Topic 3 - Forces

Topic 3 - Forces#

“Axiomata Sive Leges Motus’’, or Axioms or Laws of Motion#

Photograph of a book. The book is the original copy of Principia owned by Newton. — Fig. 3 Newton’s own copy of his Principia, with hand-written corrections for the second edition, in the Wren Library at Trinity College, Cambridge.#

In this chapter we are going to look at some of the work that Isaac Newton developed way back in the 17^th century. Unfortunately we are not going to dabble in his obsession with alchemy[1] or finding the Philosopher’s Stone, but instead will concern ourselves with a particular section of the “Philosophiæ Naturalis Principia Mathematica’’ published in 1687. In this work Newton defines three laws of motion that set the foundation for the field of classical mechanics which you will study in more depth next year.

You will already have seen these three laws of motion but there are some concepts or details that are often overlooked in pre-University courses so firstly we will review the three laws in turn before applying them to some representative examples.

I will forewarn you now that I’m going to make a point of including the original Latin of the three laws. There is some educational benefit to this, I assure you, but it is also a historical thing that usually gets lots of eyerolls from the audience. I fully intend to honour this tradition until I retire.

First Law#

Corpus omne perseverare in statu suo quiescendi vel movendi uniformiter in directum, nisi quatenus a viribus impressis cogitur statum illum mutare.

Every body persists in its state of being at rest or of moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed.

The key point that I want to emphasise with this law is a subtle one but thinking about it now will help in the future when you start thinking about special relativity. Newton makes reference to the ‘state’ of the body. This state could be at rest or a constant motion straight forward but in all cases they are simply specifics of this general ‘state’. This state is the velocity of the body which means we can rephrase the First Law as

A body acted on by no net force moves with constant velocity.

The zero velocity case is simply a unique value in an almost infinite number of possible velocities. Sometimes it is convenient to make use of the zero velocity case, particularly when you start thinking about different frames of reference.

Second Law#

Mutationem motus proportionalem esse vi motrici impressae, et fieri secundum lineam rectam qua vis illa imprimitur.

The alteration of motion is ever proportional to the motive force impress’d; and is made in the direction of the right line in which that force is impress’d.

This is by far the Law that is most frequently stated incorrectly. Or perhaps more accurately it is that the law taught to you in schools in a special case of a simplified version of the actual statement the Newton made. You will have heard people saying that the second law is simply \(F=ma\) however let us look at the original statement to see what information is missing when we look between the original and this simple-yet-incomplete definition.

What I hope you notice is that the original statement contains two key parts, conveniently separated by a semi colon. In the first part we have that the alteration of motion, or acceleration, is proportional to the applied force. This is all good and gels with our simple statement that \(F\propto a\), where the mass \(m\) is our constant of proportionality.

The information in the original definition that is lacking in our \(F=ma\) statement is that regarding the directions of the applied forces and the resulting acceleration.

When we want to combine information about the magnitude and direction of properties you should always think about vectors.

By combining the two components of the original statement for the Second Law we can give a more general statement of the law in mathematical form, namely that

(8)#\[\textbf{F}=m\textbf{a}\]

Have you spotted the ‘error’ yet?

I was being rather picky with my choice of words right after the Latin and translation text. I said that the version taught is schools is both simplified and a special case. We have taken care of the simplification by including the directions by way of vectors but the special case element needs to be addressed.

Let us look back at equation (8). I was quite correct to not change the mass into a vector, but what I have assumed is that the mass is constant. In fact I need to go back to the original statement of the Second Law and think more generally about the “alteration of motion” and consider a property that contains both the velocity and mass of the body. This is, of course, the momentum \((\textbf{p}=m\textbf{v})\) of the body.

