Forces 3. Problem solving  calculating resultant forces using graphs and vector diagrams
Doc Brown's
school physics revision notes: GCSE physics, IGCSE physics, O level
physics, ~US grades 8, 9 and 10 school science courses or equivalent for ~1416 year old
students of physics
This page will answer questions such as ...
What is a resultant force? Why is it a
vector?
How do we draw scale diagrams to deduce
a resultant force?
What do we mean by balanced and
unbalanced forces?
When dealing with acting forces, what
is an equilibrium?
See also
What are contact forces &
noncontact forces?, scalar & vector quantities, free body force diagrams
Introduction to
resultant force problem solving calculations
Forces were introduced in "What is a force?"
including contact and noncontact forces AND, importantly for this page, free
body diagrams showing multiple forces on an object.
Force data is useless without its direction
of action.
You not only need to know the value of a force in newtons (N) but the
precise direction or angle of the line of action of one force with respect to at
least one other force.
That is why force is always a vector  it
has magnitude and direction!
When a body is subjected to multiple known
forces (usually >= vectors in newtons) how can we deduce and calculate the net
resultant force and its direction?
It is possible to replace multiple forces
acting at a single point with a single force known as the resultant force.
Some resultant forces are easily deduced with
a simple addition or subtraction.
For example image an object of weight 12
N falling vertically in air.
If the air resistance opposing the fall was 5
N, what would be the resultant force? Quite simply it would be 12  5 = 7 N.
So the object would continue to fall and
accelerate due to the 7 N resultant force.
Other more complicated situations require a scale drawing showing all
the forces involved and the direction (e.g. angle) in which each individual
force acts.
From such graph work you can measure and
calculate the resultant force and its direction of action.
If the resultant force is zero,
this is described as an equilibrium situation  a position of
balance.
Examples of using graphical scale
drawings to determine a
resultant force and its direction and to test for an equilibrium situation are
explained below.
Example 1.
Two forces at 90^{o} to each other (left scaled diagram)
Suppose two forces act on an object, a 25
N force in the north direction, and 20 N force at 90^{o} to the
east.
In this case, using a scale of 2 mm = 1
N, you draw vertical line 50 mm long and connect to it a horizontal line of
40 mm.
The two lines form half a rectangle, so,
to get the resultant line, imagine the other half of the rectangle (or draw
it in faintly) and draw the resultant line diagonally across the rectangle.
In this case I made it 71 mm long, giving
a resultant force of
35.5
N.
With a protractor you can then measure
the angle of the resultant force, which is
43^{o}
from the north.
Example 2.
Three forces acting on an object (right scaled diagram)
This is a bit more tricky.
Here an object is subjected to a 200 N
force to the north, a 300 N force to the east, and a 100 N force to the
south.
This is illustrated by the 'not to scale'
sketch on the side.
You might need to draw a starting sketch
like this if all the information comes as text only!
Using a scale of 1.0 cm = 100 N.
You draw a vertical line 4.0 cm long in
the north direction.
Then at top of this line draw a
horizontal line 6.0 cm long for the east force of 300 N.
OK, so far, as in example 1., but, now
draw the line of the 3rd (south force) at the end of the east force line.
You now draw the resultant diagonal line
from the bottom of the north line to the bottom of the south line.
(You should note the reduction of the
north force is 100 N because of the opposing 100 N south force).
The resultant diagonal line is 6.7 cm
long, which equals
335 N.
With a protractor you should get a
measured angle of
7071^{o}
from the north
(I calculated it to be 70.5^{o},
using trigonometry  see
APPENDIX 1, which is
another way of solving these problems if it can be reduced to a right angled
triangle situation).
Example 3. An equilibrium or 'state of
balance' situation
In
example 3 I want you to imagine the 300 N force acting in the opposite
direction to two other forces, 225 N and 140 N at an angle to each other.
In this case there is NO net
resultant force (other than zero!). The object is then described as being in a state of
equilibrium (or balanced) and would remain stationary.
This illustrates a method of deducing a force needed
to produce an equilibrium situation involving three forces e.g. you
might be given two and have to work out the 3rd force needed to balance
the other two forces.
Example 4. Pulling an object along (but watch out for forces that
don't count!)
