In this section we will look at the effect of forces on objects and how we can make things move. This will
link together what you have learnt about motion and what you have learnt about forces.
Newton's first law (ESBKS)
Sir Isaac Newton was a scientist who lived in England (1642-1727) who was interested in the motion of
objects under various conditions. He suggested that a stationary object will remain stationary unless a
force acts on it and that a moving object will continue moving unless a force slows it down, speeds it
up or changes its direction of motion. From this he formulated what is known as Newton's first law of
motion:
Newton's first law of motion
An object continues in a state of rest or uniform motion (motion with a constant velocity) unless
it is acted on by an unbalanced (net or resultant) force.
This property of an object, to continue in its current state of motion unless acted upon by a net force,
is called inertia.
Let us consider the following situations:
An ice skater pushes herself away from the side of the ice rink and skates across the ice. She will
continue to move in a straight line across the ice unless something stops her. Objects are also like
that. If we kick a soccer ball across a soccer field, according to Newton's first law, the soccer ball
should keep on moving forever! However, in real life this does not happen. Is Newton's Law wrong? Not
really. Newton's first law applies to situations where there aren't any external forces present. This
means that friction is not present. In the case of the ice skater, the friction between the skates and
the ice is very little and she will continue moving for quite a distance. In the case of the soccer
ball, air resistance (friction between the air and the ball) and friction between the grass and the ball
is present and this will slow the ball down.
Let us look at the following two examples. We will start with a familiar example:
Seat belts:
We wear seat belts in cars. This is to protect us when the car is involved in an accident. If a car
is travelling at \(\text{120}\) \(\text{km·h$^{-1}$}\), the passengers in the car is also
travelling at \(\text{120}\) \(\text{km·h$^{-1}$}\). When the car suddenly stops a force is
exerted on the car (making it slow down), but not on the passengers. The passengers will carry on
moving forward at \(\text{120}\) \(\text{km·h$^{-1}$}\) according to Newton's first law. If
they are wearing seat belts, the seat belts will stop them by exerting a force on them and so
prevent them from getting hurt.
A spaceship is launched into space. The force of the exploding gases pushes the rocket through the
air into space. Once it is in space, the engines are switched off and it will keep on moving at a
constant velocity. If the astronauts want to change the direction of the spaceship they need to fire
an engine. This will then apply a force on the rocket and it will change its direction.
Worked example 8: Newton's first law in action
Why do passengers get thrown to the side when the car they are driving in goes around a
corner?
What happens before the car turns
Before the car starts turning both the passengers and the car are travelling at the same
velocity. (picture A)
What happens while the car turns
The driver turns the wheels of the car, which then exert a force on the car and the car
turns. This force acts on the car but not the passengers, hence (by Newton's first law) the
passengers continue moving with the same original velocity. (picture B)
Why passengers get thrown to the side?
If the passengers are wearing seat belts they will exert a force on the passengers until the
passengers' velocity is the same as that of the car (picture C). Without a seat belt the
passenger may hit the side of the car.
Textbook Exercise 2.4
If a passenger is sitting in a car and the car turns round a bend to the right, what
happens to the passenger? What happens if the car turns to the left?
Before the car starts turning both the passenger and the car are travelling at the same
velocity.
As the car turns to the right a force acts on the car but not the passengers, hence (by
Newton's first law) the passenger continues moving with the same original velocity. (In
other words the car turns, but the passenger does not).
The net result of this is that the passenger is pulled to the left as the car turns
right.
If the car instead turned to the left the passenger would be pulled to the right.
Helium is less dense than the air we breathe. Discuss why a helium balloon in a car
driving around a corner appears to violate Newton's first law and moves towards the
inside of the turn not the outside like a passenger.
As the car goes around the corner all the air keeps moving forward (it acts in the same
way that a passenger acts). This causes the air pressure on the one side of the car to
increase (this will be on the opposite side to the direction the car is turning). This
slight increase in the air pressure pushes the helium balloon to the other side of the
car.
Because of this it appears that the helium balloon does not obey Newton's first law.
Newton's second law of motion (ESBKT)
According to Newton's first law, things 'like to keep on doing what they are doing'. In other words, if
an object is moving, it tends to continue moving (in a straight line and at the same speed) and if an
object is stationary, it tends to remain stationary. So how do objects start moving?
Let us look at the example of a \(\text{10}\) \(\text{kg}\) box on a rough table. If we push lightly on
the box as indicated in the diagram, the box won't move. Let's say we applied a force of \(\text{100}\)
\(\text{N}\), yet the box remains stationary. At this point a frictional force of \(\text{100}\)
\(\text{N}\) is acting on the box, preventing the box from moving. If we increase the force, let's say
to \(\text{150}\) \(\text{N}\) and the box almost starts to move, the frictional force is \(\text{150}\)
\(\text{N}\). To be able to move the box, we need to push hard enough to overcome the friction and then
move the box. If we therefore apply a force of \(\text{200}\) \(\text{N}\) remembering that a frictional
force
of
\(\text{150}\) \(\text{N}\) is present, the 'first' \(\text{150}\) \(\text{N}\) will be used to overcome
or 'cancel' the friction and the other \(\text{50}\) \(\text{N}\) will be used to move (accelerate) the
block. In order to accelerate an object we must have a resultant force acting on the block.
Now, what do you think will happen if we pushed harder, lets say \(\text{300}\) \(\text{N}\)? Or, what do
you think will happen if the mass of the block was more, say \(\text{20}\) \(\text{kg}\), or what if it
was less? Let us investigate how the motion of an object is affected by mass and force.
A recommended experiment for formal assessment on Newton's second law of motion is also included in
this chapter. In this experiment learners will investigate the relationship between force and
acceleration (Newton's second law). You will need trolleys, different masses, inclined plane, rubber
bands, meter ruler, ticker tape apparatus, ticker timer, graph paper.
Newton's second law of motion
Aim
To investigate the relation between the acceleration of objects and the application of a constant
resultant force.
Method
A constant force of \(\text{20}\) \(\text{N}\), acting at an angle of
\(\text{60}\)\(\text{°}\) to the horizontal, is applied to a dynamics trolley.
Ticker tape attached to the trolley runs through a ticker timer of frequency
\(\text{20}\) \(\text{Hz}\) as the trolley is moving on the frictionless surface.
The above procedure is repeated 4 times, each time using the same force, but varying the
mass of the trolley as follows:
Case 1: \(\text{6,25}\) \(\text{kg}\)
Case 2: \(\text{3,57}\) \(\text{kg}\)
Case 3: \(\text{2,27}\) \(\text{kg}\)
Case 4: \(\text{1,67}\) \(\text{kg}\)
Shown below are sections of the four ticker tapes obtained. The tapes are marked with the
letters A, B, C, D, etc. A is the first dot, B is the second dot and so on. The distance
between each dot is also shown.
Instructions:
Use each tape to calculate the instantaneous velocity (in \(\text{m·s$^{-1}$}\)) of
the trolley at points B and F (remember to convert the distances to m first!). Use these
velocities to calculate the trolley's acceleration in each case.
