MKU University B.Sc., Mathematics SMTJC31 Mechanics Chapter 1

Chapter 1
Forces acting at a point

Table of Contents

Table of Contents
1 Forces acting at a point
1.1 An extended form of the parallelogram law of forces
1.2 Resolved Parts
1.3 Resultant of any number of forces acting at a point: Graphical Method
1.4 Resultant of any number of coplanar forces acting at a point: Analytical Method
1.5 Conditions of equilibrium of any number of forces acting upon a particle

1.1 An extended form of the parallelogram law of forces

Theorem 1.1.1. If forces $\lambda \overline{\mathit{OA}}$ and $\mu \overline{\mathit{OB}}$ act at a point $O$ along the lines $\mathit{OA}$ and $\mathit{OB}$, their resultant will be the force $(\lambda +\mu )\overline{\mathit{OC}}$ where $C$ is the point on $\mathit{AB}$ such that $\lambda ·\mathit{AC}=\mu .\mathit{CB}$

Proof : Forces represented by $\lambda ·\mathit{OA}$ and $\mu ·\mathit{OB}$ act along the lines $\mathit{OA}$ and $\mathit{OB}$. Take the point $C$ on $\mathit{AB}$ such that $\lambda ·\mathit{AC}=\mu ·\mathit{CB}$.

Figure 1.1:

From $\mathrm{\Delta }\mathit{OCA}$,

$$\begin{array}{llll}\hfill \overline{\mathit{OA}}& =\overline{\mathit{OC}}+\overline{\mathit{CA}}& \hfill & \\ \hfill \therefore \lambda ·\overline{\mathit{OA}}& =\lambda ·\overline{\mathit{OC}}+\lambda ·\overline{\mathit{CA}}& \hfill \text{(1.1)}\end{array}$$

From $\mathrm{\Delta }\mathit{OCB}$,

$$\begin{array}{llll}\hfill \overline{\mathit{OB}}& =\overline{\mathit{OC}}+\overline{\mathit{CB}}& \hfill & \\ \hfill \therefore \mu \overline{\mathit{OB}}& =\mu ·\overline{\mathit{OC}}+\mu \overline{\mathit{CB}}\dots \dots \dots & \hfill \text{(1.2)}\end{array}$$

Adding (1.1) and (1.2),

$$\lambda ·\overline{\mathit{OA}}+\mu \overline{\mathit{OB}}=(\lambda +\mu )·\overline{\mathit{OC}}+\lambda ·\overline{\mathit{CA}}+\mu ·\overline{\mathit{CB}}$$

(1.3)

By construction of the point $C,$ we have $\lambda ·\mathit{AC}=\mu ·\mathit{CB}$.
$\therefore$ The forces $\lambda ·\overline{\mathit{CA}}$ and $\mu ·\overline{\mathit{CB}}$ are equal and opposite forces acting at $C$.

$$\begin{array}{llll}\hfill \therefore \lambda ·\overline{\mathit{CA}}+\mu ·\overline{\mathit{CB}}& =\overline{0}& \hfill & \end{array}$$ Hence (1.3) gives

$$\lambda \overline{\mathit{OA}}+\mu ·\overline{\mathit{OB}}=(\lambda +\mu )·\overline{\mathit{OC}}$$

(1.4)

□

Corollary 1.1.2. If we put $\lambda =1=\mu$, $C$becomes the midpoint of $\mathit{AB}$. Then (1.4) gives,

$$\overline{\mathit{OA}}+\overline{\mathit{OB}}=2\overline{\mathit{OC}}$$

(1.5)

i.e., The resultant of two forces represented completely by $\overline{\mathit{OA}}$and $\overline{\mathit{OB}}$is represented by $2.\overline{\mathit{OC}}$, where $C$is the middle point of $\mathit{AB}$.

This result which will be greatly useful in the solution of a number of problems is also obvious from the parallelogram law since $\overline{\mathit{OA}}+\overline{\mathit{OB}}=\overline{\mathit{OD}}$. $C$ is the midpoint of the diagonal $\mathit{OD}$ and so $\mathit{OD}=2\overline{\mathit{OC}}$

Worked Examples

Example 1.1.3. $\mathit{ABC}$ is a triangle. $G$ is its centroid and $P$ is any point in the plane of the triangle. Show that the resultant of forces represented by $\overline{\mathit{PA}},\overline{\mathit{PB}},\overline{\mathit{PC}}$ is $3\overline{\mathit{PG}}$ and find the position of $P$, if the three forces are in equilibrium.

Solution. Let $A$ be the midpoint of $\mathit{BC}$.