Imagine a car of mass 800 kg being towed
by a breakdown truck.
The tension in the towing cable is 500 N
at the angle indicated in the diagram.
The weight of the car will be 800 x 10 =
8000 N (gravity is ~10 N/kg).
Resolve the towing force into its
vertical and horizontal components.
The graphical drawing uses a scale of 54
mm = 500 N.
This comes out of the way I've done the
diagram i.e. I didn't choose a more convenient scale!
Scaling down for the vertical force
component of length 20 mm: 500 x 20 / 54 =
185 N
Scaling down for the horizontal force
component of length 50 mm: 500 x 50 / 54 =
463 N (in the towing
direction)
Note that the weight of the car
acting in a vertical direction due to gravity does NOT affect these force calculations.
Forces that act at 90^{o} to the
direction of motion, do not contribute to the horizontal force you are calculating here
in the direction of motion.
Unless the front of the car is
physically lifted off the ground, it cannot affect the vertical component
force.
Example 5.
Multiple forces acting on an object  is the resultant force zero (equilibrium)
or otherwise
5a. An equilibrium situation, a
state of balance with no resultant force other than zero from multiple
forces
Consider four forces of 31, 36, 22 and 28
N, all acting on the same object and at four different directions.
This is shown as the left vector diagram.
To check if there is a nonzero resultant
force, you draw all the forces tiptotail to create a loop  note all the
forces are drawn clockwise to fit in with the force arrow direction.
This is the right vector diagram.
Here there is a complete loop i.e.
you end up at the starting point, this means there is no effective
resultant force and the object is in a state of equilibrium.
5b. A nonzero resultant force from
multiple forces acting on an object, not an equilibrium
Consider four forces of 32, 42, 45 and 36
N, all acting on the same object and at four different directions.
This is shown as the left vector diagram
(scale 1 mm = 1 N, 2 mm squares, 2 mm = 2 N).
Again, to check if there is a nonzero
resultant force, you draw all the forces tiptotail to create a loop  note
all the forces are drawn clockwise to fit in with the force arrow
direction.
This is the right vector diagram. There
is actually a resultant force of 10 N (10 mm) to the horizontal right.
Here there is NO complete loop
i.e. you do NOT end up at the starting point, this means there is an
effective resultant force and the object is NOT in a state of
equilibrium.
Example 6.
Examples of problem
solving to calculate a resultant force using graphs or other vector diagrams
See also
work done
calculations in Appendix 2.
Q1.
Two forces acting in the same direction (parallel) on an object
What is the resultant force on the object?
In this case you simply add up the two forces acting
from left to right:
resultant force = force 1 + force 2 = 70 N + 45 N
= 115 N
So the object experiences an acting force of
115 N to the right (vector!). 
Q2.
Two forces acting in parallel but opposite directions
What is the resultant force on the object?
Above you added the forces, but here, you subtract
one from the other.
resultant force = force 1  force 2 = 780  330 =
450 N to the right
Since the force to the right is greater than the
opposing force from the left, the net force (resultant force of 330 N)
must act from left to right. 
Q3.
Multiple force acting on an object Given
the free body diagram, what is the magnitude and direction of the
resultant force.
This
situation is not quite as complicated as it seems because acting forces 2
and 3 are equal and cancel each other out and are at 90^{o} to the line of
action of the resultant force on which they have no effect (*). Whatever the motion of the object, it will
not rise or fall.
Therefore the situation is actually the same as
example Q2. above.
resultant force = force 1  force 4 = 120  55 =
65 N horizontally to the right This
free body diagram could represent a moving vehicle.
(*) Forces that act at 90^{o} to the
direction of motion, do not contribute to the resultant force you are calculating here. 
Q4.
Resolving two forces that are not parallel
Imagine an object is subjected to a northward force
of 90 N and a due east force of 60 N.
Deduce the magnitude of the resultant force and
its angle with respect to due north.
Using graph paper and a suitable scale, you draw
the two forces out at 90^{o} to each other in the manner shown in the left
diagram ('tip to tip')  which produces a triangle when you join up the start of the
north force to the right hand tip of the east force.
