## Projectile Motion

• ### Introduction

Projectile motion is a 2D motion. Common examples are football hit by a player, bullet fired from a gun, catapult, water coming out of hose pipe, mud particles thrown away from wheels of a vehicle, a high-jump athlete etc.

Note that aeroplane/spacecraft are NOT projectiles, because the propulsive force can be adjusted after their launch. In all previous cases, once the object is launched, its motion is purely under gravitational influence.

• ### Analysis

Let the velocity of projection be $\vec V$ and the angle of projection be $\theta$. The motion can be resolved into 2 motions:

a) along horizontal direction : This is a uniform velocity motion, $V cos \ \theta$

b) along vertical direction : This is a uniformly accelerated motion, with initial velocity $V sin \ \theta$ and acceleration $\vec g$

This resolution makes the analysis simpler. We find the following factors:

1) Max. height reached $H$

At the top, vertical velocity $=0$. Using $v^2 = u^2 + 2 as$,

${H = \frac {V^2 sin^2 \ \theta}{2g}}$

2) Time of flight $T$

Time required to reach the top will be $T/2$. Using $v = u + at$,

${\frac T 2 = \frac {V sin \ \theta}{g}}$

So,

${T = \frac {2 V sin \ \theta}{g}}$

3) Horizontal range $R$

This is the distance between the point of launch and the point where the projectile lands. Using $s = ut$,

${R = V cos \ \theta \times \frac {2V sin \ \theta}{g} = \frac {V^2 sin \ 2 \theta} {g}}$

The maximum values of $R$ and $H$ can be obtained by substituting maximum values of trigonometric functions, $sin \ \theta$ and $cos \ \theta$.

4) Equation of trajectory

Using $s = ut + \frac 1 2 a t^2$,

${x = V cos \theta \ t \ and \ y = V sin \theta \ t - \frac 1 2 gt^2}$

On eliminating $t$, we get,

$y = x \ tan \theta - \frac {g x^2}{2 V^2 cos^2 \theta}$

This is of the form $y = Ax - Bx^2$, which is a parabola.

• ### Analysis of motion, when the ground is not horizontal, but inclined at an angle $\phi$ with the horizontal

There are 2 ways of analyzing this motion:

I) Resolving the acceleration vector $\vec g$ along the ground and perpendicular to the ground :

In this case, both motions are uniformly accelerated, $\vec a_x = g \ cos \ \phi$ and $\vec a_y = g \ sin \ \phi$

II) Considering the equation of ground as $y = tan \ \phi \ x$

All quantities can be found by considering equations of parabola and straight line.

1) Time of flight, ${T = \frac {2 V sin \theta}{g cos \phi}}$

2) Range along plane, ${R = \frac {2 V^2 sin \theta}{g cos^2 \phi} \times cos (\theta + \phi)}$

For maximum range, ${\theta + \frac \phi 2 = \frac {\pi}{4} = 45^o}$

3) Max. height reached (perpendicular to ground),  ${H = \frac {V^2 sin^2 \theta}{2 g cos \phi}}$

• ## Few more important points

I) Throughout this analysis, air resistance, effect of wind and effect of rotation of earth are neglected.

II) For same range, there are 2 angles of projections possible. These angles are complementary, i.e. $\theta_1 + \theta_2 = \frac \pi 2$

III) At any point, the radius of curvature is ${\rho = \frac {V_{at \ that \ point}^2}{g cos \ \zeta}}$. $\zeta$ is the angle made by velocity at that point with horizontal.

IV) Minimum radius of curvature is at the max. height

## Applications of Vectors in Geometry

Vectors are such a powerful tool in mathematics and physics, that many results can be proved very easily and intuitively.

#### Statement : If the diagonals of a parallelogram are congruent, then it is a rectangle.

Proof :

Let $ABCD$ be the parallelogram. Let $A$ be the origin. So, the position vectors will be from $A$. Diagonals are $AC$ and $BD$. We have $|\vec {AC}| = |\vec {BD}|$.

Therefore, $|\vec c| = |\vec d - \vec b|$. By parallelogram law of vector addition, $\vec c = \vec d + \vec b$. So,

${|\vec c| = |\vec d - \vec b| = |\vec d + \vec b|}$

On squaring,

${\vec d \cdot \vec d - 2 \vec d \cdot \vec b + \vec b \cdot \vec b = \vec d \cdot \vec d + 2 \vec d \cdot \vec b + \vec b \cdot \vec b}$

So,

${- 2 \vec d \cdot \vec b = 2 \vec d \cdot \vec b}$

This implies $\vec d \cdot \vec b = 0$, or $\vec b$ is perpendicular to $\vec d$. Hence, $\angle A$ is a right angle. This being a parallelogram, will have all other angles equal to $90^o$ and hence it is a rectangle.