Figure 1.2:

$$\begin{array}{llll}\hfill \overline{\mathit{PB}}+\overline{\mathit{PC}}& =2{\overline{\mathit{PA}}}^{\prime }& \hfill & \\ \hfill \overline{\mathit{PA}}+\overline{\mathit{PB}}+\overline{\mathit{PC}}& =\overline{\mathit{PA}}+2{\overline{\mathit{PA}}}^{\prime }& \hfill & \\ \hfill & =1.\overline{\mathit{PA}}+2.{\overline{\mathit{PA}}}^{\prime }& \hfill & \\ \hfill & =(1+2)\overline{\mathit{PK}}& \hfill \text{(1.6)}\end{array}$$ where $K$ is the point on ${\mathit{AA}}^{\prime }$ such that $1.\mathit{AK}=2K{A}^{\prime }$ $$i.e.,\frac{\mathit{AK}}{K{A}^{\prime }}=\frac{2}{1}$$

i.e., $K$ divides the median ${\mathit{AA}}^{\prime }$ in the ratio $2:1$.
$\therefore K$ is the same as $G$, the centroid of the $\mathrm{\Delta }$.

$\therefore \left(1.6\right)$ becomes

$$\overline{\mathit{PA}}+\overline{\mathit{PB}}+\overline{\mathit{PC}}=\overline{3\mathit{PG}}$$

If the three forces $\overline{\mathit{PA}},\overline{\mathit{PB}},\overline{\mathit{PC}}$ are in equilibrium, then their resultant should be zero.

$$\begin{array}{cc}\text{i.e.,}\overline{\mathit{PG}}\hfill & =\overline{0}\hfill \\ \therefore ,\mathit{PG}\hfill & =0\hfill \end{array}$$

i.e. $P$ must be taken at the centroid $G$ of the triangle.

Note 1.1.4. From the above, we find that if $G$ is the centroid on a $\mathrm{\Delta }\mathit{ABC},$ forces represented by $\mathit{GA},\mathit{GB},\mathit{GC}$ will be in equilibrium.

$$i.e.\overline{\mathit{GA}}+\overline{\mathit{GB}}+\overline{\mathit{GC}}=\overline{0}$$

Example 1.1.5. Five forces acting at a point are represented in magnitude and direction by the lines joining the vertices of any pentagon to the midpoints of their opposite sides. Show that they are in equilibrium.

Solution. $\mathit{ABCDE}$ is a pentagon and $P,Q,R,S,T$ are the midpoints of the sides $\mathit{CD},\mathit{DE},\mathit{EA},\mathit{AB}$ and $\mathit{BC}$ respectively.

Figure 1.3:

We have to prove that

$$\overline{\mathit{AP}}+\overline{\mathit{BQ}}+\overline{\mathit{CR}}+\overline{\mathit{DS}}+\overline{\mathit{ET}}=\overline{0}$$

Using the corollary we have,

$$\begin{array}{ccc}\hfill 2\overline{\mathit{AP}}& =(\overline{\mathit{AC}}+\overline{\mathit{AD}})\hfill & \hfill \\ \hfill i.e.,\overline{\mathit{AP}}& =\frac{1}{2}\overline{(\mathit{AC}+\overline{\mathit{AD})}}\hfill \end{array}$$

Similarly

$$\begin{array}{ccc}\hfill \overline{\mathit{BQ}}& =\frac{1}{2}(\overline{\mathit{BD}}+\overline{\mathit{BE}})\hfill & \hfill \\ \hfill \overline{\mathit{CR}}& =\frac{1}{2}(\overline{\mathit{CE}}+\overline{\mathit{CA}})\hfill \\ \hfill \overline{\mathit{DS}}& =\frac{1}{2}(\overline{\mathit{DA}}+\overline{\mathit{DB}})\hfill \\ \hfill \end{array}$$

and

$$\overline{\mathit{ET}}=\frac{1}{2}(\overline{\mathit{EB}}+\overline{\mathit{EC}})$$

Adding up, we have

$$\begin{array}{llll}\hfill \overline{\mathit{AP}}+\overline{\mathit{BQ}}+\overline{\mathit{CR}}+\mathit{DS}+\overline{\mathit{ET}}& =\frac{1}{2}[\overline{\mathit{AC}}+\overline{\mathit{AD}}+\overline{\mathit{BD}}+\overline{\mathit{BE}}+\overline{\mathit{CE}}+\overline{\mathit{CA}}+\overline{\mathit{DA}}+\overline{\mathit{DB}}+\overline{\mathit{EB}}+\overline{\mathit{EC}}]& \hfill & \\ \hfill & =\frac{1}{2}\times \overline{0}\text{(Vectors are equal and opposite)}& \hfill & \\ \hfill & =0& \hfill & \end{array}$$

Example 1.1.6. $\mathit{OA},\mathit{OB},\mathit{OC}$ are the lines of action of two forces $P$ and $Q$ and their resultant $R$ respectively. Any transversal meets the lines in $L,M$ and $N$ respectively, Prove that

$$\frac{P}{\mathit{OL}}+\frac{O}{\mathit{OM}}=\frac{R}{\mathit{ON}}$$

Solution. Let $\overline{\mathit{OA}}=P$ and $\overline{\mathit{OB}}=Q.$ Complete the parallelogram $\mathit{AOB}$, $\overline{\mathit{OC}}=R$.

