Estimating forces

Definition

Resultant force

The single force that has the same effect on an object as all of the forces acting on it combined.

Definition

Estimate

An approximate value of a quantity worked out from sensible, rounded values rather than from exact measurements.

Most road forces cannot be measured directly, but they can be estimated from sensible values of mass and acceleration.
Newton's second law links them, $force=mass×acceleration\text{force} = \text{mass} \times \text{acceleration}$ , or $F = ma$ , an equation you need to recall.
The force calculated in this way is the resultant force in newtons, $N\text{N}$ , with the mass in kilograms and the acceleration in $m/s2\text{m/s}^2$ .
The resultant force always acts in the same direction as the acceleration, so during braking it acts backwards, opposite to the motion.
An estimate needs a typical value for each quantity, chosen from everyday experience and rounded.
The answer is rounded to one or two significant figures and is often described by its order of magnitude, such as a few thousand newtons.

Typical masses

Definition

Mass

The amount of matter in an object, measured in kilograms.

Adult: about $kg70\ \text{kg}$ .
Cyclist and bicycle together: about $kg80\ \text{kg}$ .
Family car with its driver: about $1000$ to $kg1500\ \text{kg}$ , so about $kg1200\ \text{kg}$ is a good working value.
Loaded van: about $kg3000\ \text{kg}$ .
Double-decker bus: about $kg12\,000\ \text{kg}$ .
Fully loaded lorry: up to about $kg44\,000\ \text{kg}$ , the legal limit for the largest lorries in the UK.
A mass given in tonnes must be converted into kilograms, using $kg1\ \text{tonne} = 1000\ \text{kg}$ , before it is used in $F = ma$ .

Speeding up and slowing down

To estimate the force for a change in speed, first estimate the acceleration from $\dfrac{v-u}{t}$ , then multiply it by the mass.
A car of mass $kg1200\ \text{kg}$ accelerating gently at $m/s22\ \text{m/s}^2$ needs a resultant force of about $N2400\ \text{N}$ .
A loaded lorry accelerates much more slowly, at about $m/s20.5\ \text{m/s}^2$ , but its large mass means the resultant force is still about $N20\,000\ \text{N}$ .
In an emergency stop on a dry road a car decelerates at about $m/s27\ \text{m/s}^2$ , so the braking force on a $kg1200\ \text{kg}$ car is about $N8000\ \text{N}$ .
A cyclist with a total mass of $kg80\ \text{kg}$ braking at $m/s23\ \text{m/s}^2$ needs a braking force of only about $N240\ \text{N}$ , because the mass is so much smaller.

Example

Braking force on a car

A car of mass $kg1200\ \text{kg}$ slows from $m/s13\ \text{m/s}$ to rest in $s2.5\ \text{s}$ at a junction.

\frac{v-u}{t} = \frac{0\ \text{m/s} - 13\ \text{m/s}}{2.5\ \text{s}} = -5.2\ \text{m/s}^2

1200\ \text{kg} \times (-5.2\ \text{m/s}^2) = -6240\ \text{N}

The resultant force is about $N6000\ \text{N}$ , and the negative sign shows that it acts backwards, opposite to the motion.

Driving force and resistive forces

Definition

Driving force

The forward force on a vehicle produced by its engine or motor turning the wheels.

Definition

Resistive force

A force, such as friction or air resistance, that acts in the opposite direction to the motion of an object.

Definition

Air resistance

The frictional force that acts on an object moving through air, in the opposite direction to its motion.

The driving force pushes a vehicle forwards, while resistive forces such as air resistance and friction act backwards.
The resultant force is the driving force minus the resistive forces, so $F = ma$ gives the resultant force and not the driving force.
When a car travels at a steady speed its acceleration is zero, so the driving force is equal to the total resistive force.
When the car accelerates, the driving force must be larger than the resistive forces by an amount equal to $ma$ .
The resistive forces on a car are roughly $N200\ \text{N}$ at town speeds and roughly $N500\ \text{N}$ at motorway speeds, because air resistance increases with speed.