You then measure the length of the hypotenuse and
from the chosen scale you get 7.2 cm which translates to 7.2 x 15 =
108 N
Using a protractor you can then measure the angle
which I found to be 33^{o} (033^{o}) with
respect to north. I made both measurements as a student would in class
using 2 mm graph paper.
Solving
Q4. using trigonometric calculations (see
APPENDIX 1):
In GCSE exams you will have to solve it by the
graphical method described above, but you can solve it just from a
simple nonscale sketch by trigonometry using the known direction of
the two forces and the angle between them. This is really advanced A level maths using a scientific
calculator, but I decided to check my own 'honest' graph work just using
the forces and directions given and ignoring the graph completely.
tan θ = O / A = 60/90 = 0.667, tan^{1}(0.667)
= 33.7^{o} (so I made a small manual graph error of 0.7^{o})
You can work out the magnitude of the resultant
force using the accurate angle above and either of the sine or cosine
rule equations e.g. if R = resultant force (= H)
sin θ = O / H = sin (33.7^{o}) = 60/R =
0.555, R = 60/0.555 =
108 N
cos θ = A / H = cos (33.7^{o}) = 90/R =
0.832, R = 90/0.832 =
108 N
So my resultant force measurement was spot on!

Q5.
Resolving a forces into two forces at right angles to each other
Imagine a force of 156 N acting at an angle
of 51^{o} from the vertical 'north'.
Q Deduce the component vectors for due north and due
east.
You draw the resultant force line to scale at 39^{o} (90^{o}
 51^{o}) from the horizontal.
Use a convenient triangle e.g. in this case the
hypotenuse of the triangle is 7.8 cm. and the other two sides are 6.0 to
the east and 5.0 cm to the north.
I've then chosen a scale of 20.0
N/cm on 2 mm graph paper.
( Note from the scale chosen 7.8 x 20.0 = 156 N)
You can then split this 'resultant' force into two
components by draw lines for the northerly contribution
(5.0 cm) and the easterly contribution (6.0 cm). The convenient numbers
of 5.0 and 6.0 was pure coincidence!
From the scale this gives the two contributing
forces of:
5.0 x 20.0 =
100 N to the north and 6.0
x 20.0 = 120 N to the east.
A towed
object situation  just an extra little though for certain
kinds of problems you might find in an exam
If for example, this was a towing situation,
i.e. an object is being towed from left to right with the cable
acting upwards at an angle of 39^{o}, you can calculate the
effective horizontal towing force i.e. 120 N.
If this was a car, its weight acting
vertically downwards, does not affect the calculation.
Similarly, the weight of a barge being towed
at an angle from a canal bank, does not affect the calculation.
Again you can work out the answers from
trigonometry just using the angle of 39^{o} and the force of 156
N (the original information): Let N be the north force and E be the east
force
sin θ = O / H = sin (39^{o}) = 0.629 =
N/156, N = 0.629 x 156 =
98.1 N to the north
cos θ = A / H = cos (39^{o}) = 0.777
= E/156, E = 0.777 x 156 =
121.2 N to the east
So I'd made a 12% error on my graph work, and
note the trigonometrical calculations are absolutely correct based on
the original information. This level of calculation is covered by GCSE maths
courses. 
Q6.
Variations on two forces.
Situation A: A pulling force of 225 N is acting on
an object at an angle of 27^{o} to the horizontal. At the same
time it is also subjected to another pulling force at 45^{o} to
the horizontal (as shown in the diagram).
With graph paper, protractor and ruler, draw a scale diagram to deduce the magnitude and
direction of the resultant force. The graph is drawn to a scale of 1 mm
= 5 N.
This can be solved using the principle of 'parallelogram
of forces'. Its quite simple to do. The dotted line are drawn
parallel to the two known vectors and where they intersect gives you the
length of the resultant force of 60 mm.
Therefore the resultant force is 60 x 5 =
300 N horizontally to the right.
However, you can avoid the parallelogram of forces
diagram by drawing the force diagram in a simpler way as shown in
diagram B.
You draw the first force of 225 N at 27^{o}
to the horizontal (45 mm) and from its top tip draw the 2nd force
downwards at 45^{o} to the horizontal line (28 mm).
