National 5PhysicsSQARevision NotesDynamicsNewton’s LawsNewton's Laws & Unbalanced Forces

Newton's Laws & Unbalanced Forces (SQA National 5 Physics): Revision Note

Exam code: X857 75

Katie M

Written by: Katie M

Reviewed by: Leander Oates

Updated on

Newton's laws & unbalanced forces

Unbalanced forces

  • Forces acting along a straight line are said to be unbalanced if they do not cancel each other out

    • There is a resultant force on the object

    • The direction of the resultant force is in the same direction as the force with the largest size (magnitude)

Objects with unbalanced forces

Three diagrams showing boxes with forces. Left: 7 + (-3) = 4 N to the left, middle: 7 + 3 = 10 N to the right, right: 7 + (-5) = 2 N upwards.
When the forces acting on an object are unbalanced, a resultant force acts on the object

  • When the forces are acting at right angles, the force vectors can be combined to determine the magnitude and direction of the resultant force, either by

Worked Example

A force acts on an object with 60 N to the right. A second force of 100 N acts on the same object in the upward direction.

Calculate the resultant force acting on the object.

Answer:

Step 1: Draw a vector diagram

vector-diagram-forces-we1

Step 2: Calculate the magnitude of the resultant force using Pythagoras' theorem

 F space equals space square root of 60 squared space plus space 100 squared end root

F space equals space square root of 13 space 600 end root

F space equals space 117 space straight N

Step 3: Calculate the direction of the resultant vector using trigonometry

vector-diagram-forces-we2

 tan space theta space equals space opposite over adjacent

tan space theta space equals space 100 over 60

theta space equals space tan to the power of negative 1 end exponent open parentheses 100 over 60 close parentheses space equals space 59 degree

Step 4: State the final answer, complete with magnitude and direction

F space equals space 117 space straight N space at space 59 degree space from space the space horizontal

Free-body diagrams

  • Free-body diagrams are used to show the forces acting on an object

  • The force vectors can be drawn onto a picture of the object, or the object can be represented by a dot

Free-body diagram of a car

Free-body force diagram of a car showing the forward acting thrust force of the engine on the car, the backwards acting frictional force of the road on the car (tyres), the upward acting normal contact force of the road on the car, and the downward acting force of weight (gravitational pull of the Earth) on the car
Free-body force diagrams show multiple forces acting on a single object

  • The vertical forces add together

    • In the example of the car, the normal force and the weight are equal in magnitude and opposite in direction

    • Therefore, the vertical forces add up to zero

    • The forces are balanced in the vertical direction

  • The horizontal forces add together

    • In the car example, the thrust force has a greater magnitude than the frictional force and acts in the opposite direction

    • Therefore, the horizontal forces do not cancel out

    • There is an unbalanced force in the forward direction

Examiner Tips and Tricks

Force diagrams are any diagrams which show forces acting, whereas free-body force diagrams specifically show the forces acting on a single object. You don't need to know this terminology for your exam, but you do need to recognise if the forces shown in a diagram are acting on a single object or on multiple objects.

Newton's second law

  • Newton's second law of motion states:

The acceleration of an object is proportional to the unbalanced force acting on it and inversely proportional to the object's mass

  • Newton's second law of motion describes what happens when an unbalanced force acts on an object

  • An unbalanced force causes a change in the object's motion

  • This change in motion is an acceleration, which could involve

    • speeding up

    • slowing down

    • changing direction

  • When multiple forces act on an object, the object will accelerate in the direction of the unbalanced force

    • The greater the unbalanced force F, the larger the acceleration, a space proportional to space F

    • For a given force, the larger the object's mass m, the smaller the acceleration, a space proportional to space 1 over m

  • Consider two people, A and B, playing a game of tug-of-war

    • If person A pulls with 80 N to the left and person B pulls with 100 N to the right, there is an unbalanced force of 20 N to the right

    • Since person B pulled with more force than person A, the whole system (the two people and the rope) will accelerate to the right

Unbalanced forces during tug-of-war

Two people pull on a rope; Person A exerts 80N left, Person B 100N right, demonstrating unbalanced forces in a tug-of-war scenario.
A tug-of-war is an example of when forces can become unbalanced

Worked Example

Three shopping trolleys, A, B and C, are being pushed using the same force. This force causes each trolley to accelerate.

WE Newton second law, downloadable IGCSE & GCSE Physics revision notes

State and explain which trolley would have the smallest acceleration.

Answer: C

Step 1: Recall Newton's second law

  • Newton's second law states that the acceleration of an object is

    • proportional to the unbalanced force acting on it

    • inversely proportional to the object's mass

Step 2: Relate Newton's second law to the scenario

  • Since the same force is applied to each trolley, acceleration is inversely proportional to mass

a space proportional to space 1 over m

Step 3: Explain the inverse proportionality between acceleration and mass  

  • For the same force applied, a larger mass will experience a smaller acceleration

  • Therefore, trolley C will have the smallest acceleration because it has the largest mass

Calculations using Newton's second law

  • Newton's second law can be expressed as a relationship:

F space equals space m a

  • Where:

    • F = unbalanced force on the object, measured in newtons (N)

    • m = mass of the object, measured in kilograms (kg)

    • a = acceleration of the object, measured in metres per second squared (m s-2)

  • The acceleration occurs in the same direction as the unbalanced force

Relationship triangle for acceleration, mass and force

Triangle showing Newton's second law. Top: “Force (F)”. Bottom: “Mass (m)” on left, “Acceleration (a)” on right.
To use a relationship triangle, simply cover up the quantity you wish to calculate and the structure of the relationship is revealed

  • A more detailed explanation of how to use relationship triangles is covered in the revision note on motion relationships

Worked Example

A car salesman says that their best car has a mass of 910 kg and can accelerate from 0 to 27 m s-1 in 3.0 seconds.

Calculate:

a) the acceleration of the car in the first 3.0 seconds.

b) the force required to produce this acceleration.

Answer:

Part (a)

Step 1: List the known quantities

  • Initial velocity, v space equals space 0 

  • Final velocity, u space equals space 27 space straight m space straight s to the power of negative 1 end exponent 

  • Time, t space equals space 3.0 space straight s

Step 2: State the relationship between acceleration, time and velocity

a space equals space fraction numerator v space minus space u over denominator t end fraction

a space equals space fraction numerator 27 space minus space 0 over denominator 3.0 end fraction

a space equals space 9.0 space straight m space straight s to the power of negative 2 end exponent

Part (b)

Step 1: List the known quantities

  • Mass of the car, m space equals space 910 space kg 

  • Acceleration, a space equals space 9.0 space straight m space straight s to the power of negative 2 end exponent 

Step 2: Identify which law of motion to apply

  • The question involves quantities of force, mass and acceleration, so Newton's second law is required:

F space equals space m a

Step 3: Calculate the force required to accelerate the car

F space equals space 910 space cross times space 9.0

F space equals space 8190 space straight N

Step 4: Round to an appropriate amount of significant figures

  • The least precise input value is 2 s.f.

  • Therefore, the final answer can only be given to the same precision

F space equals space 8200 space straight N

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    Author
    Reviewer
    Katie M

    Author: Katie M

    Expertise: Physics Content Creator

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

    Leander Oates

    Reviewer: Leander Oates

    Expertise: Physics Content Creator

    Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.