3.3.6 Weight

What you'll learn

How to calculate the weight of a body using $W = m g$ .
How mass and weight differ.
How to include weight in a force diagram and apply Newton's second law.
How to model straight-line motion under gravity, including free fall and vertical projection.

Mass, Weight and Gravity

Mass

The mass of a body measures the amount of matter in it. Its symbol is usually $m$ , and its SI unit is the kilogram, kg.

Mass is a scalar quantity: it has magnitude but no direction. For a given body, its mass is treated as constant in A-Level mechanics problems.

Weight

Earth attracts objects towards its centre. This gravitational attraction produces a force called weight.

Definition

Weight

The weight of a body is the gravitational force acting on it. Near Earth's surface,

W = m g

where $W$ is weight in newtons, $m$ is mass in kilograms, and $g$ is the magnitude of the acceleration due to gravity.

Unless a question gives a different value, use

s−2.g=9.8\text{ m s}^{-2}.

Weight is a vector quantity because it has both magnitude and direction. It acts vertically downwards, towards the centre of Earth.

Force diagrams showing weight acting vertically downwards and a normal reaction balancing weight on a horizontal surface

Key Idea

Mass is not weight

Mass is measured in kilograms and describes the body itself. Weight is measured in newtons and is the force of gravity on that body.

Example

Calculating weight

A suitcase has mass 18 kg. Find its weight.

Identify the mass as $kgm=18\text{ kg}$ and use $s−2g=9.8\text{ m s}^{-2}$ .
Substitute into $W = m g$ :
$W=18×9.8.W=18\times 9.8.$
Calculate the force:
$N.W=176.4\text{ N}.$
The suitcase's weight is 176.4 N, acting vertically downwards.

Common Mistake

Giving weight in kilograms

Kilograms measure mass, not weight. Since weight is a force, its unit must be the newton, N.

Representing Weight in a Force Diagram

A force diagram shows the forces acting on one chosen body. Weight should be drawn as an arrow pointing vertically downwards and labelled $W$ or $m g$ .

Other forces may also act. For example:

A surface can exert a normal reaction, perpendicular to the surface.
A string can exert tension along the string.
Air resistance or another resistive force acts against the direction of motion.
A driving force or thrust acts in its stated direction.

Weight does not automatically equal the normal reaction. They are equal only when the resultant vertical force is zero and no other vertical forces act.

Example

Finding the normal reaction

A box of mass 12 kg rests on a horizontal floor. Find the normal reaction $R$ from the floor.

The box has no vertical acceleration, so its resultant vertical force is zero.
Its upward force is $R$ , while its downward weight is
$N.W=12\times 9.8=117.6\text{ N}.$
Balancing the vertical forces gives
$R - 117.6 = 0,$
so
$N.R=117.6\text{ N}.$

Using Weight with Newton's Second Law

Newton's second law states that the resultant force on a body equals its mass multiplied by its acceleration:

F = ma .

The $F$ in this equation is the resultant force, meaning the overall force after directions have been taken into account. It is not necessarily just the weight.

Before forming an equation, choose a positive direction. Forces in that direction are positive; forces in the opposite direction are negative.

Example

Descending against resistance

A package of mass 5 kg is falling vertically. Air resistance of 14 N acts upwards. Find its acceleration.

Choose vertically downwards as positive. The weight is positive and the resistance is negative.
Calculate the weight:
$N.W=5\times 9.8=49\text{ N}.$
Apply Newton's second law in the downward direction:
$49 - 14 = 5 a .$
Solve for $a$ :
$s−2.a=\frac{35}{5}=7\text{ m s}^{-2}.$
The acceleration is $s−27\text{ m s}^{-2}$ vertically downwards.

Tip

Use a direction statement

Write something such as “taking downwards as positive” before your force equation. This makes the signs in your working much easier to follow.

Common Mistake

Using weight as the resultant force

If resistance, tension, reaction or thrust also acts vertically, you must combine all the forces before using $F = ma$ .

Free Fall

A body is in free fall when gravity is the only force acting on it. Air resistance is therefore ignored in this model.