If we now reformulate the Second Law using this concept of the momentum into an equation we find that

(9)#\[\textbf{F} = \frac{\mathrm{d}\textbf{p}}{\mathrm{d}t}\]

We can now check that this more general expression does reduce to the more familar form in equation (8) under special circumstances. Let us take equation (9) and apply the product rule

\[\begin{align*} \textbf{F} &= \frac{\mathrm{d}(m\textbf{v})}{\mathrm{d}t}=m\frac{\mathrm{d}\textbf{v}}{\mathrm{d}t} + \textbf{v}\frac{\mathrm{d}m}{\mathrm{d}t} \end{align*}\]

so in the special case that the mass remains constant the right hand term become zero and therefore we find that

\[\begin{align*} \textbf{F} &= m\frac{\mathrm{d}\textbf{v}}{\mathrm{d}t}\\ &= m\textbf{a} \end{align*}\]

but only when the mass is constant. You should try to rememorise Newton’s Second Law as the force applied to a body in a given time, also known as the impulse equals the rate of change of momentum of that body.

Third Law#

Actioni contrariam semper et aequalem esse reactionem: sive corporum duorum actiones in se mutuo semper esse aequales et in partes contrarias dirigi.

To every action there is always an equal and opposite reaction: or the forces of two bodies on each other are always equal and are directed in opposite directions.

I won’t spend as much time on this Law as there is not as much to critique, but again the key point to emphasise is that in the original statement is there is mention of the magnitude and the direction of the forces. We can neaten this up slightly by introducing our good friends the vectors. If we consider a body A that is exerting a force of magnitude \(F_{AB}\) onto body B then body B exerts a force of magnitude \(F_{BA}\) onto body A, and so we can state that

\[\textbf{F}_{AB} = -\textbf{F}_{BA}\]

Summary of the three laws#

Key point

First Law: A body acted on by no net force moves with constant velocity (which may be zero).
Second Law: The general mathematical form is \(\textbf{F}=\dfrac{\mathrm{d}\textbf{p}}{\mathrm{d}t}\). This reduces to \(\textbf{F}=m\textbf{a}\) only in the case where the mass remains constant.
Third Law: If body A exerts a force on body B, then body B exerts a force on body A of the same magnitude but in the opposite direction, i.e. \(\textbf{F}_{AB} = -\textbf{F}_{BA}\)

Units of force#

All forces, regardless of their type or underlying physical cause, have units of Newtons. One Newton is equal to one kilogram meter per second squared \((\text{kg m s}^{-2})\).

Types of forces#

Below are four examples of types of force though you will come across more as you study different areas of physics. These four are chosen as they strongly relate to mechanics and will be discussed in more detail, in some cases simultaneouly where appropriate.

Weight#

All particles falling freely under gravity have the same acceleration, \(g\). This acceleration must be caused by a force acting on the particle, as required by Newton’s Second Law. This force is the weight of the particle and is equal to \(F_\text{weight} = mg\) where \(m\) is the mass of the particle.

Tension#

Imagine a particle of mass \(m\) hanging in equilibirum at the end of a string. As the particle is in equilibrium the weight \(mg\) acting downwards must be balanced by an equal and opposite force acting upwards (as per Newton’s Third Law). This upward force is the tension \(T\) in the string, and is a pulling force on the particle.

Worked Example 3.1

Question

A small object is hanging in equilibrium at the end of a vertical string. Find:

the weight of the object if its mass is \(0.2\text{ kg}\).
the mass of the object if its weight is \(4.9\text{ N}\).
the tension in the string if the mass of the object is \(1.0\text{ kg}\).
the tension in the string if the weight of the object is \(5.0\text{ N}\).

Hint

Mass and weight are not the same, although they are related. Mass is an intrinsic property of the object whereas weight is a force acting on the mass due to gravity.
In these types of questions you can assume the system is on Earth, unless told otherwise!

Solution

The weight \(W = mg = 0.2\times9.8 = 1.9(6)\text{ N}\)
Rearranging the equation for weight gives \(m = \frac{W}{g} = \frac{4.9}{9.8} = 0.5\text{ kg}\).
The object is in equilibrium meaning the tension in the string (force upward) must equal that of the weight (force downward).
Therefore \(T = W = mg = 1\times 9.8 = 9.8\text{ N}\).
The same argument applies as for part 3 of this question, so \(T = W = 5\text{ N}\).