Example

Driving force when accelerating

A car of mass $kg1200\ \text{kg}$ accelerates at $m/s21.5\ \text{m/s}^2$ while the resistive forces on it total $N400\ \text{N}$ .

\text{resultant force} = ma = 1200\ \text{kg} \times 1.5\ \text{m/s}^2 = 1800\ \text{N}

\text{driving force} = \text{resultant force} + \text{resistive forces} = 1800\ \text{N} + 400\ \text{N} = 2200\ \text{N}

The engine must provide a driving force of about $N2200\ \text{N}$ .

Forces from energy and distance

Definition

Work done

The energy transferred when a force moves an object through a distance in the direction of the force.

Definition

Kinetic energy

The energy held in the kinetic store of an object because it is moving.

When a braking distance is known instead of a time, the braking force can be estimated from energy.
The work done by the braking force equals the kinetic energy transferred, so $distance=12mv2\text{braking force} \times \text{braking distance} = \dfrac{1}{2}mv^2$ .
Rearranging gives $distance\text{braking force} = \dfrac{\frac{1}{2}mv^2}{\text{braking distance}}$ .
The same estimate can be found by using $v^2 - u^2 = 2as$ to find the deceleration and then $F = ma$ .
For a given vehicle and speed, halving the braking distance doubles the braking force needed.

Example

Braking force from a braking distance

A van of mass $kg3000\ \text{kg}$ stops from $m/s20\ \text{m/s}$ in a braking distance of $m40\ \text{m}$ .

\text{kinetic energy} = \frac{1}{2}mv^2 = \frac{1}{2} \times 3000\ \text{kg} \times (20\ \text{m/s})^2 = 600\,000\ \text{J}

\text{braking force} = \frac{600\,000\ \text{J}}{40\ \text{m}} = 15\,000\ \text{N}

The braking force is about $N15\,000\ \text{N}$ , which is about half the weight of the van.

Common Mistake

Do not use the weight of a vehicle, $m g$ , in place of the resultant force when it speeds up or slows down on a level road, because weight acts downwards, not along the road.
Do not substitute a mass in tonnes or a speed in $km/h\text{km/h}$ into $F = ma$ without converting it first.
Do not give the result of $F = ma$ as the driving force when resistive forces are acting, because it is only the resultant force.

Exam technique

Setting out an estimate

State the value you have assumed for each quantity, such as a car mass of about $kg1200\ \text{kg}$ , because sensible assumptions are credited as well as the calculation.
Recall $F = ma$ , $\dfrac{v-u}{t}$ , kinetic energy $\dfrac{1}{2}mv^2$ and work done $\text{force} \times \text{distance}$ , because you are expected to know all four.
Give the final answer to one or two significant figures with its unit, and state its direction where the question asks for it.
Check that the size is sensible, since forces on a cyclist are usually hundreds of newtons and forces on a car are usually thousands of newtons.

Self review

Recall the equation that links resultant force, mass and acceleration.
Estimate the resultant force on a $kg1200\ \text{kg}$ car accelerating at $m/s23\ \text{m/s}^2$ .
Explain why the driving force of a car is larger than the resultant force while it accelerates.
State how the driving force compares with the resistive forces when a car moves at a steady speed.
Estimate the braking force on a $kg1000\ \text{kg}$ car that stops from $m/s15\ \text{m/s}$ in $m25\ \text{m}$ .

Estimating forces

Definition

Resultant force

The single force that has the same effect on an object as all of the forces acting on it combined.

Definition

Estimate

An approximate value of a quantity worked out from sensible, rounded values rather than from exact measurements.

Most road forces cannot be measured directly, but they can be estimated from sensible values of mass and acceleration.
Newton's second law links them, $force=mass×acceleration\text{force} = \text{mass} \times \text{acceleration}$ , or $F = ma$ , an equation you need to recall.
The force calculated in this way is the resultant force in newtons, $N\text{N}$ , with the mass in kilograms and the acceleration in $m/s2\text{m/s}^2$ .
The resultant force always acts in the same direction as the acceleration, so during braking it acts backwards, opposite to the motion.
An estimate needs a typical value for each quantity, chosen from everyday experience and rounded.
The answer is rounded to one or two significant figures and is often described by its order of magnitude, such as a few thousand newtons.