Figure 1.4:

Let $\frac{\mathit{OA}}{\mathit{OL}}=\lambda$ and $\frac{\mathit{OB}}{\mathit{BM}}=\mu$. $\therefore \mathit{OA}=\mathit{\lambda OL}$ and $\mathit{OB}=\mathit{\mu OM}$.

$$\begin{array}{llll}\hfill \overline{\mathit{OA}}+\overline{\mathit{OB}}& =\lambda \overline{\mathit{OL}}+\mu \overline{\mathit{OM}}& \hfill & \\ \hfill & =(\lambda +\mu )\overline{\mathit{OK}}& \hfill \text{(1.7)}\end{array}$$

where $K$ is a point on $\mathit{LM}$. But

$$\begin{array}{lll}\hfill \overline{\mathit{OA}}+\overline{\mathit{OB}}=\overline{\mathit{OC}}& & \hfill \text{(1.8)}\end{array}$$

Hence the forces $(\lambda +\mu )\overline{\mathit{OK}}$ and $\overline{\mathit{OC}}$ must be the same.
i.e. $K$ must be a point on $\mathit{OC}$ also.
$\therefore K$ is the point of intersection of $\mathit{OC}$ and $\mathit{LM}$
i.e. $K$ is clearly the same as $N$.
Equating the magnitudes of the two equal forces on the right hand sides of (1.7) and (1.8), we get

$$\begin{array}{ccc}\hfill (\lambda +\mu )·\mathit{OK}& =\mathit{OC}\hfill & \hfill \\ \hfill \text{i.e.,}(\lambda +\mu )·\mathit{ON}& =\mathit{OC}\hfill \\ \hfill \text{or}\lambda +\mu & =\frac{\mathit{OC}}{\mathit{ON}}\hfill \end{array}$$

$$i.e.,\frac{\mathit{OA}}{\mathit{OL}}+\frac{\mathit{OB}}{\mathit{OM}}=\frac{\mathit{OC}}{\mathit{ON}}\text{or}\frac{P}{\mathit{OL}}+\frac{Q}{\mathit{OM}}=\frac{R}{\mathit{ON}}$$

Example 1.1.7. $\mathit{ABC}$ is a triangle, with a right angle at $A:\mathit{AD}$ is the perpendicular on $\mathit{BC}$. Prove that the resultant of the forces $\frac{1}{\mathit{AB}}$ acting along $\mathit{AB}$ and $\frac{1}{\mathit{AC}}$ acting along $\mathit{AC}$ is $\frac{1}{\mathit{AD}}$ acting along $\mathit{AD}$.

Solution. From Geometry, we have the following well known results

$${\mathit{AB}}^2=\mathit{BC}·\mathit{BD};{\mathit{AC}}^2=\mathit{BC}·\mathit{CD};{\mathit{AD}}^2=\mathit{BD}·\mathit{DC}$$

Figure 1.5:

The forces $\frac{1}{\mathit{AB}}$ acting along $\mathit{AB}$ and $\frac{1}{\mathit{AC}}$ acting along $\mathit{AC}$ can be considered respectively as the forces $\frac{1}{A{B}^2}·\mathit{AB}$ acting along $\mathit{AB}$ and $\frac{1}{{\mathit{AC}}^2}\mathit{AC}$

If we take $\lambda =\frac{1}{{\mathit{AB}}^2}$ and $\mu =\frac{1}{{\mathit{AC}}^2}$ then $\lambda ·A{B}^2=\mathit{\mu A}{C}^2$ (each being =1)

$$\begin{array}{llll}\hfill \lambda ·\mathit{BC}·\mathit{BD}& =\mathit{\mu BC}·\mathit{CD}& \hfill & \\ \hfill \lambda ·\mathit{BD}=\mu ·\mathit{CD}& & \hfill \end{array}$$

Hence the resultant of forces $\lambda ·\overline{\mathit{AB}}$ and $\mu ·\overline{\mathit{AC}}$ is the force $(\lambda +\mu )·\overline{\mathit{AD}}$
i.e. The resultant of forces $\frac{1}{\mathit{AB}}$ along $\mathit{AB}$ and $\frac{1}{\mathit{AC}}$ along $\mathit{AC}$ is the force $(\lambda +\mu )·\mathit{AD}$ acting along $\mathit{AD}$.