The horizontal line of 60 mm gives you the same
resultant force of
300 N.
In
situation C I want you to imagine the 300 N force acting in the opposite
direction. In this case there is NO net
resultant force (other than zero!). The object is then described as being in a state of
equilibrium (or balanced) and would remain stationary.
This illustrates a method of deducing a force needed
to produce an equilibrium situation involving three forces e.g. you
might be given two and have to work out the 3rd force needed to balance
the other two forces. more examples needed? 
Q7.
A hanging weight, tensioned horizontally
A mass of 20 kg is suspended from a beam by a wire at
angle of 32^{o} away from the vertical (see diagram). The
tension in this wire is T2. The same 20 kg weight is pulled to one side
by a horizontal string with a tension T2.
Using a graph calculate the tension force in (a)
the wire (T1) and (b) in the string (T2).
The triangle of forces is drawn on 1 mm graph paper.
Tension T3 is the weight of the object = 196 N (20 x 9.8, g = 9.8 N/kg).
This is the vertical force.
The diagonal side of the triangle is tension T1 in
the wire holding up the weight at an angle of 32^{o}.
The horizontal side of the triangle is the tension
T2 in the string pulling from the left.
(a) The graph is drawn to a scale of 2 N per mm.
The diagonal (hypotenuse) was found to be 115 mm long.
From the scale tension T1 = 115 x 2 =
230 N
(b) The opposite side of the triangle = 60 mm,
this equates to 60 x 2 = 120 N for tension T2 in the string
Trigonometric calculation check (perfect answers!):
Tension T1: cos (32^{o}) = adjacent /
hypotenuse = 0.848 = T3/T1 = 196/T1, so T1 = 196/0.848 =
231.1
N Tension T2: tan (32^{o})
= opposite / adjacent = 0.625 = T2/T3 = T2/196, so T2 = 0.625 x 196
= 122.5 N So, I think my
graph work was pretty good and an 'ok' with only a 0.5% to 2% error!

Q8.
Resolving two converging forces Two
forces, 5.0 N and 6.0 N, act on an object at an angle of 60^{o}
between the lines of action (as in the diagram).
Calculate the resultant force on the object O at
point o.
Using the principle of the parallelogram of
forces, and focus on the left of the diagram: If you draw the lines ab
(6.0 cm) parallel to action line oc and line bc (5.0 cm) parallel to
action line ao, then the diagonal of 9.5 cm gives you the resultant
distance which equates to a resultant force of 9.5 N (1 cm
= 1 N).
However, drawing true parallel lines is awkward,
so you can deduce the answer without graph paper and on plain white
paper with just a mm ruler and protractor. If you extend the line ao a
distance of 5 cm giving line od, you get exactly the same resultant
distance of 9.5 cm = resultant force of 9.5 N by joining up the relevant
tip to tip points cd.
Its simple pure geometric logic, no problem! Just
compare situations A and B in Q6. 
Q9.
Calculating a 3rd force required to establish an equilibrium
Two wires with pulling force tensions of 20 N and 24
N are pulling on a metal ring. If the angle between the lines of action
is 70^{o} calculate the force needed on the 3rd wire on the
right, to stabilise the ring to give an equilibrium situation.
Using the principle of the parallelogram of
forces, and focus on the left of the diagram: If you draw the lines ab
(6.0 cm) parallel to action line oc and line bc (5.0 cm) parallel to
action line ao, then the diagonal of 9.0 cm gives you the resultant
distance. BUT, the direction of action is opposite to that in Q8, so
this equates to a resultant force of 36.0 N (1 cm = 14N)
and the arrow points towards the right balancing and opposing the forces
acting to the left.
As in Q8, drawing true parallel lines is awkward,
so you can deduce the answer without graph paper and on plain white
paper with just a mm ruler and protractor. If you extend the line ao a
distance of 5 cm giving line od, you get exactly the same resultant
distance of 9.0 cm = resultant force of 9.0 x 4 =36 N by joining
up the relevant tip to tip points cd.
Note that although the total force to the left is
44 N, the opposing force is less (36 N), because the forces are not
acting parallel to each other. 
Q10.
A push and pull grass roller! A roller
of mass 80 kg is pushed or pulled with a force of 300 N acting at an
angle of 45^{o} to the horizontal grass surface.
Ignore friction effects in this question.
(a)
Calculate the pulling force in
the horizontal direction.