For a body of mass $m$ , taking downwards as positive gives

m g = ma .

Since $m≠0m\neq 0$ , dividing by $m$ gives

a = g .

This explains why, in the free-fall model, every body has the same downward acceleration regardless of its mass.

Key Idea

Free-fall acceleration

If weight is the only force acting, the acceleration is $g$ vertically downwards. The body's mass cancels from Newton's second law.

Example

Falling from rest

A stone is released from rest and falls freely for 3 seconds. Find its speed and the distance it falls.

Take downwards as positive. Since the stone is released, $u = 0$ , and free fall gives $s−2a=9.8\text{ m s}^{-2}$ .
Use $v = u + a t$ :
$s−1.v=0+9.8\times 3=29.4\text{ m s}^{-1}.$
Use $s=ut+12at2s=ut+\frac12at^2$ :
$m.s=0+\frac12\times 9.8\times 3^2=44.1\text{ m}.$
Its speed is 29.4 m s⁻¹ and it falls 44.1 m.

Vertical Projection

A body may initially move upwards even though its weight acts downwards. While it rises, gravity reduces its upward velocity. At the highest point, its velocity is momentarily zero, but its acceleration is still vertically downwards.

If upwards is chosen as positive, then

s−2.a=-g=-9.8\text{ m s}^{-2}.

If downwards is chosen as positive, then $a = + g$ . Either convention works, provided you use it consistently for displacement, velocity, acceleration and forces.

Example

Finding maximum height

A ball is projected vertically upwards at $s−119.6\text{ m s}^{-1}$ . Find the greatest height it reaches above its starting point, ignoring air resistance.

Take upwards as positive. Then $s−1u=19.6\text{ m s}^{-1}$ and $s−2a=-9.8\text{ m s}^{-2}$ .
At the greatest height, the instantaneous velocity is $v = 0$ . Use
$v^2=u^2+2as.$
Substitute the known values:
$0=19.6^2+2(-9.8)s.$
Rearrange:
$m.s=\frac{19.6^2}{19.6}=19.6\text{ m}.$
The greatest height above the point of projection is 19.6 m.

Common Mistake

Zero velocity does not mean zero acceleration

At the highest point of a vertical flight, the velocity is zero for an instant, but weight still acts. The acceleration remains $s−29.8\text{ m s}^{-2}$ downwards.

The Constant-Gravity Model

Using a constant value of $g$ is a mathematical model. It assumes that:

The body remains close enough to Earth's surface for gravitational acceleration to be constant.
Earth can be treated as fixed.
Air resistance is ignored unless the question includes it.
The body can usually be modelled as a particle, so its size and rotation are unimportant.

These assumptions make the acceleration constant, allowing you to use the constant-acceleration formulae.

Common Mistake

When constant-acceleration formulae apply

You may use the standard SUVAT equations only over a period in which the acceleration is constant. If a resistive force changes with speed, the acceleration may not remain constant.

Exam technique

In the exam

Draw a force diagram and include weight as $m g$ vertically downwards.
State your positive direction before applying $F = ma$ or the constant-acceleration formulae.
Use $s−2g=9.8\text{ m s}^{-2}$ unless another value is given, and distinguish carefully between mass in kg and weight in N.
Check whether gravity is the only force: if it is, the acceleration is $g$ downwards; if it is not, find the resultant force first.
At a greatest height, set the velocity to zero, not the acceleration.

Self review

Check yourself

What is the weight of a body of mass 7.5 kg, and in which direction does it act?
A falling body has weight 80 N and upward resistance 20 N. How would you form its equation of motion?
Why does a vertically projected ball still have non-zero acceleration at its highest point?

What you'll learn

How to calculate the weight of a body using $W = m g$ .
How mass and weight differ.
How to include weight in a force diagram and apply Newton's second law.
How to model straight-line motion under gravity, including free fall and vertical projection.

Mass, Weight and Gravity

Mass

The mass of a body measures the amount of matter in it. Its symbol is usually $m$ , and its SI unit is the kilogram, kg.

Mass is a scalar quantity: it has magnitude but no direction. For a given body, its mass is treated as constant in A-Level mechanics problems.

