01 · Signed addition
Add 5i − 3j N and −2i + 7j N.
Hint
Keep each component in its own column.
Worked solution
R = 3i + 4j N; |R| = 5 N.
Understand · explore · practise
Resultant force vectors and bearings
Add force components, find a resultant magnitude and compass bearing, and distinguish an equilibrant from a third-law partner.
Before you startSigned vector components, Pythagoras, trigonometry and force balance.
01 / Add components along the same pair of axes
A vector sum preserves direction as well as size.
Resultant force R = F₁ + F₂ + …
Add all i components together and all j components together. A negative component points opposite the positive axis. Do not add magnitudes unless the force directions justify it.
Head to tail, then resultantExplore
Take i east and j north. Force A = 4i + 3j N. Choose force B. Its arrow is translated to the head of A to show vector addition; this does not mean the forces act on different bodies. The green arrow runs from the start to the final endpoint.
02 / Resolve the sum before finding its magnitude
The resultant joins the start to the final endpoint.
F₁ = 7i − 2j N and F₂ = −4i + 6j N.Worked example
R = (7 − 4)i + (−2 + 6)j = 3i + 4j N
Add components independently.
|R| = √(3² + 4²) = 5 N
Use the resultant components, not the individual magnitudes.
Both components are positive
The resultant points northeast.
Watch: head-to-tail addition produces the resultant
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Three forces
Find the resultant of 2i + j N, −5i + 4j N and 3i − 2j N.
Hint
A component may cancel to zero.
Worked solution
R = (2 − 5 + 3)i + (1 + 4 − 2)j = 3j N: 3 N north.
03 / Square signed components to find the length
A magnitude cannot be negative.
03 · Northwest resultant
Find the magnitude of −6i + 8j N.
Hint
Use Pythagoras; square the negative component.
Worked solution
|R| = √(36 + 64) = 10 N.
04 · Magnitudes are not generally additive
Two perpendicular forces have magnitudes 6 N and 8 N. Find their resultant magnitude.
Hint
Draw a right triangle.
Worked solution
√(6² + 8²) = 10 N, not 14 N. The sum 14 N would apply if both forces acted in the same direction.
05 · Opposite forces
Forces of 6 N east and 8 N west act on one particle. Find their resultant.
Hint
Use signed components.
Worked solution
R = −2i N: magnitude 2 N, direction west.
04 / Measure a bearing clockwise from north
Find the quadrant before converting an acute angle.
For i east and j north, a bearing starts at north.
For R = 3i + 4j N, tan θ = 3/4 because θ is measured from the north axis. Thus the bearing is 036.9° to 1 decimal place. Measuring anticlockwise from east would give a different angle. Use degrees for compass bearings.
06 · Northwest bearing
Find the bearing of −3i + 4j N, to 1 decimal place.
Hint
First find its acute angle west of north.
Worked solution
α = tan⁻¹(3/4) = 36.869…°. Bearing = 360° − α = 323.1°.
07 · Southeast bearing
Find the bearing of 3i − 4j N, to 1 decimal place.
Hint
Start from south, then move toward east.
Worked solution
α = tan⁻¹(3/4) = 36.869…°. Bearing = 180° − α = 143.1°.
05 / Handle axis-aligned and zero resultants explicitly
Do not divide by a zero component just to find a direction.
08 · Due south
State the magnitude and bearing of −7j N.
Hint
There is no eastward component.
Worked solution
Magnitude 7 N; bearing 180°.
09 · Due east
State the magnitude and bearing of 9i N.
Hint
Bearing is measured clockwise from north.
Worked solution
Magnitude 9 N; bearing 090°.
10 · Zero resultant
Forces are 4i + 3j N and −4i − 3j N. Find the resultant magnitude and bearing.
Hint
A zero vector has no direction.
Worked solution
R = 0, magnitude 0 N. Its bearing is undefined: do not report 000°, which would mean north.
06 / Add the negative resultant to produce equilibrium
An equilibrant acts on the same particle.
Equilibrant E = −R
If the existing forces sum to R, a further force E = −R makes their total zero. E has the same magnitude as R and the opposite direction. This is not automatically a Newton’s third-law pair: third-law partners belong to the same interaction and act on different bodies.
11 · Missing balancing force
The existing resultant is 5i − 12j N. Find the equilibrant and its magnitude.
Hint
Reverse both components.
Worked solution
E = −5i + 12j N, with magnitude √(25 + 144) = 13 N.
12 · Balancing bearing
A nonzero resultant has bearing 070°. Give the bearing of its equilibrant.
Hint
Reverse direction by 180°.
Worked solution
250°. Bearings differing by 180° point in opposite directions.
07 / Use component equations to find unknown forces
A direction condition and a magnitude condition are different.
13 · Equilibrium components
Forces are ai + 2j N, 3i + bj N and −7i − 5j N. Find a and b for equilibrium.
Hint
Set both resultant components equal to zero.
Worked solution
a + 3 − 7 = 0 gives a = 4; 2 + b − 5 = 0 gives b = 3.
14 · A direction restriction
The resultant is ki − 4j N and has magnitude 5 N. Find k if it points southeast.
Hint
Magnitude gives two roots; direction selects one.
Worked solution
k² + 16 = 25 gives k = ±3. Southeast requires positive eastward component, so k = 3. k = −3 would point southwest.
Check that every force in your sum acts on the selected particle and that all components use the same axes. Resultant direction describes acceleration direction for positive mass, not necessarily the current direction of motion.
08 / Report vector, magnitude and direction separately
A complete answer says which quantity you calculated.
Add components, calculate a nonnegative magnitude, then use the signs to choose the bearing quadrant. Treat a zero vector separately. For equilibrium, negate the resultant or set both component sums to zero.
Section 1 of 8 · Add components along the same pair of axes