Friction#

When an object is in contact with a rough surface and you try to push the object across said surface then a frictional force acts to oppose the pushing force. The frictional force acts parallel to the surface, and only when the surface is perfectly smooth does the frictional force equal zero.
We will revisit friction in more detail shortly, but first we need to better understand how to resolve forces to find the resultant force (or resultant forces in Cartesian components).

Resultant forces#

We have already seen in Topic 1 that we can add two vectors together to find a single resultant vector. As forces are vector quantities the same applies here - we can add multiple forces (in vector form) to get a single resultant force vector.
Suppose that two forces, \(\textbf{F}_1\) and \(\textbf{F}_2\) are acting on a particle. The resultant force \(\textbf{R} = \textbf{F}_1 + \textbf{F}_2\), and then from Newton’s Second Law we can state that the acceleration of this particle is:

\[\textbf{a} = \frac{\textbf{R}}{m} = \frac{\textbf{F}_1 + \textbf{F}_2}{m}\]

Suppose that a third force \(\textbf{F}_3\) also acts on the particle. The resultant force is now the sum of these three forces, i.e. \(\textbf{R} = \textbf{F}_1 + \textbf{F}_2 + \textbf{F}_3\).
More generally the resultant force is the sum of all forces acting on the particle, which can be neatly written using summation notation as:

\[\textbf{R} = \sum_{i=1}^{N}\textbf{F}_i\]

Worked Example 3.2

Question

Two forces of magnitude \(5\text{ N}\) and \(6\text{ N}\) act on a particle at right angles to one another. Find the magnitude and direction of the resultant force.

Hint

When drawing the diagram for this question you can orientate it however you like so long as the two forces stated are perpendicular to one another.

This gives a right angle triangle with two known sides, and we want to find the other side and one angle.

Solution

The vector diagram shown in the Hint section is a right angle triangle with \(R\) as the hypoteneuse. Pythagoras’ Theorem gives

\[\begin{split}R^2 &= 6^2 + 5^2 = 61\\ \therefore R &= \sqrt{61} \approx 7.81\end{split}\]

Using the adjacent and opposite sides allows us to determine the angle \(\theta\):

\[\begin{split}\tan\theta &= \frac{6}{5} = 1.2\\ \therefore \theta &= \tan^{-1}\left(\frac{6}{5}\right) \approx 50.2^{\circ}\end{split}\]

where the angle \(\theta\) is with respect to the \(5\text{ N}\) force.

Physics example: Traffic Lights - weight and tension

Question

In this case we consider the type of traffic lights that are suspended from cablesold` below. We have a traffic light suspended from a single cable which in turn is suspended from two cables that make angles of \(37.0^\circ\) and \(53.0^\circ\) to the horizontal beam above. We also know that the weight of the traffic light is \(125.0\text{ N}\), and we have been tasked with calculating the tensions in each of the three cables.

Fig. 4 A hanging style set of traffic lights suspended on one vertical wire which in turn is attached to two wires that are affixed to a horizontal surface. These two wires each make a different angle to this surface and so have a different tension. The traffic light is stationary so no velocity vectors are indicated.#

Now before you jump straight in to the calculations we should take a step back and think a) about the assumptions of a system, and b) which approach we should take. The former is somewhat trivial but always worth stating: in this example I am assuming that the traffic light is stationary and therefore (from Newton’s First Law) there is zero net force in the system.

The latter point is important because taking a brief moment to pause and think about the approach gives you the time to think about which methods are valid and, of those, which are the most efficient. For the purposes of hammering this point home I’m going to solve this traffic light problem using two methods, the first being the typical approach taught in pre-University education and the second making use of the vectors that we emphasised in our earlier review of the three laws of motion.