Typical masses

Definition

Mass

The amount of matter in an object, measured in kilograms.

Adult: about $kg70\ \text{kg}$ .
Cyclist and bicycle together: about $kg80\ \text{kg}$ .
Family car with its driver: about $1000$ to $kg1500\ \text{kg}$ , so about $kg1200\ \text{kg}$ is a good working value.
Loaded van: about $kg3000\ \text{kg}$ .
Double-decker bus: about $kg12\,000\ \text{kg}$ .
Fully loaded lorry: up to about $kg44\,000\ \text{kg}$ , the legal limit for the largest lorries in the UK.
A mass given in tonnes must be converted into kilograms, using $kg1\ \text{tonne} = 1000\ \text{kg}$ , before it is used in $F = ma$ .

Speeding up and slowing down

To estimate the force for a change in speed, first estimate the acceleration from $\dfrac{v-u}{t}$ , then multiply it by the mass.
A car of mass $kg1200\ \text{kg}$ accelerating gently at $m/s22\ \text{m/s}^2$ needs a resultant force of about $N2400\ \text{N}$ .
A loaded lorry accelerates much more slowly, at about $m/s20.5\ \text{m/s}^2$ , but its large mass means the resultant force is still about $N20\,000\ \text{N}$ .
In an emergency stop on a dry road a car decelerates at about $m/s27\ \text{m/s}^2$ , so the braking force on a $kg1200\ \text{kg}$ car is about $N8000\ \text{N}$ .
A cyclist with a total mass of $kg80\ \text{kg}$ braking at $m/s23\ \text{m/s}^2$ needs a braking force of only about $N240\ \text{N}$ , because the mass is so much smaller.

Example

Braking force on a car

A car of mass $kg1200\ \text{kg}$ slows from $m/s13\ \text{m/s}$ to rest in $s2.5\ \text{s}$ at a junction.

\frac{v-u}{t} = \frac{0\ \text{m/s} - 13\ \text{m/s}}{2.5\ \text{s}} = -5.2\ \text{m/s}^2

1200\ \text{kg} \times (-5.2\ \text{m/s}^2) = -6240\ \text{N}

The resultant force is about $N6000\ \text{N}$ , and the negative sign shows that it acts backwards, opposite to the motion.

Driving force and resistive forces

Definition

Driving force

The forward force on a vehicle produced by its engine or motor turning the wheels.

Definition

Resistive force

A force, such as friction or air resistance, that acts in the opposite direction to the motion of an object.

Definition

Air resistance

The frictional force that acts on an object moving through air, in the opposite direction to its motion.

The driving force pushes a vehicle forwards, while resistive forces such as air resistance and friction act backwards.
The resultant force is the driving force minus the resistive forces, so $F = ma$ gives the resultant force and not the driving force.
When a car travels at a steady speed its acceleration is zero, so the driving force is equal to the total resistive force.
When the car accelerates, the driving force must be larger than the resistive forces by an amount equal to $ma$ .
The resistive forces on a car are roughly $N200\ \text{N}$ at town speeds and roughly $N500\ \text{N}$ at motorway speeds, because air resistance increases with speed.

Example

Driving force when accelerating

A car of mass $kg1200\ \text{kg}$ accelerates at $m/s21.5\ \text{m/s}^2$ while the resistive forces on it total $N400\ \text{N}$ .

\text{resultant force} = ma = 1200\ \text{kg} \times 1.5\ \text{m/s}^2 = 1800\ \text{N}

\text{driving force} = \text{resultant force} + \text{resistive forces} = 1800\ \text{N} + 400\ \text{N} = 2200\ \text{N}

The engine must provide a driving force of about $N2200\ \text{N}$ .

Forces from energy and distance

Definition

Work done

The energy transferred when a force moves an object through a distance in the direction of the force.

Definition

Kinetic energy

The energy held in the kinetic store of an object because it is moving.