Tabulate the mass and corresponding acceleration values as calculated in each case.
Ensure that each column and row in your table is appropriately labelled.
Draw a graph of acceleration vs. mass, using a scale of \(\text{1}\) \(\text{cm}\) =
\(\text{1}\) \(\text{m·s$^{-2}$}\) on the y-axis and \(\text{1}\) \(\text{cm}\) =
\(\text{1}\) \(\text{kg}\) on the x-axis.
Use your graph to read off the acceleration of the trolley if its mass is \(\text{5}\)
\(\text{kg}\).
Write down a conclusion for the experiment.
You will have noted in the investigation above that the heavier the trolley is, the slower it moved when
the force was constant. The acceleration is inversely proportional to the mass. In mathematical
terms:
\(a\propto \frac{1}{m}\)
In a similar investigation where the mass is kept constant, but the applied force is varied, you will
find that the bigger the force is, the faster the object will move. The acceleration of the trolley is
therefore directly proportional to the resultant force. In mathematical terms:
\(a\propto F.\)
Rearranging the above equations, we get a \(\propto\)\(\frac{F}{m}\) or
\(F = ma\).
Remember that both force and acceleration are vectors quantities. The acceleration is in the same
direction as the force that is being applied. If multiple forces are acting simultaneously then we only
need to work with the resultant force or net force.
Newton's second law of motion
If a resultant force acts on a body, it will cause the body to accelerate in the direction of the
resultant force. The acceleration of the body will be directly proportional to the resultant
force and inversely proportional to the mass of the body. The mathematical representation
is:\[\vec{F}_{net} = m\vec{a}\]
Force is a vector quantity. Newton's second law of motion should be applied to the
\(y\)- and \(x\)-directions separately. You can use the resulting \(y\)- and \(x\)-direction resultants
to calculate the overall resultant as we saw in the previous chapter.
Newton's second law can be applied to a variety of situations. We will look at the main types of
examples that you need to study.
Worked example 9: Newton's second law: Box on a surface
A \(\text{10}\) \(\text{kg}\) box is placed on a table. A horizontal force of magnitude
\(\text{32}\) \(\text{N}\) is applied to the box. A frictional force of magnitude
\(\text{7}\) \(\text{N}\) is present between the surface and the box.
Draw a force diagram indicating all of the forces acting on the box.
Calculate the acceleration of the box.
Identify the horizontal forces and draw a force diagram
We only look at the forces acting in a horizontal direction (left-right) and not vertical
(up-down) forces. The applied force and the force of friction will be included. The force of
gravity, which is a vertical force, will not be included.
Calculate the acceleration of the box
Remember that we consider the \(y\)- and \(x\)-directions separately. In
this problem we can ignore the \(y\)-direction because the box is resting on a table with
the gravitational force balanced by the normal force.
We have been given:
Applied force \({F}_{1}=\text{32}\text{ N}\)
Frictional force \({F}_{f}=-\text{7}\text{ N}\)
Mass \(m=\text{10}\text{ kg}\)
To calculate the acceleration of the box we will be using the equation
\(\vec{F}_{R}=m\vec{a}\).
Therefore: \begin{align*}
\vec{F}_R &= m\vec{a} \\
\vec{F}_1+\vec{F}_f &= (\text{10})\vec{a} \\
(\text{32}-\text{7}) &= (\text{10})\vec{a} \\
\text{25} &= (\text{10})\vec{a} \\
\vec{a} &= \text{2,5}\text{ m·s$^{-2}$}\ \text{to the left.}
\end{align*}
Worked example 10: Newton's second law: box on a surface
Two crates, \(\text{10}\) \(\text{kg}\) and \(\text{15}\) \(\text{kg}\) respectively, are
connected with a thick rope according to the diagram. A force, to the right, of
\(\text{500}\) \(\text{N}\) is applied. The boxes move with an acceleration of \(\text{2}\)
\(\text{m·s$^{-2}$}\) to the right. One third of the total frictional force is acting
on the \(\text{10}\) \(\text{kg}\) block and two thirds on the \(\text{15}\) \(\text{kg}\)
block. Calculate:
the magnitude and direction of the total frictional force present.
the magnitude of the tension in the rope at T.
Important: when you have tension in a rope in a problem like this you
need to know that both ends of the rope apply a force with the same
magnitude but opposite direction. We call this force the
tension and you should examine the force diagrams in this problem
carefully.
Evaluate what is given
To make things easier lets give the two crates labels, let us call the \(\text{10}\)
\(\text{kg}\) crate number 2 and the \(\text{15}\) \(\text{kg}\) crate number 1.
We have two crates that have overall has an acceleration that is given. The fact that the
crates are tied together with a rope means that they will both have the same acceleration.
They will also both feel the same force due to the tension in the rope.
We are told that there is friction but we are only given the relationship between the total
frictional force both crates experience and the fraction each one experiences. The total
friction, \(\vec{F}_{fT}\) will be the sum of the friction on crate 1, \(\vec{F}_{f1}\), and
the friction on crate 2, \(\vec{F}_{f2}\). We are told that
\(\vec{F}_{f1}=\frac{\text{2}}{\text{3}}\vec{F}_{fT}\) and
\(\vec{F}_{f2}=\frac{\text{1}}{\text{3}}\vec{F}_{fT}\). We know the blocks are accelerating
to the right and we know that friction will be in the direction opposite to the direction of
motion and parallel to the surface.
Draw force diagrams
The diagram for crate 1 will be:
The diagram for crate 1 (indicated by blue dotted lines) will be:
Where:
\(\vec{F}_{g1}\) is the force due to gravity on the first crate
\(\vec{N}_{1}\) is the normal force from the surface on the first crate
\(\vec{T}\) is the force of tension in the rope
\(\vec{F}_{applied}\) is the external force being applied to the crate
\(\vec{F}_{f1}\) is the force of friction on the first crate
The diagram for crate 2 (indicated by orange dashed lines) will be:
Where:
\(\vec{F}_{g2}\) is the force due to gravity on the second crate
\(\vec{N}_{2}\) is the normal force from the surface on the second crate
\(\vec{T}\) is the force of tension in the rope
\(\vec{F}_{f2}\) is the force of friction on the second crate
Apply Newton's second law of motion
The problem tells us that the crates are accelerating along the \(x\)-direction which means
that the forces in the \(y\)-direction do not result in a net force. We can treat the
different directions separately so we only need to consider the \(x\)-direction.
We are working with one dimension and can choose a sign convention to indicate the direction
of the vectors. We choose vectors to the right (or in the positive \(x\)-direction) to be
positive.