Method 1 - Resolving components

Looking at the diagram in the question I am going to split the system up into two parts, the upper and lower, and will consider the lower part first. In this part I have two components, the weight of the traffic light \(mg\) and the tension \(T_3\). Both of these are only acting in the vertical direction which means

\[T_3 = mg\]

Next I consider the upper part and here I need to split the components of the three tensions into their horizontal and vertical components. Thus my horizontal components give

(10)#\[\begin{split}T_1\cos\alpha &= T_2\cos\beta\nonumber\\ T_1 &= T_2\left(\frac{\cos\beta}{\cos\alpha}\right)\end{split}\]

Next our vertical components give

\[\begin{align*} T_3 = mg &= T_1\sin\alpha + T_2\sin\beta\\ &= T_2\left(\tan\alpha\cos\beta + \sin\beta\right)\\ \therefore T_2 &= \frac{mg}{\tan\alpha\cos\beta + \sin\beta} \approx 99.8\text{ N} \end{align*}\]

where I have substituted the expression for \(T_1\) (equation (10)) into the second line. Now that I have an expression or value for \(T_2\) I can substitute this back in to equation (10) to find

\[\begin{align*} T_1 &= T_2\left(\frac{\cos\beta}{\cos\alpha}\right)\\ &= \left(\frac{mg}{\tan\alpha\cos\beta + \sin\beta}\right)\left(\frac{\cos\beta}{\cos\alpha}\right) \\ &\approx 75.2\text{ N} \end{align*}\]

And there we have it. We have found that \(T_1=75.2\text{ N}\), \(T_2=99.8\text{ N}\) and \(T_3=125.0\text{ N}\) to the correct number of significant figures.

Method 2 - Vectors

The method above does work but it involved an unneccesary amount of work. By remembering that we are working with vectors we can avoid splitting our forces into their components of direction and magnitude (or in this case their horizontal and vertical), and just work with the vectors themselves.

We have assumed that the system is not accelerating which means there is no net force. Under this assumption the sum of the forces in the system should equal zero which, when expressed as a diagram, means the three vectors form a closed triangle when placed end to end. We will start by sketching this:

Fig. 5 Three arrows indicate the vector forces, one due to gravity pointing directly down, and the two force vectors point at angles with respect to the horizontal. The three vector arrows form a closed triangle because the traffic light is stationary and thus there is zero net force.#

When presented in this form I hope it is fairly clear that we have a triangle with all lengths and angles defined (even if we do not yet know all the values). Note that I’m borrowing the result from the previous method that \(T_3=mg\), but this is pretty obvious when you think about the system.

The fact that this is a triangle of vectors makes no difference; we can apply the geometry rules we all know and love to solve for the unknowns, in this case \(T_1\) and \(T_2\).

I will make use of the sine rules of triangles, which if you cannot remember it off the bat is defined as

\[ \frac{mg}{\sin\theta'} = \frac{T_1}{\sin\beta'}=\frac{T_2}{\sin\alpha'}.\]

We know all of the parameters in the left hand side so will use this part twice, first to find \(T_1\) and again to find \(T_2\). For \(T_1\) this gives

\[\begin{align*} \frac{mg}{\sin\theta'}&=\frac{T_1}{\sin\beta'}\\ T_1&=mg\left(\frac{\sin\theta'}{\sin\beta'}\right)\\ &\approx0.602mg = 75.2\text{ N} \end{align*}\]

Similar steps will give you \(T_2 = 99.8\text{ N}\). These are the same as we found using Method 1, as we would expect, but the vector method is much more efficient

Resolving a contact force into normal (and frictional) components.#

Normal contact force#

Think of a particle at rest on a horizontal surface. If the surface was removed the particle would accelerate downwards due to gravity, but as it is at rest there must be some equal and opposite force acting upwards due to the surface. This force must balance the weight of the particle such that there is no resultant force and therefore no acceleration.
This force from the surface is called the normal reaction force, normal contact force, or simply the normal force and is often denoted with either \(\textbf{N}\) for normal or \(\textbf{R}\) for reaction. A key point here is that this force acts normal or perpendicular to the surface, and for instances where the surface is at an angle to the horizontal (e.g. in Worked Example 3.4 below) this normal force will not be directed vertically upwards.

If we now try to push the particle horizontally across the surface then, up to some maximum applied force, the particle does not move. There is an additional frictional force that acts to balance the pushing force. Let us have a closer look now at this frictional force.