When a braking distance is known instead of a time, the braking force can be estimated from energy.
The work done by the braking force equals the kinetic energy transferred, so $distance=12mv2\text{braking force} \times \text{braking distance} = \dfrac{1}{2}mv^2$ .
Rearranging gives $distance\text{braking force} = \dfrac{\frac{1}{2}mv^2}{\text{braking distance}}$ .
The same estimate can be found by using $v^2 - u^2 = 2as$ to find the deceleration and then $F = ma$ .
For a given vehicle and speed, halving the braking distance doubles the braking force needed.

Example

Braking force from a braking distance

A van of mass $kg3000\ \text{kg}$ stops from $m/s20\ \text{m/s}$ in a braking distance of $m40\ \text{m}$ .

\text{kinetic energy} = \frac{1}{2}mv^2 = \frac{1}{2} \times 3000\ \text{kg} \times (20\ \text{m/s})^2 = 600\,000\ \text{J}

\text{braking force} = \frac{600\,000\ \text{J}}{40\ \text{m}} = 15\,000\ \text{N}

The braking force is about $N15\,000\ \text{N}$ , which is about half the weight of the van.

Common Mistake

Do not use the weight of a vehicle, $m g$ , in place of the resultant force when it speeds up or slows down on a level road, because weight acts downwards, not along the road.
Do not substitute a mass in tonnes or a speed in $km/h\text{km/h}$ into $F = ma$ without converting it first.
Do not give the result of $F = ma$ as the driving force when resistive forces are acting, because it is only the resultant force.

Exam technique

Setting out an estimate

State the value you have assumed for each quantity, such as a car mass of about $kg1200\ \text{kg}$ , because sensible assumptions are credited as well as the calculation.
Recall $F = ma$ , $\dfrac{v-u}{t}$ , kinetic energy $\dfrac{1}{2}mv^2$ and work done $\text{force} \times \text{distance}$ , because you are expected to know all four.
Give the final answer to one or two significant figures with its unit, and state its direction where the question asks for it.
Check that the size is sensible, since forces on a cyclist are usually hundreds of newtons and forces on a car are usually thousands of newtons.

Self review

Recall the equation that links resultant force, mass and acceleration.
Estimate the resultant force on a $kg1200\ \text{kg}$ car accelerating at $m/s23\ \text{m/s}^2$ .
Explain why the driving force of a car is larger than the resultant force while it accelerates.
State how the driving force compares with the resistive forces when a car moves at a steady speed.
Estimate the braking force on a $kg1000\ \text{kg}$ car that stops from $m/s15\ \text{m/s}$ in $m25\ \text{m}$ .

8.1.8 Estimating forces in road situations

Estimating forces

Typical masses

Speeding up and slowing down

Braking force on a car

Driving force and resistive forces

Driving force when accelerating

Forces from energy and distance

Braking force from a braking distance

Setting out an estimate

Review 8.1.8 Estimating forces in road situations by teaching Genie

8.1.8 Estimating forces in road situations

Estimating forces

Typical masses

Speeding up and slowing down

Braking force on a car

Driving force and resistive forces

Driving force when accelerating

Forces from energy and distance

Braking force from a braking distance

Setting out an estimate

Review 8.1.8 Estimating forces in road situations by teaching Genie

8.1.8 Estimating forces in road situations

Estimating forces

Typical masses

Speeding up and slowing down

Braking force on a car

Driving force and resistive forces

Driving force when accelerating

Forces from energy and distance

Braking force from a braking distance

Setting out an estimate

Review 8.1.8 Estimating forces in road situations by teaching Genie

8.1.8 Estimating forces in road situations Revision Guide

8.1.8 Estimating forces in road situations

Estimating forces

Typical masses

Speeding up and slowing down

Braking force on a car

Driving force and resistive forces

Driving force when accelerating

Forces from energy and distance

Braking force from a braking distance

Setting out an estimate

Review 8.1.8 Estimating forces in road situations by teaching Genie

8.1.8 Estimating forces in road situations Revision Guide