01 · Three forces
Add (2, −1, 4), (−3, 5, 0) and (1, −2, −6), all in newtons.
Hint
Use three component sums.
Worked solution
R = (0, 2, −2) N; |R| = √8 = 2√2 N.
Understand · explore · practise
Forces and equilibrium in 3D
Add force vectors, apply Newton’s second law component by component, find an equilibrant and include weight with the correct vertical sign.
Before you startVector addition, magnitude, unit vectors and basic Newton’s second law.
01 / Choose axes and list every force on the particle
Resultant force is the vector sum of all forces.
R = F₁ + F₂ + …
Use one consistent coordinate system and express all forces in newtons.
Force has magnitude and direction. Adding only the magnitudes usually gives the wrong resultant. Take positive z vertically upwards when weight is present. A particle model ignores rotational effects; it is the appropriate model for the calculations here.
Include weight in the resultantExplore
Use g = 9.8 m s⁻² and positive z upwards. This particle model includes only the selected applied force and weight.
F = (0, 0, 9.8) N; W = (0, 0, −9.8) N.
R = (0, 0, 0) N.
a = (0, 0, 0) m s⁻²; |a| = 0 m s⁻².
Equilibrium: resultant zero. Velocity remains constant; it need not be zero.
Blue: applied force. Amber: weight. Green: resultant. A zero resultant has no green arrow. Diagram lengths are projected; calculations use the full 3D components.
02 / Add corresponding components with their signs
Opposing components cancel algebraically.
F₁ = (5, −2, 4) N and F₂ = (−1, 3, −6) N.Worked example
R = (5 − 1, −2 + 3, 4 − 6) = (4, 1, −2) N
Add each coordinate separately.
|R| = √(4² + 1² + (−2)²) = √21 N
Find the magnitude after adding vectors.
The z component is negative
The resultant has a downward component if z is upwards.
02 · Opposite forces
Two forces (3, 4, 0) N and (−3, −4, 0) N act. Is the resultant magnitude 10 N?
Hint
Add the vectors before taking magnitude.
Worked solution
No. R = 0 and |R| = 0 N. Their individual magnitudes are 5 N, but they cancel.
03 / Divide the resultant by the mass
R = ma is a vector equation.
a = R/m
Mass m is positive and measured in kg; acceleration is in m s⁻².
A 2 kg particle has resultant (4, 1, −2) N.Worked example
a = (2, 1/2, −1) m s⁻²
Divide every component by 2.
|a| = √21/2 m s⁻²
The acceleration magnitude is |R|/m.
a points in the same direction as R
Positive mass changes scale, not direction.
03 · Acceleration
A 4 kg particle has resultant (8, −12, 4) N. Find a and |a|.
Hint
Divide by 4, then use the magnitude formula.
Worked solution
a = (2, −3, 1) m s⁻²; |a| = √14 m s⁻².
04 · Required resultant
A 3 kg particle has acceleration (−1, 2, 2) m s⁻². Find its resultant force and magnitude.
Hint
Multiply by the mass.
Worked solution
R = (−3, 6, 6) N; |R| = 9 N.
04 / Equilibrium requires all resultant components to vanish
Zero acceleration permits constant nonzero velocity.
R = 0 ⇔ a = 0
For a particle of positive constant mass in an inertial frame.
F₁ = (3, −2, 5) N and F₂ = (−1, 4, −3) N. Find F₃ for equilibrium.Worked example
F₁ + F₂ = (2, 2, 2) N
First add known forces.
F₃ = −(F₁ + F₂) = (−2, −2, −2) N
The equilibrant is the negative of their resultant.
Total = (0, 0, 0) N
Check every component.
Watch: cancel the resultant
Pause, replay or seek freely. The notes explain the same idea and stay in view.
05 · Equilibrant
Find the equilibrant of (4, −1, 2) N and (−2, −3, 1) N.
Hint
Negate their sum.
Worked solution
The sum is (2, −4, 3) N, so the equilibrant is (−2, 4, −3) N.
06 · Moving equilibrium
A particle moves with constant velocity (5, 0, −1) m s⁻¹. What is its resultant force?
Hint
Acceleration is the rate of change of velocity.
Worked solution
Zero. Constant velocity has zero acceleration, even though the particle is moving.
05 / Use the required acceleration to find an unknown force
The missing force is ma minus the known force sum.
A 2 kg particle has acceleration (1, −2, 3) m s⁻². One force is (5, 1, −2) N; find the other force G.Worked example
R = ma = (2, −4, 6) N
Find the required resultant.
G = R − (5, 1, −2)
Subtract the known force.
G = (−3, −5, 8) N
Add back to check the resultant.
07 · Unknown force
m = 3 kg, a = (2, 0, −1) m s⁻². Known forces sum to (1, 4, 2) N. Find the missing force.
Hint
Compute ma − Fknown.