Friction and the coefficient of friction#

First let us visualise the system.

We have a particle of weight \(mg\) acting down, a normal force \(\textbf{N}\) acting upwards, a pushing force \(\textbf{P}\) acting to the right and a frictional force \(\textbf{F}_f\). For as long as the particle remains stationary then

\[\begin{split}\textbf{P} &= -\textbf{F}_f\\ \text{or } |\textbf{P}| &= |\textbf{F}_f|\end{split}\]

What this means is that the friction force for a particular surface is not constant. It increases as the applied force \(\textbf{P}\) increases, at least up to a value \(\textbf{F}_\text{max}\) beyond which it cannot increase. At this point the particle is just about to start accelerating because the applied force is equal to the maximum resistive frictional force.
This maximum force is given by

\[F_\text{max} = \mu R\]

where \(\mu\) is a constant of proportionality called the coefficient of friction

For a stationary particle, \(0\leq F_f\leq\mu R\). Once the particle beins to accelerate the frictional force opposing the relative motion remains at the constant value \(F_\text{max} = \mu R\).

Key summary points:

Friction acts to oppose relative motion.
Until it reaches its limiting value the magnitude of the frictional force is just sufficient to prevent relative motion.
When the limiting value is reached \(F_\text{max} = \mu R\).
For all rough surfaces, \(0\leq F_f\leq\mu R\).
For a smooth surface, \(F=0\).
When the particle begins to slide the frictional force has the limiting value \(\mu R\) and acts in the opposite direction of relative motion.

Worked Example 3.3 - Friction on a horizontal surface

Question

A particle of weight \(30\text{ N}\) rests on a horizontal plane. The coefficient of friction between the particle and the place is \(\mu=0.3\). A horizontal force of magnitude \(P\text{ N}\) is applied to the particle.
Find the value of \(P\) for which the particle just begins to slide.

Hint

Draw the force diagram, and consider splitting the system into different (perpendicular) components.

Solution

Resolving horizontal forces gives

\[P = F_\text{max} = \mu R\]

Vertical components gives

\[mg = 30 = R\]

Substituting this value for \(R\) into the expression for the horizontal forces gives

\[\begin{split}P &= \mu \times R\\ &= 0.3 \times 30 = 9\text{ N}\end{split}\]

Worked Example 3.4 - Friction on an inclined surface

Question

A particle of mass \(4\text{ kg}\) rests in limiting equilibrium on a rough plane inclined at \(30^\circ\) to the horizontal. Find the coefficient of friction between the particle and the plane.

Hint

Draw a force diagram, and remember to split the forces into perpendicular components. Keep in mind that, for an inclined surface, the normal force is not directly upwards.

Options are to resolve into vertical and horizontal components, or parallel and perpendicular to the plane components.

Solution

Note that \(\textbf{F}_f\) acts up the plane, since without friction the particle would accelerate down the plane.

Resolving perpendicular to the plane gives:

\[\begin{split}R &= 4g\cos 30\\ &= 4g\times \frac{\sqrt{3}}{2} = 2\sqrt{3}g\end{split}\]

Resolving parallel to the plane gives:

\[\begin{split}F_f &= 4g\sin 30\\ &= 4g\times\frac{1}{2} = 2g\end{split}\]

At equilibrium \(F=\mu R\) and so:

\[\begin{split}2g &= \mu\times 2\sqrt{3}g\\ \therefore \mu &= \frac{1}{\sqrt{3}} \approx 0.58\end{split}\]

You should retry this worked example but instead of resolving into parallel and perpendicular to the inclined surface you should find the horizontal and vertical components of all forces involved. The answer will be the same.

Vertical motion#

In the previous sections we have considered systems known as statics, meaning that while there are forces acting on the object there is no net force. The object is in equilibrium and does not accelerate, though we did start thinking about when the pushing force for example could overcome the frictional force and so the object being pushed starts to move. We will now move into the realms of dynamics and consider how applied forces cause an object to move and accelerate.