
  1 mark · Resolve a vector into components
6. ✓
  Net y-component $F_{3 y} = 0$ N
  $F_{y} = 606, 22 - 514, 23 = 91, 99$ N upward
  1 mark · Find a resultant in one dimension
7. ✓
  Magnitude of the resultant $F_{n e t}^{2} = F_{y}^{2} + F_{x}^{2}$
  $F_{n e t}^{2} = (312, 84)^{2} + (91, 99)^{2}$
  $F_{n e t} = 326, 08$ N
  1 mark · Find resultant of perpendicular vectors
8. ✓
  Angle $tan θ = \frac{91 , 99}{312 , 84}$
  1 mark · Find resultant of perpendicular vectors
9. ✓
  Direction of the resultant $θ = 16, 39°$ above the positive x-axis
  Direction (memo): $73, 7°$
  1 mark · Find resultant of perpendicular vectors

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