Statement : The diagonals of a kite are at right angles.

Proof :

Let $ABCD$ be the kite, with $A$ as the origin. Clearly $BC$ and $DC$ are equal sides. In terms of vectors,

${|\vec {BC}| = |\vec {DC}| \ \ i.e. \ \ |\vec c - \vec b| = |\vec c - \vec d|}$

On squaring,

${\vec c \cdot \vec c - 2 \vec c \cdot \vec b + \vec b \cdot \vec b = \vec c \cdot \vec c - 2 \vec c \cdot \vec d + \vec d \cdot \vec d}$

Canceling common terms,

${- 2 \vec c \cdot \vec b = -2 \vec c \cdot \vec d}$

Note that $AB$ and $AD$ are equal. So, $b^2 = d^2$. So,

${\vec c \cdot (\vec b - \vec d) = 0}$

Hence, $\vec {AC} \cdot \vec {DB} = 0$. Hence the diagonals are perpendicular.

Statement : If in a tetrahedron, edges in each of the two pairs of opposite edges are perpendicular then the edges in the third pair are also perpendicular.

Proof:

In a tetrahedron, each triangle shares an edge with the other. Considering any 2 triangular faces, we are left with only 1 edge. The pair of common edge and the uncommon edge is said to be a pair of opposite edges. Let $OABC$ be a tetrahedron. So, $(OA,BC)$, $(AB, OC)$ and $(OB,AC)$ are the pairs of opposite edges. Let any 2 of them be perpendicular.

$\vec {OA} \cdot \vec {BC} = 0$ and $\vec {OC} \cdot \vec {AB} =0$. Therefore,

${\vec a \cdot (\vec c - \vec b) = 0 \ \ and \ \ \vec c \cdot {\vec b - \vec a} = 0}$

Expanding the brackets and then adding the equations,

${\vec a \cdot \vec c - \vec a \cdot \vec b + \vec c \cdot \vec b - \vec c \cdot \vec a =0}$

Or

${- \vec a \cdot \vec b + \vec c \cdot \vec b = 0}$

This gives $\vec b \cdot (\vec c - \vec a ) = 0$ i.e. $\vec {OB} \cdot \vec {AC} = 0$. Hence $OB$ is perpendicular to $BC$ and these 2 form the third pair.

• ## Introduction

Curvilinear stands for along a curve. This is the most general specification of motion. It can either be two-dimensional or three-dimensional.

Examples of 2D motions are motion of a projectile and circular motion. These involve motions along planar curves.

Examples of 3D motions are a roller coaster, satellite launching etc. They follow a space curve. These are slightly complicated to analyze as 3 space variables are involved.

• ## Description

We defined the quantities distance $s$, displacement $\vec r$, position vector $\vec {r_p}$, average speed $v_{avg}$, instantaneous speed $v_{inst}$, average velocity $\vec v_{avg}$, instantaneous velocity $\vec v_{inst}$ and instantaneous acceleration $\vec a_{inst}$ and jerk $\vec j$ ” for rectilinear motion. Their definitions are valid for curvilinear motion as well. We will use them extensively.

• ## Specifying the Quantities – Coordinate Systems

The quantities are specified by considering certain reference (the frame). There are 3 general systems of specifications:

• ### Rectangular Coordinate System

It involves 3 mutually perpendicular axes, $X$, $Y$ and $Z$. The axes intersect at a point, known as origin. These axes are fixed. Thus, the unit vectors $\hat i, \hat j$ and $\hat k$ are fixed. (THIS IS IMPORTANT)

Any vector can be resolved into its components by projecting it on respective axes. Thus,

${\vec r = x \hat i + y \hat j + z \hat k}$

${\vec v = \frac {dx}{dt} \hat i + \frac {dy}{dt} \hat j + \frac {dz}{dt} \hat k}$

${\vec a = \frac {d^2x}{dt^2} \hat i + \frac {d^2y}{dt^2} \hat j + \frac {d^2z}{dt^2} \hat k}$

When motion is 2-dimensional, $\hat k$ component is absent.