We can now apply Newton's second law of motion to the first crate because we know the
acceleration and we know all the forces acting on the crate. Using positive to indicate a
force to the right we know that \({F}_{res1} = F_{applied}-{F}_{f1}-T\)
\begin{align*}
\vec{F}_{res1} &=m_1\vec{a} \\
F_{applied}-{F}_{f1}-T &=m_1a \\
F_{applied}-\frac{\text{2}}{\text{3}}{F}_{fT}-T &=m_1a \\
(500)-\frac{\text{2}}{\text{3}}{F}_{fT}-T &=(\text{15})(2) \\
-T &=(\text{15})(2) -(500)+\frac{\text{2}}{\text{3}}{F}_{fT}
\end{align*}
Now apply Newton's second law of motion to the second crate because we know the acceleration
and we know all the forces acting on the crate. We know that \({F}_{res2} = T-{F}_{f2}\).
Note that tension is in the opposite direction.
\begin{align*}
\vec{F}_{res2} &=m_2\vec{a} \\
T-{F}_{f2} &=m_2a \\
T - \frac{\text{1}}{\text{3}}{F}_{fT} &=m_2a \\
T &=(\text{10})(2) +\frac{\text{1}}{\text{3}}{F}_{fT}
\end{align*}
Solve simultaneously
We have used Newton's second law of motion to create two equations with two unknowns, this
means we can solve simultaneously. We solved for \(T\) in the equations above but one
carries a negative sign so if we add the two equations we will subtract out the value of the
tension allowing us to solve for \({F}_{fT}\):
\begin{align*}
(T) + (-T) &= ((\text{10})(2) +\frac{\text{1}}{\text{3}}{F}_{fT}) + ((\text{15})(2)
-(500)+\frac{\text{2}}{\text{3}}{F}_{fT}) \\
0 & = \text{20} + \text{30} - \text{500} + \frac{\text{1}}{\text{3}}{F}_{fT} +
\frac{\text{2}}{\text{3}}{F}_{fT} \\
0 & = -\text{450} + {F}_{fT} \\
{F}_{fT} & = \text{450}\text{ N}
\end{align*}
We can substitute the magnitude of \({F}_{fT}\) into the equation for crate 2 to determine
the magnitude of the tension:
\begin{align*}
T &=(\text{10})(2) +\frac{\text{1}}{\text{3}}{F}_{fT} \\
T &=(\text{10})(2) +\frac{\text{1}}{\text{3}}(\text{450}) \\
T &= \text{20} +\text{150} \\
T &= \text{170}\text{ N}
\end{align*}
Quote final answers
The total force due to friction is \(\text{450}\) \(\text{N}\) to the left. The magnitude of
the force of tension is \(\text{170}\) \(\text{N}\).
Worked example 11: Newton's second law: box on a surface (Alternative Method)
Two crates, \(\text{10}\) \(\text{kg}\) and \(\text{15}\) \(\text{kg}\) respectively, are
connected with a thick rope according to the diagram. A force, to the right, of
\(\text{500}\) \(\text{N}\) is applied. The boxes move with an acceleration of \(\text{2}\)
\(\text{m·s$^{-2}$}\) to the right. One third of the total frictional force is acting
on the \(\text{10}\) \(\text{kg}\) block and two thirds on the \(\text{15}\) \(\text{kg}\)
block. Calculate:
the magnitude and direction of the total frictional force present.
the magnitude of the tension in the rope at T.
Important: when you have tension in a rope in a problem like this you
need to know that both ends of the rope apply a force with the same
magnitude but opposite direction. We call this force the
tension and you should examine the force diagrams in this problem
carefully.
Draw a force diagram
Always draw a force diagram although the question might not ask for it. The acceleration of
the whole system is given, therefore a force diagram of the whole system will be drawn.
Because the two crates are seen as a unit, the force diagram will look like this:
Calculate the frictional force
To find the frictional force we will apply Newton's second law. We are given the mass
(\(\text{10}\) + \(\text{15}\) \(\text{kg}\)) and the acceleration (\(\text{2}\)
\(\text{m·s$^{-2}$}\)). Choose the direction of motion to be the positive direction (to
the right is positive).
The frictional force is \(\text{450}\) \(\text{N}\) opposite to the direction of motion (to
the left).
Find the tension in the rope
To find the tension in the rope we need to look at one of the two crates on their own. Let's
choose the \(\text{10}\) \(\text{kg}\) crate. Firstly, we need to draw a force diagram:
Figure 2.5:
Force diagram of \(\text{10}\) \(\text{kg}\) crate.
The frictional force on the \(\text{10}\) \(\text{kg}\) block is one third of the total,
therefore:
The negative answer here means that the force is in the direction opposite to the motion, in
other words to the left, which is correct. However, the question asks for the magnitude of
the force and your answer will be quoted as \(\text{170}\) \(\text{N}\).
Worked example 12: Newton's second law: man pulling a box
A man is pulling a \(\text{20}\) \(\text{kg}\) box with a rope that makes an angle of
\(\text{60}\)\(\text{°}\) with the horizontal. If he applies a force of magnitude
\(\text{150}\) \(\text{N}\) and a frictional force of magnitude \(\text{15}\) \(\text{N}\)
is present, calculate the acceleration of the box.
Draw a force diagram or free body diagram
The motion is horizontal and therefore we will only consider the forces in a horizontal
direction. Remember that vertical forces do not influence horizontal motion and vice versa.
Calculate the horizontal component of the applied force
We first need to choose a direction to be the positive direction in this problem. We choose
the positive \(x\)-direction (to the right) to be positive.
The applied force is acting at an angle of \(\text{60}\)\(\text{°}\) to the horizontal.
We can only consider forces that are parallel to the motion. The horizontal component of the
applied force needs to be calculated before we can continue:
\begin{align*}
F_x & = F_{applied}\cos(\theta) \\
&= \text{150}\cos(\text{60}\text{°}) \\
& = \text{75}\text{ N}
\end{align*}
Calculate the acceleration
To find the acceleration we apply Newton's second law:
\begin{align*}
F_R&=ma \\
F_x+F_f&=(\text{20})a \\
(\text{75}) + (-\text{15})&= (\text{20})a\\
a & = \frac{\text{60}}{\text{20}} \\
a & = \text{3}\text{ m·s$^{-2}$}
\end{align*}The acceleration is \(\text{3}\) \(\text{m·s$^{-2}$}\) to the right.
Worked example 13: Newton's second law: truck and trailer
A \(\text{2 000}\) \(\text{kg}\) truck pulls a \(\text{500}\) \(\text{kg}\) trailer with
a constant acceleration. The engine of the truck produces a thrust of \(\text{10 000}\)
\(\text{N}\). Ignore the effect of friction. Calculate the:
acceleration of the truck; and
tension in the tow bar T between the truck and the trailer, if the tow bar makes an
angle of \(\text{25}\)\(\text{°}\) with the horizontal.
Figure 2.6:
Truck pulling a trailer.
Draw a force diagram
Draw a force diagram indicating all the forces on the system as a whole:
Figure 2.7:
Free body diagrams for truck pulling a trailer.
Apply Newton's second law of motion
We choose the positive \(x\)-direction to be the positive direction. We only need to consider
the horizontal forces. Using only the horizontal forces means that we first need to note
that the tension acts at an angle to the horizontal and we need to use the horizontal
component of the tension in our calculations.