$$\begin{array}{llll}\hfill \text{Magnitude of the resultant}& =(\lambda +\mu )\mathit{AD}& \hfill & \\ \hfill & =\left(\frac{1}{{\mathit{AB}}^2}+\frac{1}{{\mathit{AC}}^2}\right)\mathit{AD}& \hfill & \\ \hfill & =\left(\frac{A{C}^2+A{B}^2}{A{B}^2·A{C}^2}\right)\mathit{AD}& \hfill & \\ \hfill & =\frac{{\mathit{BC}}^2}{{\mathit{AB}}^2{\mathit{AC}}^2}\mathit{AD}=\frac{{\mathit{BC}}^2}{\mathit{BC}·\mathit{BD}·\mathit{BC}·\mathit{CD}}\mathit{AD}& \hfill & \\ \hfill & =\frac{1}{\mathit{BD}·\mathit{CD}}\mathit{AD}=\frac{1}{{\mathit{AD}}^2}\mathit{AD}=\frac{1}{\mathit{AD}}& \hfill & \end{array}$$ and this acts along $\mathit{AD}$.

Example 1.1.8. $P$ is a point in the plane of the triangle $\mathit{ABC}$ and $I$ is the in centre. Show that the resultant of forces represented by $\mathit{PA}·\sin A,\mathit{PB}·\sin B$ and $\mathit{PC}·\sin C$ is $4\mathit{PI}·\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

Solution. Let $\mathit{AD}$ be the internal bisector of $\mathit{\angle A}$ and $I$ the incenter.

Figure 1.6:

Then we know that

$$\begin{array}{llll}\hfill \frac{\mathit{BD}}{\mathit{DC}}& =\frac{\mathit{AB}}{\mathit{AC}}=\frac{c}{b}=\frac{\sin C}{\sin B}& \hfill & \\ \hfill \sin B·\mathit{BD}& =\sin C·\mathit{DC}& \hfill \text{(1.9)}\end{array}$$

Also, as $\mathit{BI}$ bisects $\mathit{\angle B}$,

$$\frac{\mathit{AI}}{\mathit{ID}}=\frac{\mathit{AB}}{\mathit{BD}}$$

As $\mathit{CI}$ bisects $\mathit{\angle C}$,

$$\frac{\mathit{AI}}{\mathrm{\Pi }D}=\frac{\mathit{AC}}{\mathit{CD}}$$

$$\begin{array}{llll}\hfill \therefore \frac{\mathit{AI}}{\mathit{ID}}& =\frac{\mathit{AB}}{\mathit{BD}}=\frac{\mathit{AC}}{\mathit{CD}}& \hfill & \\ \hfill \therefore \frac{\mathit{AI}}{\mathit{ID}}\text{is also}\frac{\mathit{AB}+\mathit{AC}}{\mathit{BD}+\mathit{CD}}& =\frac{\mathit{AB}+\mathit{AC}}{\mathit{BC}}& \hfill & \\ \hfill & =\frac{c+b}{a}& \hfill & \\ \hfill & =\frac{\sin B+\sin C}{\sin A}& \hfill & \\ \hfill i.e.,\mathit{AI}\sin A& =(\sin B+\sin C)\mathit{ID}& \hfill \text{(1.10)}\end{array}$$ Now, since $\sin B·\mathit{BD}=\sin C·\mathit{DC}$ from (1.9) $$\begin{array}{lll}\hfill \overline{\mathit{PB}}\sin B+\overline{\mathit{PC}}\sin C& =(\sin B+\sin C)\overline{\mathit{PB}}& \hfill \text{(1.11)}\end{array}$$

$$\begin{array}{llll}\hfill \overline{\mathit{PA}}\sin A+\overline{\mathit{PB}}\sin B+\overline{\mathit{PC}}\sin C& =\overline{\mathit{PA}}\sin A+(\sin B+\sin C)\overline{\mathit{CD}}& \hfill & \\ \hfill & =(\sin A+\sin B+\sin C)\overline{\mathit{PI}}& \hfill \text{(1.12)}\end{array}$$ Since $\mathit{AI}\sin A=(\sin B+\sin C)\mathit{ID}$ from (1.10).
But we know that in a $\mathrm{\Delta }$,

$$\sin A+\sin B+\sin C=4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$$

(1.13)

Using (1.13) in (1.12), we have the required resultant $=4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}·\overline{\mathit{PI}}$.

1.2 Resolved Parts

Theorem 1.2.1. The algebraic sum of the resolved parts of two forces in any direction is equal to the resolved part of the resultant in the same direction.