• ### Specifying Radial and Transverse Components (Polar Coordinates)

The radial direction is that direction to which the radius vector (or position vector) $\vec r$ is directed. The unit vector along radial direction is denoted by $\hat r$. The transverse direction is obtained, when $\hat r$ is rotated through $90^o$. Thus,

${\vec r = r \hat r}$

${\vec v = \frac {d}{dt} (r \hat r) = \frac {dr}{dt} \hat r + r \frac {d}{dt} \hat r = (\dot r) \hat r + (r \dot \theta) \hat s}$

$\hat s$ is the unit vector in transverse direction. Similarly,

${\vec a = \frac {d \vec v}{dt} = [\ddot r - r(\dot \theta)^2]\hat r + [2 \dot r \dot \theta + r \ddot \theta] \hat s}$

$\theta$ denotes the angular position of the particle. Hence, $\dot \theta$ is the rate of rotation and $\ddot \theta$ is the rate of change of rate of rotation.

Thus, both velocity and accelerations have radial and transverse components.

Note: Radial component of velocity describes the rate of movement of an object from an observer. In astronomy, the radial velocity describes how quick is the star receding the earth. Transverse component of velocity describes the rate of movement of an object perpendicular to the observer. For an observer observing a star, this is perpendicular to his line of sight.

These 2 apparent components tell us the actual relative velocity of star w.r.t earth.

• ### Specifying Normal and Tangential Components

In this type, a vector is resolved along 2 components, one along the tangent to the path and the other normal to the path. The unit vector along tangent is $\hat e_T$ and along normal is $\hat e_N$. Unlike $\hat i , \hat j$ and $\hat k$, the unit vectors $\hat e_T, \hat e_N$ may change their directions. (THIS IS IMPORTANT).

The velocity vector $\vec v$ is always tangential to the path. Hence,

${\vec v = v \hat e_T + 0 \hat e_N}$

${\vec a = \frac {d \vec v}{dt} = \frac {d}{dt} (v \hat e_T) = \frac {dv}{dt} \hat e_T + v \frac {d \hat e_T}{dt} .... by \ Chain \ Rule}$

The time derivative of $\hat e_T$ is not equal to zero, as it changes its direction. It is equal to $\frac {v}{\rho} e_N$. Therefore,

${\vec a = \frac {dv}{dt} \hat e_T + \frac {v^2}{\rho} \hat e_N}$

$\rho$ is the radius of curvature of the curve at that point.

Note: Since the velocity is always tangential to the path, it has only 1 component in this representation. Normal component is zero.

• ## Concept of Radius of Curvature

The term radius is well defined for circles. We also say that for a straight line, the radius is infinite. For any other curve, radius of curvature at a point is the radius of that circle, which best fits the curve or, which shares common tangent with the curve.

Mathematically, if $y=f(x)$ is the equation of curve, then radius of curvature is given by

${\rho = \frac {\Big[1+ \big( \frac {dy}{dx} \big)^2 \Big]^{3/2}} {\big( \frac {d^2y}{dx^2}\big) }}$

• ## Introduction

Circular motion is a common type of motion and is a part of a more general set of curvilinear motions. In such motions, the forces acting on a body have a component directed along the radius of curvature of the curve towards the center of curvature.

When the motion is circular, this force is directed along the center of the circle and is along the radius of the circle.

The specific name given to that component of force is the centripetal force. So, it is a real force.

Uniform circular motion is the motion, where the magnitude of velocity of the rotating object, i.e. $|\vec v|$ is constant. The only component of acceleration is because of the change in direction of the velocity. For example, a stone attached to a string and revolved in a horizontal circle.

The centripetal force in above case is due to the tension in the string. Its magnitude is

${\vec F_{centripetal} = \frac {mv^2}{r} = mr \omega^2}$

This is equal to the tension in the string.

The centripetal acceleration is directed towards the center, and is given by

${\vec a_{centripetal} = \frac {\vec F_{centripetal}}{m} = \frac {v^2}{r} = r \omega^2}$

The weight of stone DOES NOT play any role.

• ## Vertical Circular Motion

In vertical circular motion, the resultant acceleration is a result of 2 accelerations:

1) due to a continuous change in the direction of velocity of the rotating body and

2) due to acceleration due to gravity

At any instant, the gravity vector $\vec g$ is always directed vertically downward. The acceleration due to change in direction is directed along the radius of the circle and is known as the centripetal acceleration.

Thus, the body is under the influence of 2 forces, viz. the tension $T$ and its weight $mg$.

Note the directions of real forces at each of the 4 points $A,B,C$ and $D$.

• ### Analysis from Stone’s Frame of Reference

For 2 cases discussed above (viz. horizontal and vertical circular motions), when you refer to the ground as the frame of reference, the Newton’s 2nd law of motion governs the motion.

However, for the same cases, when observed from the frame of reference of stone (which is stationary from its frame), one has to account for an additional fictitious force, to bring the stone to standstill. This force is known as the centrifugal force.