The horizontal component has a magnitude of \(T\cos(\text{25}\text{°})\).
In the absence of friction, the only force that causes the system to accelerate is the thrust
of the engine. If we now apply Newton's second law of motion to the truck we have:
\begin{align*}
\vec{F}_{Rtruck} &= m_{truck}\vec{a}\ \text{(we use signs to indicate direction)} \\
{F}_{engine} - T\cos(\text{25}\text{°}) &= (\text{2 000})a \\
(\text{10 000}) - T\cos(\text{25}\text{°}) &= (\text{2 000})a \\
a & = \frac{(\text{10 000}) - T\cos(\text{25}\text{°})}{(\text{2 000})}
\end{align*}
We now apply the same principle to the trailer (remember that the direction of the tension
will be opposite to the case of the truck):
\begin{align*}
\vec{F}_{Rtrailer} &= m_{trailer}\vec{a}\ \text{(we use signs to indicate direction)} \\
T\cos(\text{25}\text{°}) &= (\text{500})a \\
a & = \frac{T\cos(\text{25}\text{°})}{(\text{500})}
\end{align*}
We now have two equations and two unknowns so we can solve simultaneously. We subtract the
second equation from the first to get:
\begin{align*}
(a) - (a) &=(\frac{(\text{10 000}) -
T\cos(\text{25}\text{°})}{(\text{2 000})}) -
(\frac{T\cos(\text{25}\text{°})}{(\text{500})}) \\
0 & = (\frac{(\text{10 000}) - T\cos(\text{25}\text{°})}{(\text{2 000})})
- (\frac{T\cos(\text{25}\text{°})}{(\text{500})})\\
& \text{(multiply through by \text{2 000})} \\
0 & = (\text{10 000}) - T\cos(\text{25}\text{°}) -
4T\cos(\text{25}\text{°}) \\
\text{5}T\cos(\text{25}\text{°}) & = (\text{10 000}) \\
T & = \frac{(\text{10 000})}{\text{5}\cos(\text{25}\text{°})} \\
T & = \text{2 206,76}\text{ N}
\end{align*}
Now substitute this result back into the second equation to solve for the magnitude of \(a\)
\begin{align*}
a & = \frac{T\cos(\text{25}\text{°})}{(\text{500})}\\
& = \frac{(\text{2 206,76})\cos(\text{25}\text{°})}{(\text{500})}\\
& = \text{4,00}\text{ m·s$^{-2}$}
\end{align*}
temp text
Object on an inclined plane
In an earlier section we looked at the components of the gravitational force parallel and
perpendicular to the slope for objects on an inclined plane. When we look at problems on an inclined
plane we need to include the component of the gravitational force parallel to the slope.
Think back to the pictures of the book on a table, as one side of the table is lifted higher the book
starts to slide. Why? The book starts to slide because the component of the gravitational force
parallel to the surface of the table gets larger for the larger angle of inclination. This is like
the applied force and it eventually becomes larger than the frictional force and the book
accelerates down the table or inclined plane.
The force of gravity will also tend to push an object 'into' the slope. This is the component of the
force perpendicular to the slope. There is no movement in this direction as this force is balanced
by the slope pushing up against the object. This “pushing force” is the normal force (N)
which we have already learnt about and is equal in magnitude to the perpendicular component of the
gravitational force, but opposite in direction.
Do not use the abbreviation \(W\) for weight as it is used to abbreviate 'work'. Rather use the
force of gravity \({F}_{g}\) for weight.
Worked example 14: Newton's second law: box on inclined plane
A body of mass \(M\) is at rest on an inclined plane due to friction.
What of the following option is the magnitude of the frictional force acting on the body?
\(F_g\)
\(F_g\cos(θ)\)
\(F_g\sin(θ)\)
\(F_g\tan(θ)\)
Analyse the situation
The question asks us to determine the magnitude of the frictional force. The body is said to
be at rest on the plane, which means that it is not moving and therefore the acceleration is
zero. We know that the frictional force will act parallel to the slope. If there were no
friction the box would slide down the slope so friction must be acting up the slope. We also
know that there will be a component of gravity perpendicular to the slope and parallel to
the slope. The free body diagram for the forces acting on the block is:
Determine the magnitude of the frictional force
We can apply Newton's second law to this problem. We know that the object is not moving so
the resultant acceleration is zero. We choose up the slope to be the positive direction.
Therefore:
\begin{align*}
\vec{F}_R &= m\vec{a}\; \text{using signs for direction}\\
F_f - F_g\sin(\theta) &= m (0)\\
F_f - F_g\sin(\theta) &= m (0)\\
F_f & = F_g\sin(\theta)
\end{align*}
Quote your final answer
The force of friction has the same magnitude as the component of the force of gravitation
parallel to the slope, \(F_g\sin(\theta)\).
Worked example 15: Newton's second law: object on an incline
A force of magnitude \(T=\text{312} \text{N}\) up an incline is required to keep a body
at rest on a frictionless inclined plane which makes an angle of
\(\text{35}\)\(\text{°}\) with the horizontal. Calculate the magnitudes of the force
due to gravity and the normal force, giving your answers to three significant figures.
Find the magnitude of \(\vec{F}_g\)
We are usually asked to find the magnitude of \(\vec{T}\), but in this case \(\vec{T}\) is
given and we are asked to find \(\vec{F}_g\). We can use the same equation. \(T\) is the
force that balances the component of \(\vec{F}_g\) parallel to the plane (\({F}_{gx}\)) and
therefore it has the same magnitude.
We can apply Newton's second law to this problem. We know that the object is not moving so
the resultant acceleration is zero. We choose up the slope to be the positive direction.
Therefore:
\begin{align*}
\vec{F}_R &= m\vec{a}\; \text{using signs for direction}\\
T - F_g\sin(\theta) &= m (0)\\
F_g & = \frac{T}{\sin(\theta)} \\
& = \frac{\text{312}}{\sin(\text{35}\text{°})} \\
& = \text{543,955}\text{ N}
\end{align*}
Find the magnitude of \(\vec{N}\)
We treat the forces parallel and perpendicular to the slope separately. The block is
stationary so the acceleration perpendicular to the slope is zero. Once again we can apply
Newton's second law of motion. We choose the direction of the normal force as the positive
direction.
\begin{align*}
\vec{F}_R &= m\vec{a}\; \text{using signs for direction}\\
N - F_g\cos(\theta) &= m (0)\\
N & = F_g\cos(\theta)
\end{align*}
We could substitute in the value of \(F_g\) calculated earlier. We would like to illustrate
that there is another approach to adopt to ensure you get the correct answer even if you
made a mistake calculating \(F_g\). \(F_g\cos(\theta)\) can also be determined with the use
of trigonometric ratios. We know from the previous part of the question that \(T =
F_g\sin(\theta)\). We also know that
\begin{align*}
\tan(\theta) &=\frac{F_g\sin(\theta)}{F_g\cos(\theta)}\\
&=\frac{T}{N}\\
N&=\frac{T}{\tan(\theta)}\\
& = \frac{\text{312}}{\tan(\text{35}\text{°})} \\
& = \text{445,58}\text{ N}
\end{align*}
Note that the question asks that the answers be given to 3 significant figures. We therefore
round \(\vec{N}\) from \(\text{445,58}\) \(\text{N}\) up to \(\text{446}\) \(\text{N}\)
perpendicular to the surface upwards and \(\vec{T}\) from \(\text{543,955}\) \(\text{N}\) up
to \(\text{544}\) \(\text{N}\) parallel to the plane, up the slope.
temp text
Lifts and rockets
So far we have looked at objects being pulled or pushed across a surface, in other words motion
parallel to the surface the object rests on. Here we only considered forces parallel to the surface,
but we can also lift objects up or let them fall. This is vertical motion where only vertical forces
are being considered.