Proof :

Figure 1.7:

Let $\mathit{AB}$ and $\mathit{AD}$ represent completely the forces $P$ and $Q$ and $\mathit{AX}$ be the direction in which the forces are to be resolved. Complete the parallelogram $\mathit{ABCD}$ so that the resultant $R$ is represented by $\mathit{AC}$.
Draw $\mathit{BL}$, $\mathit{DN}$ and $\mathit{CM}$ perpendiculars to $\mathit{OX}$ and $\mathit{BK}\perp \mathit{CM}$.
Then $\mathit{AL},\mathit{AN}$ and $\mathit{AM}$ are the resolved parts of the forces $P,Q,R$ along $\mathit{AX}$.
From the Figure 1.7, $\mathit{AD}$ makes an obtuse angle with $\mathit{AX}$ and so the resolved part of $Q$ is $-\mathit{AN}$.
To Prove : $\mathit{AL}+\mathit{AN}=\mathit{AM}$.
The triangles $\mathit{DAN}$ and $\mathit{CBK}$ are congruent and hence $\mathit{AN}=\mathit{BK}$.

$$\therefore \mathit{AL}\pm \mathit{AN}=\mathit{AL}\pm \mathit{BK}=\mathit{AL}\pm \mathit{LM}=\mathit{AM}.$$

Obviously the above theorem can be extended to the resultant of any number of forces acting at a point.
Suppose $P_1,P_2,P_3$ are three forces acting at $O$.
Let $R_1$, be the resultant of $P_1$ and $P_2$ and $R_2$ be the resultant of $R_1$ and $P_3$.
Applying the theorem to the two sets of three forces $P_1,P_2,R_1$, and $R_1,P_3,R_2$, we have

$$\begin{array}{lll}\hfill \text{resolved part of}{R}_1\text{along}\mathit{OX}& =\text{resolved part of}{P}_1+\text{resolved part of}{P}_2& \hfill \text{(1.14)}\\ \hfill \text{and resolved part of}{R}_2\text{along}\mathit{OX}& =\text{resolved part of}{R}_1+\text{resolved part of}{P}_3& \hfill \text{(1.15)}\end{array}$$

Combining (1.14) and (1.15), we have resolved part of $R_2$ = resolved part of $P_1$ + resolved part of $P_2$ + resolved part of $P_3$ and so on. □

Hence in a generalised form, we have the theorem:

Theorem 1.2.2. The algebraic sum of the resolved parts of a number of forces in any direction is equal to the resolved part of the resultant in the same direction.

1.3 Resultant of any number of forces acting at a point: Graphical Method

Figure 1.8:

Let $P,Q,R,S$ be the forces acting at $O$.
Take a point $A$ and draw lines $\mathit{AB},\mathit{BC},\mathit{CD}$ and $\mathit{DE}$ to represent successively the forces $P,Q,R$ and $S$ in magnitude and direction.
Compounding the forces by vector law, step by step, we have

$$\begin{array}{llll}\hfill P+Q& =\overline{\mathit{AB}}+\overline{\mathit{BC}}=\overline{\mathit{AC}}& \hfill & \\ \hfill P+Q+R& =\overline{\mathit{AC}}+\overline{\mathit{CD}}=\overline{\mathit{AD}}& \hfill & \\ \hfill \text{and}P+Q+R+S& =\overline{\mathit{AD}}+\overline{\mathit{DE}}=\overline{\mathit{AE}}& \hfill & \end{array}$$

Hence the required resultant is represented in magnitude and direction by the line $\mathit{AE}$. The same construction will apply for any number of forces. The Figure 1.8 $\mathit{ABCDE}$ is said to be the force-polygon.

Example 1.3.1. $\mathit{ABCD}$ is a quadrilateral and forces acting at a point are represented in direction and magnitude by $\mathit{BA},\mathit{BC},\mathit{CD}$ and $\mathit{DA}$. Find their resultant.

Solution. We have

$$\begin{array}{llll}\hfill \overline{\mathit{BC}}+\overline{\mathit{CD}}+\overline{\mathit{DA}}& =\overline{\mathit{BA}}& \hfill & \\ \hfill \overline{\mathit{BA}}+(\overline{\mathit{BC}}+\overline{\mathit{CD}}+\overline{\mathit{DA}})& =\overline{\mathit{BA}}+\overline{\mathit{BA}}=2\overline{\mathit{BA}}& \hfill & \end{array}$$

Hence the resultant is $2\mathit{BA}$, both in magnitude and direction.

1.4 Resultant of any number of coplanar forces acting at a point: Analytical Method

Figure 1.9:

Let forces $P_1,P_2,P_3,\dots P_n$ act at $O$. Through $O$, draw two lines $\mathit{OX}$ and $\mathit{OY}$ at right angles to each other in the place of the forces. Let the lines of action of $P_1,P_2,\dots ,P_n$ make angles ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n$ with $\mathit{OX}$. Let $R$ be the resultant inclined at an angle $\theta$ to $\mathit{OX}$. Then,