The scale drawing on the right shows you how
to obtain the horizontal pulling force.
At 45^{o} on the graph paper, the
diagonal line for 300 N equals 56.5 mm.
The horizontal line has a length of 40 mm, so
scaling down, gives ..
300 x 40 / 56.5 =
212 N
Note that, neglecting friction effects,
the downward force of 800 N of the weight of the roller has no
effect on the calculation.
This is because it acts at 90^{o}
to the direction of the motion and doesn't contribute to the
resultant force you are calculating here.
(b) Calculate the vertical force of the roller on the
grass when it is pushed.
You first need to calculate the vertical force
involved due to the person pushing or pulling the roller (call it Fp).
You can do this simply with a graph as illustrated
above, or a trigonometric
calculation shown below.
sin (45^{o}) = opposite/hypotenuse = 0.707
= Fp/300, so Fp = 0.707 x 300 = 212 N in a vertical direction.
You also need the weight of the roller = m x g =
80 x 10 = 800 N (taking gravity as 10 N/kg)
When pushing the roller you are increasing the
overall downward force of the roller, therefore
total vertical force exerted by the roller =
weight of roller + vertical pushing force
= 800 + 212 =
1012 N, the
vertical normal contact force due to the roller.
(c) Calculate the vertical force of the roller on the
grass when it is pushed.
When pulling the roller you are decreasing the
overall downward force of the roller, therefore
total vertical force exerted by the roller =
weight of roller  vertical pulling force
= 800  212 =
588 N, the
vertical normal contact force due to the roller.
AND, now you can see why its easier to pull a
roller than push it and how accurate was your graph work! 
Q11.
A suspended microphone! A microphone of
mass 750 g is suspended by a wire at 25^{o} to the vertical
(tension T2, diagram) and pulled to the right by a horizontal cord
(tension T1, diagram).
(a) Calculate the tension in the wire holding up
the microphone.
(b) What tension must be applied to the horizontal
cord to maintain the microphone in a stable position?
You can do this simply with a graph as in Q7, but
I'll leave you to practice that and I'll go straight for trigonometric
calculations, also at the end of Q7 and as in Q11. above.
750 g = 0.75 g, so m = m x g = 0.75 x 10 = 7.5
N for the weight of the microphone.
The diagram shows how the forces will operate.
(a) The tension in the wire T2
cos (25^{o}) = adjacent/hypotenuse =
0.9063 = 7.5/T2, so T2 = 7.5/0.9063 =
wire tension = 8.27 N
(3 sf, 2 dp)
Note that the tension on the wire is greater
than the weight of the microphone because it is being pulled both
downwards and to one side.
(b) Force T1 needed to stabilise microphone
tan 25^{o}) = opposite/adjacent =
0.466 = T1/7.5, so T1 = 0.466 x 7.5 =
cord tension required =
3.50 N (3 sf)
How accurate was your graph work! 
Q12 The motor
of a ferry boat provides driving force of 250 N perpendicular to the
bank of a river. If the flow of the river acts in the same direction
as the river bank with a force of 100 N ...
(a) Calculate the magnitude and direction of the resultant force.
By drawing a graph diagram or from a trigonometric calculation
you should get a resultant force of
269 N.
Ignoring the rudder! the boat will travel away from the bank at
an angle of
68^{o}
to it (or 22^{o} from the perpendicular line from the bank).
(b) With reference to your answer to (b), how should the ferry be
steered to minimise the distance travelled to cross the river?
To counteract the flow of the river, the boat should be steered
at an angle opposite to the direction indicated above in (a). This
means working against the flow of the river to counteract the
enforced sideways movement of the boat.
You don't save energy but you can reduce the distance travelled.

Q13 The
engine of a doubledecker bus weighing 20 000 N
generates a driving force of 2000 N against opposing friction forces of 1500 N.
(a)
Draw a free body diagram of the forces involved and explain their
origin.
F1 = the contact force of the road
material pushing up
F2 = contact force from the weight of the
bus
F3 = forces of resistance due to friction
 air resistance, wheels on road, moving parts of bus etc.
F4 = driving force generated by the engine
of the bus
(b) Calculate the resultant force
Unless the bus is moving up and down F1
will equal F2, so no resultant in the vertical direction.
