Free body diagrams Β· Calculating resultant Β· Balanced and unbalanced forces Β· Force triangles
AQA GCSE Physics 4.5 | Year 11 Higher Tier
π Draw and interpret free body diagrams showing all forces acting on an object
β Calculate the resultant of two or more forces acting along the same line
βοΈ Distinguish between balanced forces (equilibrium) and unbalanced forces
π Use Pythagoras' theorem to find the resultant of two perpendicular forces
π Use trigonometry to find the angle (direction) of a resultant force
πΊ Construct and use scale force triangles to find unknown forces
What is a Force?
A force is a push or pull that acts on an object due to an interaction with another object. Forces are vectors β they have both magnitude (size) and direction. The SI unit of force is the Newton (N).
Vector quantity: A quantity that has both magnitude AND direction. Force, velocity, acceleration, and displacement are all vectors. Contrast with scalars (mass, speed, temperature) which have magnitude only.
Common forces you will encounter include:
Weight (W) β gravitational pull of the Earth, always acting downward. W = mg
Normal contact force (N) β perpendicular to a surface, pushing objects apart
Friction (f) β opposes relative motion between surfaces, acts along the surface
Air resistance / drag (D) β opposes motion through a fluid
Tension (T) β pulling force transmitted along a rope, string, or cable
Upthrust (U) β upward force exerted by a fluid on a submerged object
Applied force (F) β any deliberate push or pull from a person or engine
Symbol
Quantity
SI Unit
F
Force (general)
Newton (N)
W
Weight
Newton (N)
m
Mass
kilogram (kg)
g
Gravitational field strength
N/kg
ΞΈ
Angle of resultant
degrees (Β°)
Free Body Diagrams
A free body diagram (FBD) is a simplified sketch of an object showing all the forces acting ON that object. Each force is drawn as an arrow starting at (or from) the object, pointing in the direction the force acts, with the arrow length roughly proportional to the force's magnitude.
Rules for drawing free body diagrams:
Draw the object as a simple box or dot
Draw every force arrow FROM the object's centre or surface
Label each force with its name and value (if known)
Make arrow lengths proportional to magnitude
Only include forces ON the object β NOT forces the object exerts on other things
Example β Book on a table: Weight (W) acts downward. Normal contact force (N) acts upward. If the book is stationary, N = W and the forces are balanced.
Example β Car accelerating: Weight acts down. Normal contact force acts up. Driving force acts forward. Friction/air resistance acts backward. If the driving force is greater than friction, the forces are unbalanced and the car accelerates.
Example β Skydiver in freefall: Weight acts downward. Air resistance acts upward. Initially weight > air resistance, so the skydiver accelerates downward. As speed increases, air resistance increases until it equals weight β this is terminal velocity.
Newton's Third Law reminder: Forces always come in equal and opposite pairs acting on DIFFERENT objects. The normal force from the table on the book is NOT Newton's Third Law pair of the book's weight β the pair of weight is the gravitational pull of the book on the Earth.
Resultant Forces β Collinear (Same Line)
The resultant force is the single force that has the same effect as all the individual forces acting together. It is found by combining (adding) all forces, taking direction into account.
Resultant force: The vector sum of all forces acting on an object. It is the single equivalent force that produces the same acceleration as the combination of all individual forces.
For forces acting along the same straight line (collinear forces), choose a positive direction (e.g. right or up) and add forces in that direction while subtracting forces in the opposite direction.
F_resultant = Fβ + Fβ + Fβ + β¦ (taking direction into account)
Balanced forces (equilibrium): When the resultant force is zero, the object is in equilibrium. According to Newton's First Law, it will either remain stationary or continue moving at constant velocity in a straight line.
Unbalanced forces: When the resultant force is non-zero, the object will accelerate in the direction of the resultant force, as described by Newton's Second Law: F = ma.
F = ma
where F = resultant force (N), m = mass (kg), a = acceleration (m/sΒ²)
Sign convention example: A car experiences a driving force of 3000 N forward and friction of 1200 N backward. Taking forward as positive: F_resultant = 3000 β 1200 = +1800 N (forward). The car accelerates forward.
Resultant of Perpendicular Forces
When two forces act at right angles (90Β°) to each other, the resultant cannot be found by simple addition. Instead, we use Pythagoras' theorem to find the magnitude and trigonometry to find the direction.
The resultant is represented as the hypotenuse of a right-angled triangle formed by the two component forces. The angle ΞΈ is measured from the larger force (or from a defined reference direction such as horizontal).
Always state both the magnitude (in N) AND the direction (in degrees from a reference) when giving a resultant force. A vector answer without direction is incomplete!
Example: A force of 6 N acts eastward and a force of 8 N acts northward. The resultant = β(6Β² + 8Β²) = β(36 + 64) = β100 = 10 N. The angle from east = arctan(8/6) = 53.1Β° north of east.
For forces NOT at right angles, a scale drawing (force triangle or parallelogram) is used, which is covered in the next section.
Force Triangles and Scale Drawings
When forces are at angles that are NOT 90Β°, we use scale drawings to find the resultant. The most useful method is the triangle of forces (also called the tip-to-tail method or vector triangle).