$$\begin{array}{llll}\hfill R\cos \theta & =\text{resolved part of the resultant along}\mathit{OX}& \hfill & \\ \hfill & =\text{algebraic sum of the resolved parts of}{P}_1,P_2,\dots P_n,\text{along}\mathit{OX}.& \hfill & \\ \hfill & =P_1\cos {\theta }_1+P_2\cos {\theta }_2+\cdots +P_n\cos {\theta }_n,& \hfill & \\ \hfill R\cos \theta & =X\text{(say)}& \hfill \text{(1.16)}\\ \hfill R\sin \theta & =\text{resolved part of the resultant along}\mathit{OY}& \hfill & \\ \hfill & =\text{algebraic sum of the resolved parts of}{P}_1,P_2,\dots P_n,\text{along}\mathit{OY}.& \hfill & \\ \hfill & =P_1\sin {\theta }_1+P_2\sin {\theta }_2+\cdots +P_n\sin {\theta }_n,& \hfill & \\ \hfill R\sin \theta & =Y\text{(say)}& \hfill \text{(1.17)}\end{array}$$

Squaring (1.16) and (1.17) and adding, we have

$$\begin{array}{lll}\hfill R^2& =X^2+Y^2\text{i.e.,}R=\sqrt{X^2+Y^2}& \hfill \text{(1.18)}\\ \multicolumn{3}{c}{\text{Dividing (}\text{1.17}\text{) by (}\text{1.16}\text{), we get}}\\ \\ \hfill \tan \theta & =\frac{Y}{X}\theta ={\tan }^{-1}\left(\frac{Y}{X}\right)& \hfill \text{(1.19)}\end{array}$$

Equations (1.18) and (1.19) give respectively the magnitude and direction of the resultant.

Example 1.4.1. $\mathit{ABCDEF}$ is a regular hexagon and at $A$, act forces represented by $\overline{\mathit{AB}},2\overline{\mathit{AC}},3\overline{\mathit{AD}},4\overline{\mathit{AE}}$ and $5\overline{\mathit{AF}}$. Show that the magnitude of the resultant is $\mathit{AB}·\sqrt{351}$ and that it makes an angle ${\tan }^{-1}\left(\frac{7}{\sqrt{3}}\right)$ with $\mathit{AB}$.

Solution.

Figure 1.10:

Let $a$ be the side of the hexagon.

Each interior angle of a regular hexagon $=120^{◦}$.

$$\mathit{\angle CAB}=\mathit{\angle ACB}=\mathit{\angle FAE}=\mathit{\angle FEA}=30^{◦}$$

From the isosceles $\mathrm{\Delta }\mathit{ABC}$.

$$\mathit{AC}=2\mathit{AB}\cos 30^{◦}=2a\frac{\sqrt{3}}{2}=a\sqrt{3}=\mathit{AE}.$$

From the Figure 1.10, $\mathit{\angle AED}=90^{◦}$.

$$\begin{array}{llll}\hfill A{D}^2& =A{E}^2+E{D}^2=3a^2+a^2=4a^2& \hfill & \\ \hfill \mathit{AD}& =2a& \hfill & \end{array}$$

Since the vertices of a regular hexagon lie on a circle,

$$\begin{array}{llll}\hfill \mathit{\angle DAB}+\mathit{\angle DCB}& =180^{◦}& \hfill & \\ \hfill \mathit{\angle DAB}& =180^{◦}-120^{◦}=60^{◦}& \hfill & \\ \hfill \mathit{\angle DAC}& =30^{◦}\text{and}\mathit{\angle EAD}=30^{◦}& \hfill & \end{array}$$

The magnitudes of the forces acting at $A$ are $a,2a\sqrt{3},6a,4a\sqrt{3}$ and $5a$ as shown in the Figure 1.10.

Take $\mathit{AB}$ and $\mathit{AE}$ as axes of $x$ and $y$ and let $R$ be the resultant inclined at an angle $\theta$ to $\mathit{AB}$.

Resolving the forces along $\mathit{AB}$ and $\mathit{AE}$, we have

$$\begin{array}{llll}\hfill R\cos \theta & =a+2a\sqrt{3}\cos 30^{◦}+6a\cos 60^{◦}+5a\cos 120^{◦}& \hfill & \\ \hfill & =a+2a\sqrt{3}\frac{\sqrt{3}}{2}+6a\frac{1}{2}-5a\frac{1}{2}& \hfill & \\ \hfill & =\frac{9a}{2}& \hfill \text{(1.20)}\\ \multicolumn{3}{c}{\text{and}}\\ \\ \hfill R\sin \theta & =2\mathit{ay}\sqrt{3}\cos 60^{◦}+6a\cos 30^{◦}+4a\sqrt{3}+5a\cos 30^{◦}& \hfill & \\ \hfill & =5a\sqrt{3}+\frac{11a\sqrt{3}}{2}& \hfill & \\ \hfill & =\frac{21a\sqrt{3}}{2}& \hfill \text{(1.21)}\end{array}$$