Let us consider a \(\text{500}\) \(\text{kg}\) lift, with no passengers, hanging on a cable. The
purpose of the cable is to pull the lift upwards so that it can reach the next floor or lower the
lift so that it can move downwards to the floor below. We will look at five possible stages during
the motion of the lift and apply our knowledge of Newton's second law of motion to the situation.
The 5 stages are:
A stationary lift suspended above the ground.
A lift accelerating upwards.
A lift moving at a constant velocity.
A lift decelerating (slowing down).
A lift accelerating downwards (the cable snaps!).
We choose the upwards direction to be the positive direction for this discussion.
Stage 1:
The \(\text{500}\) \(\text{kg}\) lift is stationary at the second floor of a tall building.
The lift is not accelerating. There must be a tension \(\vec{T}\) from the cable acting on the lift
and there must be a force due to gravity, \(\vec{F}_g\). There are no other forces present and we
can draw the free body diagram:
We apply Newton's second law to the vertical direction:
\begin{align*}
\vec{F}_R & = m_{\text{lift}} \vec{a}\ \text{(we use signs to indicate direction)} \\
T - F_g & = m_{\text{lift}} (0) \\
T & = F_g
\end{align*}
The forces are equal in magnitude and opposite in direction.
Stage 2:
The lift moves upwards at an acceleration of \(\text{1}\) \(\text{m·s$^{-2}$}\).
If the lift is accelerating, it means that there is a resultant force in the direction of the motion.
This means that the force acting upwards is now greater than the force due to gravity \(\vec{F}_g\)
(down). To find the magnitude of the \(\vec{T}\) applied by the cable we can do the following
calculation: (Remember we have chosen upwards as positive.)
We apply Newton's second law to the vertical direction:
\begin{align*}
\vec{F}_R & = m_{\text{lift}} \vec{a}\ \text{(we use signs to indicate direction)} \\
T - F_g & = m_{\text{lift}} (\text{1}) \\
T & = F_g + m_{\text{lift}} (\text{1})
\end{align*}
The answer makes sense as we need a bigger force upwards to cancel the effect of gravity as well as
have a positive resultant force.
Stage 3:
The lift moves at a constant velocity.
When the lift moves at a constant velocity, the acceleration is zero,
\begin{align*}
\vec{F}_R & = m_{\text{lift}} \vec{a}\ \text{(we use signs to indicate direction)} \\
T - F_g & = m_{\text{lift}} (0) \\
T & = F_g
\end{align*}
The forces are equal in magnitude and opposite in direction. It is common mistake to
think that because the lift is moving there is a net force acting on it. It is only if it is
accelerating that there is a net force acting.
Stage 4:
The lift slows down at a rate of \(\text{2}\) \(\text{m·s$^{-2}$}\). The lift was moving upwards
so this means that it is decelerating or accelerating in the direction opposite to the direction of
motion. This means that the acceleration is in the negative direction.
\begin{align*}
\vec{F}_R & = m_{\text{lift}} \vec{a}\ \text{(we use signs to indicate direction)} \\
T - F_g & = m_{\text{lift}} (-\text{2}) \\
T & = F_g -\text{2}m_{\text{lift}}
\end{align*}
As the lift is now slowing down there is a resultant force downwards. This means that the force
acting downwards is greater than the force acting upwards.
This makes sense as we need a smaller force upwards to ensure that the resultant force is downward.
The force of gravity is now greater than the upward pull of the cable and the lift will slow down.
Stage 5:
The cable snaps.
When the cable snaps, the force that used to be acting upwards is no longer present. The only force
that is present would be the force of gravity. The lift will fall freely and its acceleration.
Apparent weight
Your weight is the magnitude of the gravitational force acting on your body. When you stand in a lift
that is stationery and then starts to accelerate upwards you feel you are pressed into the floor
while the lift accelerates. You feel like you are heavier and your weight is more. When you are in a
stationery lift that starts to accelerate downwards you feel lighter on your feet. You feel like
your weight is less.
Weight is measured through normal forces. When the lift accelerates upwards you feel a greater normal
force acting on you as the force required to accelerate you upwards in addition to balancing out the
gravitational force.
When the lift accelerates downwards you feel a smaller normal force acting on you. This is because a
net force downwards is required to accelerate you downwards. This phenomenon is called
apparent weight because your weight didn't actually change.
temp text
Rockets
As with lifts, rockets are also examples of objects in vertical motion. The force of gravity pulls
the rocket down while the thrust of the engine pushes the rocket upwards. The force that the engine
exerts must overcome the force of gravity so that the rocket can accelerate upwards. The worked
example below looks at the application of Newton's second law in launching a rocket.
Worked example 16: Newton's second law: rocket
A rocket (of mass \(\text{5 000}\) \(\text{kg}\)) is launched vertically upwards into
the sky at an acceleration of \(\text{20}\) \(\text{m·s$^{-2}$}\). If the magnitude of
the force due to gravity on the rocket is \(\text{49 000}\) \(\text{N}\), calculate the
magnitude and direction of the thrust of the rocket's engines.
We are asked to find the thrust of the rocket engine \(\vec{F}\).
Find the thrust of the engine
We will apply Newton's second law:
\begin{align*}
\vec{F}_R & = m\vec{a}\ \text{(using signs to indicate direction)} \\
F - F_g & = (\text{5 000})(\text{20}) \\
F - (\text{49 000}) & = (\text{5 000})(\text{20}) \\
F &= \text{149 000}\text{ N}
\end{align*}
Quote your final answer
The force due to the thrust is \(\text{149 000}\) \(\text{N}\) upwards.
Worked example 17: Rockets
How do rockets accelerate in space?
Gas explodes inside the rocket.
This exploding gas exerts a force on each side of the rocket
(as shown in the picture below of the explosion chamber inside the
rocket).
Due to the symmetry of the situation, all the forces exerted
on the rocket are balanced by forces on the opposite side, except for the force
opposite the open side. This force on the upper surface is unbalanced.
This is therefore the resultant force acting on the rocket and it
makes the rocket accelerate forwards.