Weight

Earth attracts objects towards its centre. This gravitational attraction produces a force called weight.

Definition

Weight

The weight of a body is the gravitational force acting on it. Near Earth's surface,

W = m g

where $W$ is weight in newtons, $m$ is mass in kilograms, and $g$ is the magnitude of the acceleration due to gravity.

Unless a question gives a different value, use

s−2.g=9.8\text{ m s}^{-2}.

Weight is a vector quantity because it has both magnitude and direction. It acts vertically downwards, towards the centre of Earth.

Force diagrams showing weight acting vertically downwards and a normal reaction balancing weight on a horizontal surface

Key Idea

Mass is not weight

Mass is measured in kilograms and describes the body itself. Weight is measured in newtons and is the force of gravity on that body.

Example

Calculating weight

A suitcase has mass 18 kg. Find its weight.

Identify the mass as $kgm=18\text{ kg}$ and use $s−2g=9.8\text{ m s}^{-2}$ .
Substitute into $W = m g$ :
$W=18×9.8.W=18\times 9.8.$
Calculate the force:
$N.W=176.4\text{ N}.$
The suitcase's weight is 176.4 N, acting vertically downwards.

Common Mistake

Giving weight in kilograms

Kilograms measure mass, not weight. Since weight is a force, its unit must be the newton, N.

Representing Weight in a Force Diagram

A force diagram shows the forces acting on one chosen body. Weight should be drawn as an arrow pointing vertically downwards and labelled $W$ or $m g$ .

Other forces may also act. For example:

A surface can exert a normal reaction, perpendicular to the surface.
A string can exert tension along the string.
Air resistance or another resistive force acts against the direction of motion.
A driving force or thrust acts in its stated direction.

Weight does not automatically equal the normal reaction. They are equal only when the resultant vertical force is zero and no other vertical forces act.

Example

Finding the normal reaction

A box of mass 12 kg rests on a horizontal floor. Find the normal reaction $R$ from the floor.

The box has no vertical acceleration, so its resultant vertical force is zero.
Its upward force is $R$ , while its downward weight is
$N.W=12\times 9.8=117.6\text{ N}.$
Balancing the vertical forces gives
$R - 117.6 = 0,$
so
$N.R=117.6\text{ N}.$

Using Weight with Newton's Second Law

Newton's second law states that the resultant force on a body equals its mass multiplied by its acceleration:

F = ma .

The $F$ in this equation is the resultant force, meaning the overall force after directions have been taken into account. It is not necessarily just the weight.

Before forming an equation, choose a positive direction. Forces in that direction are positive; forces in the opposite direction are negative.

Example

Descending against resistance

A package of mass 5 kg is falling vertically. Air resistance of 14 N acts upwards. Find its acceleration.

Choose vertically downwards as positive. The weight is positive and the resistance is negative.
Calculate the weight:
$N.W=5\times 9.8=49\text{ N}.$
Apply Newton's second law in the downward direction:
$49 - 14 = 5 a .$
Solve for $a$ :
$s−2.a=\frac{35}{5}=7\text{ m s}^{-2}.$
The acceleration is $s−27\text{ m s}^{-2}$ vertically downwards.

Tip

Use a direction statement

Write something such as “taking downwards as positive” before your force equation. This makes the signs in your working much easier to follow.

Common Mistake

Using weight as the resultant force

If resistance, tension, reaction or thrust also acts vertically, you must combine all the forces before using $F = ma$ .

Free Fall

A body is in free fall when gravity is the only force acting on it. Air resistance is therefore ignored in this model.

For a body of mass $m$ , taking downwards as positive gives

m g = ma .

Since $m≠0m\neq 0$ , dividing by $m$ gives

a = g .

This explains why, in the free-fall model, every body has the same downward acceleration regardless of its mass.

Key Idea

Free-fall acceleration

If weight is the only force acting, the acceleration is $g$ vertically downwards. The body's mass cancels from Newton's second law.

Example

Falling from rest

A stone is released from rest and falls freely for 3 seconds. Find its speed and the distance it falls.