Triangle of forces: To add two forces graphically, draw the first force vector to scale, then draw the second force vector starting from the tip (end) of the first. The resultant is the vector drawn from the tail of the first to the tip of the second β closing the triangle.
Steps for a scale force drawing:
Choose a suitable scale (e.g. 1 cm = 10 N)
Draw the first force vector as an arrow of the correct length and direction
From the tip of the first arrow, draw the second force vector (correct length and direction)
Draw the closing line from the start of the first arrow to the tip of the second β this is the resultant
Measure the length of the resultant and convert back using your scale
Measure the angle with a protractor
For an object in equilibrium (zero resultant force), three forces form a closed triangle β the tip of the last vector meets the tail of the first, with no closing gap.
Parallelogram of forces: An alternative graphical method where both forces are drawn from the same point, then a parallelogram is completed. The resultant is the diagonal of the parallelogram from the origin point.
For Higher tier GCSE, you should be able to: (a) construct a triangle of forces to scale, (b) measure the resultant and its angle, and (c) check whether a set of three forces is in equilibrium by checking if they form a closed triangle.
For three forces in equilibrium:
Fβ + Fβ + Fβ = 0 (vector sum)
The three force vectors form a closed triangle.
Example 1: A box of mass 5 kg sits on a horizontal surface. A person pushes the box horizontally with a force of 40 N. Friction acts on the box with a magnitude of 15 N opposing the motion. (a) Draw a free body diagram. (b) Calculate the resultant horizontal force. (c) Calculate the acceleration of the box. (g = 10 N/kg)
1Free body diagram: Draw the box. Four forces act: Weight W = mg = 5 Γ 10 = 50 N downward. Normal contact force N = 50 N upward. Applied force F = 40 N to the right. Friction f = 15 N to the left.
2Vertical forces: N (up) = W (down) = 50 N. These are balanced β no vertical acceleration.
3Horizontal resultant: Take right as positive. F_R = 40 β 15 = +25 N (to the right)
4Acceleration: Using F = ma β a = F_R Γ· m = 25 Γ· 5 = 5 m/sΒ²
Resultant horizontal force = 25 N to the right
Acceleration = 5 m/sΒ² to the right
Example 2: A boat is pulled by two tugboats. Tugboat A pulls with a force of 12 000 N due north. Tugboat B pulls with a force of 5 000 N due east. Calculate the magnitude and direction of the resultant force on the boat.
1Identify the geometry: The two forces are perpendicular (north and east), so we use Pythagoras' theorem.
3Direction of resultant:
tan ΞΈ = F_east Γ· F_north = 5 000 Γ· 12 000 = 0.4167
ΞΈ = arctan(0.4167) = 22.6Β° east of north
4Check: The angle makes sense β the larger force is northward, so the resultant should be closer to north than to east. 22.6Β° from north confirms this. β
Resultant force = 13 000 N at 22.6Β° east of north
Example 3: A skydiver of mass 80 kg reaches terminal velocity. (a) State what is meant by terminal velocity. (b) Calculate the weight of the skydiver. (c) State the magnitude of the air resistance force at terminal velocity. (d) On a free body diagram, describe the lengths of the force arrows. (g = 10 N/kg)
1Terminal velocity definition: Terminal velocity is the constant maximum velocity reached when the driving force (weight) is exactly balanced by the resistive force (air resistance), giving a zero resultant force and therefore zero acceleration.
2Weight:
W = mg = 80 Γ 10 = 800 N (downward)
3Air resistance at terminal velocity:
At terminal velocity, resultant force = 0 N.
Therefore: Air resistance = Weight = 800 N (upward)
4Free body diagram description: Two arrows of equal length β weight arrow pointing downward (800 N) and air resistance arrow pointing upward (800 N). Equal lengths represent equal magnitudes β balanced forces β zero resultant.
Weight = 800 N downward
Air resistance = 800 N upward
Resultant force = 0 N β constant velocity (terminal velocity)
Example 4: Three forces act on an object. Force P = 60 N acts horizontally to the right. Force Q = 80 N acts vertically upward. Force R is the third force that keeps the object in equilibrium. Find the magnitude and direction of force R.
1Find the resultant of P and Q:
Since P and Q are perpendicular:
F_PQ = β(60Β² + 80Β²) = β(3600 + 6400) = β10000 = 100 N
2Direction of F_PQ:
tan ΞΈ = 60 Γ· 80 = 0.75
ΞΈ = arctan(0.75) = 36.9Β° from vertical (upward-right direction)
3Find force R for equilibrium:
For equilibrium, the vector sum of all three forces = 0.
Therefore R must be equal in magnitude but exactly opposite in direction to F_PQ.
R = 100 N acting at 36.9Β° from vertically downward, toward the lower-left.
4Check using components:
Horizontal: 60 + 0 + R_x = 0 β R_x = β60 N β
Vertical: 0 + 80 + R_y = 0 β R_y = β80 N β
|R| = β(60Β² + 80Β²) = 100 N β
Force R = 100 N acting at 36.9Β° from vertically downward (pointing lower-left, opposite to P + Q)
Question 1: A car experiences a driving force of 4500 N forward and air resistance of 1500 N backward. What is the resultant force on the car?