Textbook Exercise 2.5
A tug is capable of pulling a ship with a force of \(\text{100}\) \(\text{kN}\). If
two such tugs are pulling on one ship, they can produce any force ranging from a
minimum of \(\text{0}\) \(\text{kN}\) to a maximum of \(\text{200}\) \(\text{kN}\).
Give a detailed explanation of how this is possible. Use diagrams to support your
result.
We start off with the two tug boats pulling in opposite directions:
The resultant force is \(\text{0}\) \(\text{kN}\) since the tug boats are pulling
with equal forces in opposite directions.
If the two tugboats pull in the same direction then we get:
The resultant force is \(\text{200}\) \(\text{kN}\) since the tug boats are pulling
with equal forces in the same direction.
To get any force between these two extremes, one tugboat will have to be pulling on
the ship at a different angle to the second tugboat, for example:
Note that the resultant force in this situation is less than \(\text{200}\)
\(\text{kN}\) (You can check this by using any of the vector addition techniques).
A car of mass \(\text{850}\) \(\text{kg}\) accelerates at \(\text{2}\)
\(\text{m·s$^{-2}$}\). Calculate the magnitude of the resultant force that is
causing the acceleration.
\begin{align*}
F & = ma \\
& = (\text{850})(\text{2}) \\
& = \text{1 700}\text{ N}
\end{align*}
Find the force needed to accelerate a \(\text{3}\) \(\text{kg}\) object at
\(\text{4}\) \(\text{m·s$^{-2}$}\).
\begin{align*}
F & = ma \\
& = (\text{3})(\text{4}) \\
& = \text{12}\text{ N}
\end{align*}
Calculate the acceleration of an object of mass \(\text{1 000}\) \(\text{kg}\)
accelerated by a force of magnitude \(\text{100}\) \(\text{N}\).
\begin{align*}
F & = ma \\
a & = \frac{F}{m} \\
& = \frac{\text{100}}{\text{1 000}} \\
& = \text{0,1}\text{ m·s$^{-2}$}
\end{align*}
An object of mass \(\text{7}\) \(\text{kg}\) is accelerating at \(\text{2,5}\)
\(\text{m·s$^{-2}$}\). What resultant force acts on it?
\begin{align*}
F & = ma \\
& = (\text{7})(\text{2,5}) \\
& = \text{17,5}\text{ N}
\end{align*}
Find the mass of an object if a force of \(\text{40}\) \(\text{N}\) gives it an
acceleration of \(\text{2}\) \(\text{m·s$^{-2}$}\).
\begin{align*}
F & = ma \\
m & = \frac{F}{a} \\
& = \frac{\text{40}}{\text{2}} \\
& = \text{20}\text{ kg}
\end{align*}
Find the acceleration of a body of mass \(\text{1 000}\) \(\text{kg}\) that has
a force with a magnitude of \(\text{150}\) \(\text{N}\) acting on it.
\begin{align*}
F & = ma \\
a & = \frac{F}{m} \\
& = \frac{\text{150}}{\text{1 000}} \\
& = \text{0,15}\text{ m·s$^{-2}$}
\end{align*}
Find the mass of an object which is accelerated at \(\text{3}\)
\(\text{m·s$^{-2}$}\) by a force of magnitude \(\text{25}\) \(\text{N}\).
\begin{align*}
F & = ma \\
m & = \frac{F}{a} \\
& = \frac{\text{25}}{\text{3}} \\
& = \text{8,33}\text{ kg}
\end{align*}
Determine the acceleration of a mass of \(\text{24}\) \(\text{kg}\) when a force of
magnitude \(\text{6}\) \(\text{N}\) acts on it. What is the acceleration if the
force were doubled and the mass was halved?
We first find the acceleration:
\begin{align*}
F & = ma \\
a & = \frac{F}{m} \\
& = \frac{\text{6}}{\text{24}} \\
& = \text{0,25}\text{ m·s$^{-2}$}
\end{align*}
If the force is doubled and the mass is halved then the acceleration will be four
times this amount: \(\text{1}\) \(\text{m·s$^{-2}$}\).
You can check this by doubling the force and halving the mass.
A mass of \(\text{8}\) \(\text{kg}\) is accelerating at \(\text{5}\)
\(\text{m·s$^{-2}$}\).
Determine the resultant force that is causing the acceleration.
\begin{align*}
F & = ma \\
& = (\text{8})(\text{5}) \\
& = \text{40}\text{ N}
\end{align*}
What acceleration would be produced if we doubled the force and reduced the mass
by half?
If we doubled the force and halved the mass we would get four times the
acceleration or \(\text{20}\) \(\text{m·s$^{-2}$}\). You can check this by
carrying out the calculation using double the force and half the mass.
A motorcycle of mass \(\text{100}\) \(\text{kg}\) is accelerated by a resultant force
of \(\text{500}\) \(\text{N}\). If the motorcycle starts from rest:
What is its acceleration?
\begin{align*}
F & = ma \\
a & = \frac{F}{m} \\
& = \frac{\text{500}}{\text{100}} \\
& = \text{5}\text{ m·s$^{-2}$}
\end{align*}
How fast will it be travelling after \(\text{20}\) \(\text{s}\)?
We can use the equations of motion (recall from grade 10: motion in one
dimension) to determine how fast it will be travelling:
A force of \(\text{200}\) \(\text{N}\), acting at \(\text{60}\)\(\text{°}\) to
the horizontal, accelerates a block of mass \(\text{50}\) \(\text{kg}\) along a
horizontal plane as shown.
Calculate the component of the \(\text{200}\) \(\text{N}\) force that accelerates
the block horizontally.
A toy rocket experiences a force due to gravity of magnitude \(\text{4,5}\)
\(\text{N}\) is supported vertically by placing it in a bottle. The rocket is then
ignited. Calculate the force that is required to accelerate the rocket vertically
upwards at \(\text{8}\) \(\text{m·s$^{-2}$}\).
A constant force of magnitude \(\text{70}\) \(\text{N}\) is applied vertically to a
block as shown. The block experiences a force due to gravity of \(\text{49}\)
\(\text{N}\). Calculate the acceleration of the block.
A student experiences a gravitational force of magnitude \(\text{686}\) \(\text{N}\)
investigates the motion of a lift. While he stands in the lift on a bathroom scale
(calibrated in newton), he notes three stages of his journey.
For \(\text{2}\) \(\text{s}\) immediately after the lift starts, the scale
reads \(\text{574}\) \(\text{N}\).
For a further \(\text{6}\) \(\text{s}\) it reads \(\text{686}\) \(\text{N}\).
For the final \(\text{2}\) \(\text{s}\) it reads \(\text{854}\) \(\text{N}\).
Answer the following questions:
Is the motion of the lift upward or downward? Give a reason for your answer.
Stage 1: Downwards. The scale reads less than the gravitational force that he
experiences.
Stage 2: Stationary. The scale reads the same as the gravitational force that he
experiences.
Stage 3: Upwards. The scale reads more than the gravitational force that he
experiences.
Write down the magnitude and the direction of the resultant force acting on the
student for each of the stages 1, 2 and 3.
