Grade 11 Physical Sciences · Mechanics

Find resultant of perpendicular vectors

Find the size and direction of the resultant of two vectors at right angles using Pythagoras and trig (3 N and 4 N give 5 N).

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Question WE10 · Siyavula Grade 11 Physical Sciences, 1.2 Resultant of perpendicular vectors, Worked examples 10 and 11 · 4 marks

WE10 A force of

40

N in the positive

x

-direction acts simultaneously (at the same time) to a force of

30

N in the positive

y

-direction. Calculate the magnitude and the direction of the resultant force.(4)

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

More past paper questions: find resultant of perpendicular vectors

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3. Question 7.2.3 · May/June 2025 · Paper 1 · National (DBE) · 5 marks
7.2 Sphere S is attached to the ceiling at point P by a light inextensible string. When sphere R, on an insulated stand, is brought close to sphere S they repel and sphere S comes to rest with the horizontal distance between the centres of the spheres equal to $0, 03$ m.

Sphere R is directly below point P. The string makes an angle θ with the vertical, as shown in the diagram below.

Answer the questions that follow.
Sphere R has a charge of $+ 6$ nC. Sphere S has a charge $+ 3$ nC and a mass of $0, 15$ g.
7.2.3 Calculate angle θ.(5)
Show the memo (5 ticks)Memo
1. ✓
  Formula $F = \frac{k Q _{1} Q _{2}}{r ^{2}}$
  1 mark · Apply Coulomb's law
2. ✓
  Substitution $F = \frac{( 9 \times 1 0 ^{9} ) ( 6 \times 1 0 ^{- 9} ) ( 3 \times 1 0 ^{- 9} )}{( 0 , 03 ) ^{2}}$
  $= 1, 8 \times 1 0^{- 4}$ N
  1 mark · Apply Coulomb's law
3. ✓
  Weight of S $w = m g = (1, 5 \times 1 0^{- 4}) (9, 8) = 1, 47 \times 1 0^{- 3}$ N
  1 mark · Calculate weight
4. ✓
  Trig ratio $tan θ = \frac{F _{E}}{m g} = \frac{1 , 8 \times 1 0 ^{- 4}}{1 , 47 \times 1 0 ^{- 3}}$
  1 mark · Find resultant of perpendicular vectors
5. ✓
  Answer $θ = 6, 98°$
  1 mark · Find resultant of perpendicular vectors
4. Question 7.2.2 · May/June 2024 · Paper 1 · National (DBE) · 6 marks
7.2 Two identical neutral polystyrene balls A and B are suspended from a ceiling by insulated, light inextensible strings of equal length, as shown in the diagram below.

Ball B is then given an initial negative charge, $Q_{B}$ , of unknown magnitude. The balls attract each other, touch and then repel each other. The balls come to rest with their centres $10$ cm apart so that each string makes an angle of $9°$ with the vertical. See the diagrams below.

Answer the questions that follow.
7.2.2 Calculate the magnitude of the initial charge $Q_{B}$ given to ball B if the mass of each ball was $0, 012$ kg.(6)
Show the memo (6 ticks)Memo
1. ✓
  Correct trig expression equated to $F_{E}$ $F_{E} = w tan θ$
  
  Any one: $F_{E} = m g tan θ$ or $F_{E} = \frac{w}{t a n ( 90° - θ )}$
  1 mark · Find resultant of perpendicular vectors
2. ✓
  Substitution of mg with angle $9°$ $F_{E} = (0, 012) (9, 8) (tan 9°)$
  $= 0, 0186$ N
  1 mark · Find resultant of perpendicular vectors
3. ✓
  Coulomb's formula $F_{E} = \frac{k Q _{A} Q _{B}}{r ^{2}}$
  1 mark · Apply Coulomb's law
4. ✓
  Substitution into Coulomb's formula $0, 0186 = \frac{( 9 \times 1 0 ^{9} ) ( Q ^{2} )}{0 , 1 ^{2}}$
  $Q = 1, 44 \times 1 0^{- 7}$ C
  1 mark · Apply Coulomb's law
5. ✓
  Calculating 2Q $Q_{B} = 2 (1, 44 \times 1 0^{- 7})$
  1 mark · Calculate shared charge and electrons transferred
6. ✓
  Final answer $Q_{B} = 2, 88 \times 1 0^{- 7}$ C
  1 mark · Calculate shared charge and electrons transferred
5. Question WE8 · Siyavula Grade 11 Physical Sciences, 1.2 Resultant of perpendicular vectors, Worked example 8 · 4 marks
WE8 A number of tugboats are trying to manoeuvre a submarine in the harbour but they are not working as a team. Each tugboat is exerting a different force on the submarine. Given the following force vectors, determine the resultant force:
- $F_{1} = 3, 4$ kN in the positive $x$ -direction
- $F_{2} = 4000$ N in the positive $y$ -direction
- $F_{3} = 300$ N in the negative $y$ -direction
- $F_{4} = 7$ kN in the negative $y$ -direction(4)
Show the memo (4 ticks)Memo
1. ✓
  Consistent units $F_{2} = 4000 \times 1 0^{- 3}$ kN $= 4$ kN
  $F_{3} = 300 \times 1 0^{- 3}$ kN $= 0, 3$ kN
  1 mark · Convert units
2. ✓
  Resultant in the y-directionPositive $y$ up: $R_{y} = 4 - 0, 3 - 7 = - 3, 3$ kN
  $R_{y} = 3, 3$ kN in the negative $y$ -direction; $R_{x} = F_{1} = 3, 4$ kN in the positive $x$ -direction
  1 mark · Find a resultant in one dimension
3. ✓
  Magnitude of the resultantDraw $R_{x}$ and $R_{y}$ head-to-tail (scale $1$ kN : $1$ cm) and measure the resultant.
  $R = 4, 7$ kN
  1 mark · Find resultant of perpendicular vectors
4. ✓
  DirectionThe angle with the $x$ -axis is $44°$ ( $tan^{- 1} \frac{3 , 3}{3 , 4} = 44°$ ).
  $R$ is $4, 7$ kN at $- 44°$ from the positive $x$ -direction.
  1 mark · Find resultant of perpendicular vectors
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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