Squaring (1.20) and (1.21) and adding,

$$\begin{array}{llll}\hfill R^2& ={\left(\frac{9a}{2}\right)}^2+{\left(\frac{21a\sqrt{3}}{2}\right)}^2& \hfill & \\ \hfill & =\frac{81a^2}{4}+\frac{441}{4}\times 3a^2& \hfill & \\ \hfill & =\frac{1404a^2}{4}=351a^2& \hfill & \\ \hfill R& =a\sqrt{351}=\mathit{AB}\sqrt{351}& \hfill & \end{array}$$

Dividing (1.21) by (1.20),

$$\tan \theta =\frac{21a\sqrt{3}}{2}\times \frac{2}{9a}=\frac{21\sqrt{3}}{9}=\frac{7\sqrt{3}}{3}=\frac{7}{\sqrt{3}}$$

Hence the resultant is a force of magnitude $\mathit{AB}\sqrt{351}$, in a direction making an angle ${\tan }^{-1}\left(\frac{7}{\sqrt{3}}\right)$ with $\mathit{AB}$.

1.5 Conditions of equilibrium of any number of forces acting upon a particle

Forces acting at a point are in equilibrium when their resultant is zero. i.e., number of forces acting at a point of a rigid body or on a particle, in order that the body, or the particle may be at rest.

Geometrical or graphical conditions

If forces acting at a point are represented in magnitude and direction by lines forming the successive sides of a polygon, then for equilibrium, the polygon must be closed. When there are only three forces acting on a particle, the conditions of equilibrium are often most easily found by applying Lami’s theorem.

Analytical Conditions

If we resolve the forces in any two directions at right angles and the sums of the components in these directions be $X$ and $Y$, the resultant $R$ is given by $R^2=X^2+Y^2$.
If the forces are in equilibrium, $R=0$.
Then $X^2+Y^2=0$.
Now, the sum of the squares of two real quantities cannot be zero unless each quantity is separately zero.
$\therefore X=0$ and $Y=0$.

Hence, if any number of forces acting at a point are in equilibrium, the algebraic sums of the resolved parts of the forces in any two perpendicular directions must be zero separately.

Conversely if the algebraic sum of the resolved parts of the forces acting at a point in any two perpendicular directions are zero separately, the forces will be in equilibrium.

This is because when $X=0$ and $Y=0$, we must have $R=0$.

Example 1.5.1. Forces acting at a point represented in magnitude and direction by $\overline{\mathit{AB}},2\overline{\mathit{BC}},2\overline{\mathit{CD}},\overline{\mathit{DA}}$ and $\overline{\mathit{DB}}$ where $\mathit{ABCD}$ is a square. Show that the forces are in equilibrium.

Solution.

$$\begin{array}{llll}\hfill \overline{\mathit{AB}}+2\overline{\mathit{BC}}+2\overline{\mathit{CD}}+\overline{\mathit{DA}}+\overline{\mathit{DB}}& =(\overline{\mathit{AB}}+\overline{\mathit{BC}}+\overline{\mathit{CD}}+\overline{\mathit{DA}})+(\overline{\mathit{BC}}+\overline{\mathit{CD}}+\overline{\mathit{DB}})& \hfill & \\ \hfill & =\overline{0}& \hfill & \end{array}$$

Since the forces in the first set of brackets are in equilibrium by the polygon of forces (square $\mathit{ABCD}$) and the forces in the second set are in equilibrium by the triangle of forces (triangle $\mathit{BCD}$)]

Example 1.5.2. $\mathit{ABCD}$ and $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$ are parallelograms. Prove that forces $\overline{A{A}^{\prime }},\overline{B^{\prime }B},\overline{C{C}^{\prime }}$ and $\overline{D^{\prime }D}$ acting at a point will keep it at rest.

Solution. Let $G$ and $G^{\prime }$ be the points of intersection of the diagonals. By the polygon of forces, from the quadrilateral $\mathit{AG}{G}^{\prime }{A}^{\prime }$.

$$\begin{array}{llll}\hfill \overline{A{A}^{\prime }}& =\overline{\mathit{AG}}+\overline{G{G}^{\prime }}+\overline{G^{\prime }{A}^{\prime }}& \hfill & \\ \hfill \overline{B^{\prime }B}& =\overline{B^{\prime }{G}^{\prime }}+\overline{G^{\prime }G}+\overline{\mathit{GB}}& \hfill & \\ \hfill \overline{C{C}^{\prime }}& =\overline{\mathit{CG}}+\overline{G{G}^{\prime }}+\overline{G^{\prime }{C}^{\prime }}& \hfill & \\ \hfill \overline{D^{\prime }D}& =\overline{D^{\prime }{G}^{\prime }}+\overline{G^{\prime }G}+\overline{\mathit{GD}}& \hfill & \end{array}$$ Adding up, $$\overline{A{A}^{\prime }}+\overline{B^{\prime }B}+\overline{C{C}^{\prime }}+\overline{D^{\prime }D}=0$$

The concerned vectors in the right side are equal and opposite.