Stage 1: \(\text{574}\) \(\text{N}\) downwards
Stage 2: The resultant force on the student is zero.
Stage 3: \(\text{854}\) \(\text{N}\) upwards
A car of mass \(\text{800}\) \(\text{kg}\) accelerates along a level road at
\(\text{4}\) \(\text{m·s$^{-2}$}\). A frictional force of \(\text{700}\)
\(\text{N}\) opposes its motion. What force is produced by the car's engine?
Two objects, with masses of \(\text{1}\) \(\text{kg}\) and \(\text{2}\) \(\text{kg}\)
respectively, are placed on a smooth surface and connected with a piece of string. A
horizontal force of \(\text{6}\) \(\text{N}\) is applied with the help of a spring
balance to the \(\text{1}\) \(\text{kg}\) object. Ignoring friction, what will the
force acting on the \(\text{2}\) \(\text{kg}\) mass, as measured by a second spring
balance, be?
The force acting on the \(\text{2}\) \(\text{kg}\) block is \(\text{6}\)
\(\text{N}\). Since the surface is assumed to be frictionless, the applied force on
the \(\text{1}\) \(\text{kg}\) block is equal to the force experienced by the
\(\text{2}\) \(\text{kg}\) block.
A rocket of mass \(\text{200}\) \(\text{kg}\) has a resultant force of
\(\text{4 000}\) \(\text{N}\) upwards on it.
What is its acceleration on the Earth, where it experiences a gravitational force
of \(\text{1 960}\) \(\text{N}\)?
The force on the rocket is in the upwards direction, while the force due to
gravity is in the downwards direction. Taking upwards as positive:
\begin{align*}
F_{R} & = ma \\
F_{g} + F_{\text{rocket}} & = ma \\
-\text{1 960} + \text{4 000} & = ma \\
a & = \frac{\text{2 040}}{\text{200}} \\
& = \text{10,2}\text{ m·s$^{-2}$}
\end{align*}
What driving force does the rocket engine need to exert on the back of the rocket
on the Earth?
On Earth the rocket engines need to overcome the gravitational force and so need
to exert a force of \(\text{1 960}\) \(\text{N}\) or greater.
A car going at \(\text{20}\) \(\text{m·s$^{-1}$}\) accelerates uniformly and
comes to a stop in a distance of \(\text{20}\) \(\text{m}\).
What is its acceleration?
We can use the equations of motion to find the acceleration:
If the car is \(\text{1 000}\) \(\text{kg}\) how much force do the brakes
exert?
\begin{align*}
F & = ma \\
& = (\text{1 000})(\text{10}) \\
& = \text{10 000}\text{ N}
\end{align*}
A block on an inclined plane experiences a force due to gravity, \(\vec{F}_g\) of
\(\text{300}\) \(\text{N}\) straight down. If the slope is inclined at
\(\text{67,8}\)\(\text{°}\) to the horizontal, what is the component of the
force due to gravity perpendicular and parallel to the slope? At what angle would
the perpendicular and parallel components of the force due to gravity be equal?
For the two components to be equal the angle must be \(\text{45}\)\(\text{°}\).
(\(\sin (45) = \cos(45)\)).
A block on an inclined plane is subjected to a force due to gravity, \(\vec{F}_g\) of
\(\text{287}\) \(\text{N}\) straight down. If the component of the gravitational
force parallel to the slope is \(\vec{F}_{gx}\)=\(\text{123,7}\) \(\text{N}\) in the
negative \(x\)-direction (down the slope), what is the incline of the slope?
A block on an inclined plane experiences a force due to gravity, \(\vec{F}_g\) of
\(\text{98}\) \(\text{N}\) straight down. If the slope is inclined at an unknown
angle to the horizontal but we are told that the ratio of the components of the
force due to gravity perpendicular and parallel to the slope is 7:4. What is the
angle the incline makes with the horizontal?
We first write down the equations for the parallel and perpendicular components:
Recall from trigonometry that \(\frac{\sin \theta}{\cos \theta} = \tan \theta\).
Two crates, \(\text{30}\) \(\text{kg}\) and \(\text{50}\) \(\text{kg}\) respectively,
are connected with a thick rope according to the diagram. A force, to the right, of
\(\text{1 500}\) \(\text{N}\) is applied. The boxes move with an acceleration
of \(\text{2}\) \(\text{m·s$^{-2}$}\) to the right. The ratio of the frictional
forces on the two crates is the same as the ratio of their masses. Calculate:
the magnitude and direction of the total frictional force present.
Let the \(\text{50}\) \(\text{kg}\) crate be crate 1 and the \(\text{30}\)
\(\text{kg}\) crate be crate 2.
To find the frictional force we will apply Newton's second law. We are given the
mass (\(\text{30}\) + \(\text{50}\) \(\text{kg}\)) and the acceleration
(\(\text{2}\) \(\text{m·s$^{-2}$}\)). Choose the direction of motion to be
the positive direction (to the right is positive).
\begin{align*}
F_{R} & = ma \\
T + F_{f} & = (30)(2) \\
T + -\text{502,5} & = 60 \\
T & = \text{562,5}\text{ N}
\end{align*}
Two crates, \(\text{30}\) \(\text{kg}\) and \(\text{50}\) \(\text{kg}\) respectively,
are connected with a thick rope according to the diagram. If they are dragged up an
incline such that the ratio of the parallel and perpendicular components of the
gravitational force on each block are 3 : 5. The boxes move with an acceleration of
\(\text{7}\) \(\text{m·s$^{-2}$}\) up the slope. The ratio of the frictional
forces on the two crates is the same as the ratio of their masses. The magnitude of
the force due to gravity on the \(\text{30}\) \(\text{kg}\) crate is \(\text{294}\)
\(\text{N}\) and on the \(\text{50}\) \(\text{kg}\) crate is \(\text{490}\)
\(\text{N}\). Calculate:
the magnitude and direction of the total frictional force present.
Let crate 1 be the \(\text{50}\) \(\text{kg}\) crate and crate 2 be the
\(\text{30}\) \(\text{kg}\) crate. We will choose up the slope as positive.
We need to find all the forces that act parallel to the slope on each box in turn
and use this to solve simultaneously for the frictional force.
We draw free body diagrams for each crate:
First we find the gravitational force parallel to the slope for each box:
Newton's third law of Motion deals with the interaction between pairs of objects. For example, if you
hold a book up against a wall you are exerting a force on the book (to keep it there) and the book is
exerting a force back at you (to keep you from falling through the book). This may sound strange, but if
the book was not pushing back at you, your hand would push through the book! These two forces (the force
of the hand on the book (\({F}_{1}\)) and the force of the book on the hand (\({F}_{2}\))) are called an
action-reaction pair of forces. They have the same magnitude, but act in opposite directions and act on
different objects (the one force is onto the book and the other is onto your hand).
There is another action-reaction pair of forces present in this situation. The book is pushing against
the wall (action force) and the wall is pushing back at the book (reaction). The force of the book on
the wall (\({F}_{3}\)) and the force of the wall on the book ( \({F}_{4}\)) are shown in the diagram.