Take downwards as positive. Since the stone is released, $u = 0$ , and free fall gives $s−2a=9.8\text{ m s}^{-2}$ .
Use $v = u + a t$ :
$s−1.v=0+9.8\times 3=29.4\text{ m s}^{-1}.$
Use $s=ut+12at2s=ut+\frac12at^2$ :
$m.s=0+\frac12\times 9.8\times 3^2=44.1\text{ m}.$
Its speed is 29.4 m s⁻¹ and it falls 44.1 m.

Vertical Projection

If upwards is chosen as positive, then

s−2.a=-g=-9.8\text{ m s}^{-2}.

If downwards is chosen as positive, then $a = + g$ . Either convention works, provided you use it consistently for displacement, velocity, acceleration and forces.

Example

Finding maximum height

A ball is projected vertically upwards at $s−119.6\text{ m s}^{-1}$ . Find the greatest height it reaches above its starting point, ignoring air resistance.

Take upwards as positive. Then $s−1u=19.6\text{ m s}^{-1}$ and $s−2a=-9.8\text{ m s}^{-2}$ .
At the greatest height, the instantaneous velocity is $v = 0$ . Use
$v^2=u^2+2as.$
Substitute the known values:
$0=19.6^2+2(-9.8)s.$
Rearrange:
$m.s=\frac{19.6^2}{19.6}=19.6\text{ m}.$
The greatest height above the point of projection is 19.6 m.

Common Mistake

Zero velocity does not mean zero acceleration

At the highest point of a vertical flight, the velocity is zero for an instant, but weight still acts. The acceleration remains $s−29.8\text{ m s}^{-2}$ downwards.

The Constant-Gravity Model

Using a constant value of $g$ is a mathematical model. It assumes that:

The body remains close enough to Earth's surface for gravitational acceleration to be constant.
Earth can be treated as fixed.
Air resistance is ignored unless the question includes it.
The body can usually be modelled as a particle, so its size and rotation are unimportant.

These assumptions make the acceleration constant, allowing you to use the constant-acceleration formulae.

Common Mistake

When constant-acceleration formulae apply

You may use the standard SUVAT equations only over a period in which the acceleration is constant. If a resistive force changes with speed, the acceleration may not remain constant.

Exam technique

In the exam

Draw a force diagram and include weight as $m g$ vertically downwards.
State your positive direction before applying $F = ma$ or the constant-acceleration formulae.
Use $s−2g=9.8\text{ m s}^{-2}$ unless another value is given, and distinguish carefully between mass in kg and weight in N.
Check whether gravity is the only force: if it is, the acceleration is $g$ downwards; if it is not, find the resultant force first.
At a greatest height, set the velocity to zero, not the acceleration.

Self review

Check yourself

What is the weight of a body of mass 7.5 kg, and in which direction does it act?
A falling body has weight 80 N and upward resistance 20 N. How would you form its equation of motion?
Why does a vertically projected ball still have non-zero acceleration at its highest point?

What you'll learn

Mass, Weight and Gravity

Mass

Weight

Representing Weight in a Force Diagram

Using Weight with Newton's Second Law

Free Fall

Vertical Projection

The Constant-Gravity Model

Review 3.3.6 Weight by teaching Genie

3.3.6 Weight

What you'll learn

Mass, Weight and Gravity

Mass

Weight

Representing Weight in a Force Diagram

Using Weight with Newton's Second Law

Free Fall

Vertical Projection

The Constant-Gravity Model

Review 3.3.6 Weight by teaching Genie

3.3.6 Weight

What you'll learn

Mass, Weight and Gravity

Mass

Weight

Representing Weight in a Force Diagram

Using Weight with Newton's Second Law

Free Fall

Vertical Projection

The Constant-Gravity Model

Review 3.3.6 Weight by teaching Genie

3.3.6 Weight Revision Guide

3.3.6 Weight

What you'll learn

Mass, Weight and Gravity

Mass

Weight

Representing Weight in a Force Diagram

Using Weight with Newton's Second Law

Free Fall

Vertical Projection

The Constant-Gravity Model

Review 3.3.6 Weight by teaching Genie

3.3.6 Weight Revision Guide