Question 2: A box of mass 10 kg is in equilibrium on a horizontal surface. What is the magnitude of the normal contact force?
Question 3: Two forces act at right angles: 9 N and 12 N. Calculate the magnitude of the resultant force. Enter your answer in N.
Question 4: A skydiver is falling and accelerating downward. Which of the following correctly describes the forces acting?
Question 5: An object has a resultant force of 0 N acting on it. What can you conclude about its motion?
Challenge 1 (6 marks): A cyclist of mass 70 kg (including the bicycle) accelerates from rest. The cyclist exerts a driving force of 250 N. Air resistance is 30 N and friction from the road is 20 N. (a) Draw and label a free body diagram. (b) Calculate the resultant force. (c) Calculate the acceleration. (d) Explain how the forces change as the cyclist reaches top speed.
(a) Free body diagram: Box (cyclist). Weight = 700 N downward. Normal contact force = 700 N upward. Driving force = 250 N forward. Air resistance = 30 N backward. Road friction = 20 N backward.
(c) Acceleration: F = ma β a = F Γ· m = 200 Γ· 70 = 2.86 m/sΒ² (forward)
(d) As speed increases: Air resistance increases (proportional to speedΒ²). Road friction also increases slightly. The total resistive force increases, so the resultant force decreases. The acceleration decreases. Eventually, at top speed, driving force = total resistance (250 N), resultant = 0 N, acceleration = 0, constant velocity.
Challenge 2 (5 marks): A ship is acted on by two tugboat forces: Tug A pulls with 15 000 N at 0Β° (east). Tug B pulls with 20 000 N at 90Β° (north). A current exerts a force of 5 000 N at 180Β° (west). (a) Find the net eastward force. (b) Find the net northward force. (c) Calculate the magnitude of the overall resultant force. (d) Calculate the direction of the resultant to the nearest degree.
(a) Net eastward force: 15 000 (east) β 5 000 (west) = 10 000 N eastward
(b) Net northward force: 20 000 N northward (no opposing north-south force)
(d) Direction:
tan ΞΈ = 10 000 Γ· 20 000 = 0.5
ΞΈ = arctan(0.5) = 26.6Β° east of north
Resultant = 22 400 N at 26.6Β° east of north
Challenge 3 (5 marks): Three forces act on an object in equilibrium. Force A = 50 N horizontally to the right. Force B = 50 N vertically upward. Force C is unknown. (a) Explain what "in equilibrium" means in terms of forces. (b) Calculate the magnitude of force C. (c) Calculate the direction of force C. (d) Describe how you would verify this using a scale drawing / force triangle.
(a) Equilibrium: The vector sum of all forces acting on the object is zero (resultant = 0 N). The object is either stationary or moving at constant velocity in a straight line.
(b) Magnitude of C:
C must cancel the combined effect of A and B.
Resultant of A and B = β(50Β² + 50Β²) = β(2500 + 2500) = β5000 = 70.7 N
For equilibrium: |C| = 70.7 N
(c) Direction of C:
A and B point right and up. C must point in the opposite direction: lower-left.
tan ΞΈ = 50 Γ· 50 = 1 β ΞΈ = 45Β°
C acts at 45Β° below the horizontal (225Β° from east / lower-left diagonal)
(d) Scale drawing verification:
Choose scale: 1 cm = 10 N. Draw A (5 cm, right). From its tip, draw B (5 cm, upward). From the tip of B, draw C (7.07 cm, toward lower-left at 45Β°). The tip of C should arrive back exactly at the starting point of A, forming a closed triangle β confirming equilibrium.
Challenge 4 β Extended Response (6 marks): A parachutist jumps from an aircraft. Describe and explain, with reference to forces and Newton's Laws, the complete motion from the moment of jumping until landing safely. Include how the free body diagram changes at each stage.
Stage 1 β Initial freefall (parachute closed):
Weight acts downward. Air resistance acts upward but is initially very small (low speed). Resultant force is downward β parachutist accelerates downward (Newton's 2nd Law: F = ma). FBD: weight arrow much longer than air resistance arrow.
Stage 2 β Building speed in freefall:
As speed increases, air resistance increases (drag β vΒ²). Resultant force decreases (weight stays constant, resistance grows). Acceleration decreases. FBD: arrows become more equal in length. Eventually air resistance = weight β resultant = 0 β terminal velocity β 55 m/s.
Stage 3 β Parachute opens:
Parachute massively increases surface area β air resistance suddenly increases greatly. Air resistance >> weight β resultant force is now UPWARD. Parachutist decelerates rapidly. FBD: air resistance arrow now much longer than weight arrow.
Stage 4 β New terminal velocity:
As speed decreases, air resistance decreases until air resistance = weight again. New terminal velocity β 5β6 m/s. FBD: equal arrows again. Parachutist lands safely at this low speed.
Key physics: Newton's 1st Law β zero resultant = constant velocity (terminal velocity). Newton's 2nd Law β unbalanced force causes acceleration or deceleration. At all terminal velocities, the object is in equilibrium (resultant = 0).