However, there is a resultant force to the
right of F4  F3 = 2000  1500 =
500 N
(c) Calculate the acceleration of the bus
(this requires much more advanced knowledge
Assume gravity force is 10 N/kg.
W = m x g, m = W / g, and from Newton's 2nd Law: F = ma, a =
F/m
F = resultant force = 500 N
weight = mass x g, mass = weight /g = 20
000/10 = 2000 kg
a = F/m = 500 / 2000 =
0.25 m/s^{2}
See
4. Newton's First, Second and Third Laws of
Motion, inertia and F = ma calculations
gcse physics revision notes 


TOP OF PAGE
APPENDIX 1:
right angle triangle rules
Important formulae in the trigonometry of a
right angled triangle:
for angle θ shown on the right diagram
tangent rule: tan θ = ^{opposite} / _{adjacent}
sine rule: sin θ = ^{opposite} / _{hypotenuse}
cosine rule: cos θ = ^{adjacent} / _{hypotenuse}
APPENDIX
2 Calculating work done from a resultant force
If a source of energy is available, you can
calculate the work done from the acting force and the distance the force acts
through.
work done (joules) = force (newtons) x
distance along the line of action of the force (metres)
W (J) = F (N) x d (m), F =
W/d, d = W/F
Q1
If you drag a heavy box with a force of 200 N across a floor for 3 m, what
work is done?
work done = 200 x 3 =
600 J
Q2 If a machine part
does 500 J of work moving linearly 2.5 m, what force was applied by the
machine?
work done = force x distance,
rearranging, force (N) = work done (J) ÷ distance (m)
force = 500 ÷ 2.6 =
200 N
Q3 Part of a machine requires a continuous resultant force of
500 N from a motor to move it in a linear direction.
(a) How much work is done in moving
it a distance of 50 m?
work done = force x distance =
500 x 50 =
25000 J (25 kJ)
Q4 A toy model car has a clockwork motor, whose spring can
store 8.75 J of elastic potential energy.
On release the clockwork motor can
deliver a continuous force of 2.5 N.
How far will the car travel in one
go?
energy store = total work done =
force x distance
distance = energy store / force =
8.75 / 2.5 =
3.5
m
These examples were 'borrowed' from
Types
of energy stores, mechanical work done and power calculations
TOP OF PAGE
Forces revision notes index
FORCES 1. What are contact forces &
noncontact forces?, scalar & vector quantities, free body force diagrams
FORCES
2. Mass and the effect of gravity force on it  weight, (mention of work done and
GPE)
FORCES 3. Calculating resultant forces using vector
diagrams and work done
FORCES
4.
Elasticity and energy stored in a spring
FORCES 5. Turning forces and moments  from spanners
to wheelbarrows and equilibrium situations
FORCES 6. Pressure in liquid fluids and hydraulic
systems
FORCES 7. Pressure & upthrust in liquids, why do
objects float or sink in a fluid?, variation of atmospheric pressure with
height
resulting
forces and work done calculations IGCSE revision
notes working out resultant forces KS4 physics Science notes on working out
resultant forces GCSE physics guide
notes on working out resultant forces for schools colleges academies science course tutors images
pictures diagrams for working out resultant forces science revision notes on
working out resultant forces for revising physics modules physics topics notes to help on understanding of
working out resultant forces university courses in physics
careers in science physics jobs in the engineering industry
technical laboratory assistant
apprenticeships engineer internships in physics USA US grade 8 grade 9 grade10 AQA
GCSE 91 physics science revision notes on working out resultant
forces GCSE
notes on working out resultant forces Edexcel GCSE 91 physics science revision notes on
working out resultant forces for OCR GCSE 91 21st century
physics science notes on working out resultant forces OCR GCSE
91 Gateway physics science
revision notes on working out resultant forces WJEC gcse science CCEA/CEA gcse science
how to calculate the work done from a resultant force
TOP OF PAGE
















Website content © Dr
Phil Brown 2000+. All copyrights reserved on revision notes, images,
quizzes, worksheets etc. Copying of website material is NOT
permitted. Exam revision summaries & references to science course specifications
are unofficial. 