Worked solution
(6, 0, −3) − (1, 4, 2) = (5, −4, −5) N.
06 / Removing a balancing force leaves its negative as the resultant
Use the equilibrium equation before changing the model.
Three forces F₁, F₂, F₃ balance. F₃ = (−4, 2, −6) N is removed from a 2 kg particle.Worked example
F₁ + F₂ = −F₃ = (4, −2, 6) N
The remaining forces no longer sum to zero.
a = (2, −1, 3) m s⁻²
Divide the new resultant by 2.
It accelerates opposite to the removed force
Do not retain the removed force in the new sum.
08 · Remove one force
A 5 kg particle is in equilibrium. A force (10, −5, 0) N is removed. Find the new acceleration.
Hint
The remaining resultant is the negative of the removed force.
Worked solution
a = (−10, 5, 0)/5 = (−2, 1, 0) m s⁻².
07 / Weight belongs in the vertical force component
With positive z upwards, W = −mgk.
W = (0, 0, −mg)
Use the stated value of g. Weight is a force, not the mass.
A 2 kg particle is acted on by applied force (4, −3, 24.6) N and its weight. Use g = 9.8 m s⁻².Worked example
W = (0, 0, −19.6) N
Weight acts vertically down.
R = (4, −3, 5) N
Include both forces.
a = (2, −1.5, 2.5) m s⁻²
The vertical acceleration is upwards.
09 · Vertical balance
What upward force balances the weight of a 3 kg particle when g = 9.8?
Hint
The magnitude is mg.
Worked solution
29.4 N upwards, or (0, 0, 29.4) N.
10 · Gravity alone
For a freely falling particle under weight alone, find a with positive z upwards.
Hint
Divide (0,0,−mg) by m.
Worked solution
a = (0, 0, −g) m s⁻², independent of mass in this model.
08 / A specified direction must be converted to a unit vector
A force of magnitude T along d is T d/|d|.
A 15 N force acts in the direction (2, −1, 2).Worked example
|d| = √(4 + 1 + 4) = 3
Normalise the direction.
Unit direction = (2/3, −1/3, 2/3)
It has magnitude one.
F = (10, −5, 10) N
Its magnitude is 15 N.
11 · Force from direction
A 20 N force acts in direction (0, 3, 4). Find its vector.
Hint
The direction magnitude is 5.
Worked solution
F = 20(0,3/5,4/5) = (0,12,16) N.
12 · Opposite direction
Find a 6 N force acting opposite to (1, 2, 2).
Hint
Normalise, then negate.
Worked solution
F = −6(1,2,2)/3 = (−2,−4,−4) N.
09 / An unknown mass or parameter must satisfy every component
One equation produces a candidate; the others test it.
A particle has R = (6, −3, 9) N and a = (2, −1, 3) m s⁻².Worked example
6 = 2m gives m = 3 kg
Use a nonzero component.
−3 = −m and 9 = 3m both hold
All components give the same positive mass.
The data are consistent
A contradictory component would rule out the model.
13 · Inconsistent data
Can R = (6, −3, 8) N and a = (2, −1, 3) m s⁻² correspond to one mass?
Hint
The first component gives m = 3.
Worked solution
No. Then the third force component would be 9 N, not 8 N.
14 · Balance parameters
Forces (p, 2, −1), (3, q, 4), (−5, −6, r) N balance. Find p, q, r.
Hint
Set each component sum equal to zero.
Worked solution
p = 2, q = 4, r = −3.
10 / Acceleration direction does not determine current motion direction
Use velocity to decide whether the particle is ascending.
At an instant v = (0, 0, −5) m s⁻¹ and a = (0, 0, 2) m s⁻².Worked example
The vertical velocity is negative
The particle is descending at this instant.
The vertical acceleration is positive
Its downward speed is decreasing.
Upward resultant does not mean upward motion
Force determines acceleration, not velocity directly.
15 · Initially level flight
An aircraft has zero initial vertical velocity and constant positive vertical acceleration. What happens immediately afterwards?
Hint
Use vz = uz + azt with uz = 0.
Worked solution
For t > 0, vz > 0, so it begins ascending. If initial vertical velocity were negative, positive acceleration alone would not justify this conclusion.
11 / Keep the force sum, mass and units visible
Calculate the resultant before the acceleration.
- Define the positive axes.
- List all forces, including weight if relevant.
- Add vectors component by component.
- Use R = ma with positive mass.
- Check every component of unknown-parameter solutions.
- Equilibrium means zero acceleration, not necessarily rest.
- Use velocity to interpret ascent or descent.
16 · Final check
A 2 kg particle has applied force (6, 0, 19.6) N and weight. Use g = 9.8. Find its acceleration.
Hint
First subtract 19.6 N in the vertical component.
Worked solution
R = (6,0,0) N; a = (3,0,0) m s⁻².
Section 1 of 11 · Choose axes and list every force on the particle