Figure 2.8:
Newton's action-reaction pairs.
Newton's third law of motion
If body A exerts a force on body B, then body B exerts a force of equal magnitude on body A, but
in the opposite direction.
These action-reaction pairs have several properties:
the same type of force acts on the objects,
the forces have the same magnitude but opposite direction, and
the forces act on different objects.
Newton's action-reaction pairs can be found everywhere in life where two objects interact with one
another. The following worked examples will illustrate this:
Worked example 18: Newton's third law - seat belt
Dineo is seated in the passenger seat of a car with the seat belt on. The car suddenly stops and
he moves forwards (Newton's first law - he continues in his state of motion) until the seat belt
stops him. Draw a labelled force diagram identifying two action-reaction pairs in this
situation.
Draw a force diagram
Start by drawing the picture. You will be using arrows to indicate the forces so make your
picture large enough so that detailed labels can also be added. The picture needs to be
accurate, but not artistic! Use stick-men if you have to.
Label the diagram
Take one pair at a time and label them carefully. If there is not enough space on the drawing,
then use a key on the side.
Worked example 19: Newton's third law: forces in a lift
Tammy travels from the ground floor to the fifth floor of a hotel in a lift moving at constant
velocity. Which ONE of the following statements is TRUE about the magnitude of the force exerted
by the floor of the lift on Tammy's feet? Use Newton's third law to justify your answer.
It is greater than the magnitude of Tammy's weight.
It is equal in magnitude to the force Tammy's feet exert on the floor of the lift.
It is equal to what it would be in a stationary lift.
It is greater than what it would be in a stationary lift.
Analyse the situation
This is a Newton's third law question and not Newton's second law. We need to focus on the
action-reaction pairs of forces and not the motion of the lift. The following diagram will show
the action-reaction pairs that are present when a person is standing on a scale in a lift.
Figure 2.9:
Newton's action-reaction pairs in a lift.
In this question statements are made about the force of the floor (lift) on Tammy's feet. This
force corresponds to \({F}_{2}\) in our diagram. The reaction force that pairs up with this one
is \({F}_{1}\), which is the force that Tammy's feet exerts on the floor of the lift. The
magnitude of these two forces are the same, but they act in opposite directions.
Choose the correct answer
It is important to analyse the question first, before looking at the answers. The answers might
confuse you if you look at them first. Make sure that you understand the situation and know what
is asked before you look at the options.
The correct answer is number \(\text{2}\).
Worked example 20: Newton's third law: book and wall
Bridget presses a book against a vertical wall as shown in the photograph.
Draw a labelled force diagram indicating all the forces acting on the book.
State, in words, Newton's third law of Motion.
Name the action-reaction pairs of forces acting in the horizontal plane.
Draw a force diagram
A force diagram will look like this:
Note that we had to draw all the forces acting on the book and not the action-reaction pairs.
None of the forces drawn are action-reaction pairs, because they all act on the same object (the
book). When you label forces, be as specific as possible, including the direction of the force
and both objects involved, for example, do not say gravity (which is an incomplete answer) but
rather say 'Downward (direction) gravitational force of the Earth (object) on the book
(object)'.
State Newton's third law
If body A exerts a force onto body B, then body B will exert a force equal in magnitude, but
opposite in direction, onto body A.
Name the action-reaction pairs
The question only asks for action-reaction forces in the horizontal plane. Therefore:
Pair 1: Action: Applied force of Bridget on the book; Reaction: The force of the book on the
girl.
Pair 2: Action: Force of the book on the wall; Reaction: Force of the wall on the book.
Note that a Newton's third law pair will always involve the same combination of words, like 'book
on wall' and 'wall on book'. The objects are 'swapped around' in naming the pairs.
temp text
Balloon rocket
Aim
In this experiment for the entire class, you will use a balloon rocket to investigate Newton's
third law. A fishing line will be used as a track and a plastic straw taped to the balloon will
help attach the balloon to the track.
Apparatus
You will need the following items for this experiment:
balloons (one for each team)
plastic straws (one for each team)
tape (cellophane or masking)
fishing line, \(\text{10}\) metres in length
a stopwatch - optional (a cell phone can also be used)
a measuring tape - optional
Method
Divide into groups of at least five.
Attach one end of the fishing line to the blackboard with tape. Have one teammate hold
the other end of the fishing line so that it is taut and roughly horizontal. The line
must be held steady and must not be moved up or down during the
experiment.
Have one teammate blow up a balloon and hold it shut with his or her fingers. Have
another teammate tape the straw along the side of the balloon. Thread the fishing line
through the straw and hold the balloon at the far end of the line.
Let go of the rocket and observe how the rocket moves forward.
Optionally, the rockets of each group can be timed to determine a winner of the fastest
rocket.
Assign one teammate to time the event. The balloon should be let go when the time
keeper yells “Go!” Observe how your rocket moves toward the
blackboard.
Have another teammate stand right next to the blackboard and yell
“Stop!” when the rocket hits its target. If the balloon does not
make it all the way to the blackboard, “Stop!” should be called when
the balloon stops moving. The timekeeper should record the flight time.
Measure the exact distance the rocket travelled. Calculate the average speed at
which the balloon travelled. To do this, divide the distance travelled by the
time the balloon was “in flight.” Fill in your results for Trial 1
in the Table below.
Each team should conduct two more races and complete the sections in the Table
for Trials 2 and 3. Then calculate the average speed for the three trials to
determine your team's race entry time.
Results
Distance (m)
Time (s)
Speed (\(\text{m·s$^{-1}$}\))
Trial 1
Trial 2
Trial 3
Average:
Conclusions
The winner of this race is the team with the fastest average balloon speed.
The Saturn V (pronounced “Saturn Five”) was an American human-rated expendable rocket used by
NASA's Apollo and Skylab programs from 1967 until 1973. A multistage liquid-fuelled launch vehicle, NASA
launched 13 Saturn Vs from the Kennedy Space Center, Florida with no loss of crew or payload. It remains
the tallest, heaviest, and most powerful rocket ever brought to operational status and still holds the
record for the heaviest launch vehicle payload.
Textbook Exercise 2.6
A fly hits the front windscreen of a moving car. Compared to the magnitude of the force
the fly exerts on the windscreen, the magnitude of the force the windscreen exerts on
the fly during the collision, is:
zero.
smaller, but not zero.
bigger.
the same.
the same
Which of the following pairs of forces correctly illustrates Newton's third law?
A or B
Forces in equilibrium (ESBKW)
At the beginning of this chapter it was mentioned that resultant forces cause objects to accelerate in a
straight line. If an object is stationary or moving at constant velocity then either,
no forces are acting on the object, or
the forces acting on that object are exactly balanced.
In other words, for stationary objects or objects moving with constant velocity, the resultant force
acting on the object is zero.
Equilibrium
An object in equilibrium has both the sum of the forces acting on it equal to zero.