Example 1.5.3. $E$ is the middle point of the side $\mathit{CD}$ of a square $\mathit{ABCD}$. Forces $16,20,4\sqrt{5},12\sqrt{2}$ kg. wt. act along $\mathit{AB},\mathit{AD},\mathit{EA},\mathit{CA}$ in the directions indicated by the order of the letters. Show that they are in equilibrium.

Solution.

Take $\mathit{AB}$ and $\mathit{AD}$ as axes of $X$ and $Y$.
Produce $\mathit{EA}$ to $F$ and let $\mathit{\angle BAF}=\theta$.
Produce $\mathit{CA}$ to $G$.

$$\mathit{\angle BAG}=\mathit{\angle BAC}+\mathit{\angle CAG}=45^{◦}+180^{◦}=225^{◦}$$

Let $R$ be the resultant of the forces inclined at an angle $\theta$ to $\mathit{AB}$.
Resolving the forces along $\mathit{AB}$ and $\mathit{AD}$, we have

$$\begin{array}{llll}\hfill R\cos \theta & =16+12\sqrt{2}\cos 225^{◦}+4\sqrt{5}\cos \theta & \hfill & \\ \hfill & =16+12\sqrt{2}\cos (180^{◦}+45^{◦})+4\sqrt{5}\cos \theta & \hfill & \\ \hfill & =16+12\sqrt{2}\times -\cos 45^{◦}+4\sqrt{5}\cos \theta & \hfill & \\ \hfill & =16-12\sqrt{2}\times \frac{1}{\sqrt{2}}+4\sqrt{5}\cos \theta & \hfill & \\ \hfill & =4+4\sqrt{5}\cos \theta & \hfill \text{(1.22)}\\ \hfill \mathit{\angle BAE}& =\mathit{\angle BAF}-\mathit{\angle EAF}=\theta -180^{◦}& \hfill & \\ \hfill \text{and}\mathit{\angle DEA}& =\text{alt}\mathit{\angle BAE}=\theta -180^{◦}& \hfill & \\ \hfill \text{From the}\mathrm{\Delta }\mathit{AED}\text{,}A{E}^2& =A{D}^2+D{E}^2& \hfill & \\ \hfill & =a^2+\frac{a^2}{4}\text{aeing the side of the square}& \hfill & \\ \hfill & =\frac{5a^2}{4}& \hfill & \\ \hfill \mathit{AE}& =\frac{a\sqrt{5}}{2}& \hfill & \\ \hfill \cos (\theta -180^{◦})& =\frac{\left(\frac{a}{2}\right)}{\left(\frac{a\sqrt{5}}{2}\right)}=\frac{1}{\sqrt{5}}& \hfill & \\ \hfill \cos (180^{◦}-\theta )& =\frac{1}{\sqrt{5}}=-\cos \theta & \hfill & \\ \hfill \cos \theta & =-\frac{1}{\sqrt{5}}& \hfill \text{(1.23)}\end{array}$$

From

$$\begin{array}{lll}\hfill R\cos \theta & =4+4\sqrt{5}\times -\frac{1}{\sqrt{5}}=0& \hfill \text{(1.24)}\\ \hfill R\sin \theta & =12\sqrt{2}\sin 225^{◦}+4\sqrt{5}\sin \theta +20& \hfill & \\ \hfill & =12\sqrt{2}\sin (180^{◦}-45^{◦})+4\sqrt{5}\sin \theta +20& \hfill & \\ \hfill & =-12\sqrt{2}\sin 45^{◦}+4\sqrt{5}\sin \theta +20& \hfill & \\ \hfill & =-12\sqrt{2}\frac{1}{\sqrt{2}}+4\sqrt{5}\sin \theta +20& \hfill & \\ \hfill & =8+4\sqrt{5}\sin \theta & \hfill \text{(1.25)}\end{array}$$

From the $\mathrm{\Delta }\mathit{AED}$

$$\begin{array}{llll}\hfill \sin (\theta -180^{◦})& =\frac{\mathit{AD}}{\mathit{AE}}=\frac{a}{\left(\frac{a\sqrt{5}}{2}\right)}=\frac{2}{\sqrt{5}}& \hfill & \\ \hfill -\sin (180^{◦}-\theta )& =\frac{2}{\sqrt{5}}\text{or}\sin \theta =-\frac{2}{\sqrt{5}}& \hfill & \end{array}$$

From (1.25),

$$\begin{array}{llll}\hfill R\sin \theta & =8+4\sqrt{5}\times -\frac{2}{\sqrt{5}}& \hfill & \\ \hfill & =8-8=0& \hfill \text{(1.26)}\end{array}$$

Squaring (1.24) and (1.26) and adding,

$$R^2=0+0=0$$

The forces are in equilibrium.