Thus, we can write,

${\vec T + m \vec g = \vec F_{centrifugal}, \ vertical \ motion}$

and

${\vec T = \vec F_{centrifugal}, \ horizontal \ motion}$

• ## D’Alembert’s Principle and Circular Motion

The principle converts a problem of dynamics into a problem of statics. i.e. now we assume the rotating stone as our frame of reference. By adding the inertial force (which is the centrifugal force), we get,

${\vec T + m \vec g = \frac {m v^2}{r}}$

OR

${\vec T = \frac {mv^2}{r} - m \vec g}$

• ## Tensions at Top and Bottom Positions

At the top,

${T = \frac {mv^2}{r} - mg}$

At the bottom,

${T = \frac {mv^2}{r}+mg}$

Note that the directions of forces have been taken into consideration.

So, the least speed required at the top, s.t. the string does not slack ($T \to 0$) will be $v_{top} = \sqrt {gr}$.

Using the principle of conservation of energy, we can write,

${\frac 1 2 m v^2_{bottom} = \frac 1 2 m v^2_{top} + mg (2r)}$

$r$ is the radius of the circle, as shown.

• ## The Origin of Friction

Friction is defined as the resistive force to motion, offered by the contacting surfaces. Another common observation is generation of heat from the contacting surfaces. The applied kinetic energy is converted to heat. The work done is $0$, till the friction is overcome.

The origin of friction is in the nature of surfaces, surface irregularities, roughness, deformation of surface etc. It ultimately boils down to the electromagnetic forces between charged particles. However, it is impractical as well as uneconomical to compute the magnitude of frictional force by this method. Hence, empirical methods are common in studying friction.

Tribology is a subject under mechanical engineering, where friction is studied extensively.

• ## Terms

I) Dry Friction – Surfaces are dry, magnitude of frictional force is generally high.

II) Fluid Friction – Relatively smaller magnitudes, used for lubrication. Exists between a solid and a liquid as well as between 2 liquids. Drag is a kind of fluid friction, where a fluid tries to move across a solid.

III) Limiting Friction – The maximum resistance offered due to friction. If the applied force is more than limiting friction, the bodies will move relative to each other.

IV) Static Friction – Friction between stationary surfaces

V) Kinetic Friction – Friction between moving (sliding/ rolling) surfaces

• ## Analysis of Friction

We will consider the case of limiting friction on a block under the influence of self-weight. Let $P$ be the applied force, $R$ the normal reaction, $W$ the weight and $F_{fr}$ the limiting frictional force. The body will be in equilibrium under these forces.

The coefficient of static friction, $\mu_s$ is defined by

${\mu_s = \frac {F_{fr}}{R}}$

If the applied force leads to motion, we consider the coefficient of kinetic friction. The kinetic friction is generally lower than static friction.

${\mu_d = \frac {F_{fr}^{kinetic}}{R}}$

• ## Angle and Cone of Friction

The angle of friction $\phi$ is the inverse tangent of the coefficient of friction. The cone of friction is an imaginary cone with $\phi$ as the semi-cone angle, normal reaction $R$ as height and the resultant of $R$ and $F_{fr}$ as the slant height.

The significance of cone of friction is as follows:

Let $P$ be the resultant of all external forces acting on the body, excluding the reaction and friction. If the body is in static equilibrium, then $P$ must lie inside the cone. If it doesn’t, the body will no longer be in equilibrium.

• ## Laws of Friction

These are generalized statements, based on observations:

1) Frictional force acts in a direction opposite to the movement of the body.

2) For static friction, the maximum resistive force is given by $\mu_s R$

3) For kinetic friction, the maximum resistive force is given by $\mu_d R$

4) The frictional force is independent of the area in contact, however, it depends on the nature of surfaces.

Now we will see few applications of friction with corresponding mathematical analyses. These are

I) Belt Drives

II) Band Brakes and

III) Wedges

• ## Belt Friction

Pulleys and belts are often used to transmit power from one point to another. (In flour mills, you must have seen a belt connecting the mill and the motor). In the absence of friction, the rotation of pulleys will not lead to movement of belt. Hence, friction is very important in power transmission.

Depending on the applications, the belts are chosen. The problems involve calculating the tensions on tight $(T_1)$ and slack $(T_2)$ sides. The relationship between $T_1$ and $T_2$ depends on the coefficient of friction $\mu$ \$between the belt material and pulley and the angle of lap $\beta$. The angle of lap is the angle made by an arc along the pulley, where the belt is lapped on the pulley. It is measured in radians (Always!).

### I) Flat Belts

The cross-section of a flat-belt is rectangular. For flat belts, the relationship between $T_1$ and $T_2$ is given by

${\frac {T_1}{T_2} = e^{\mu \beta}}$

### II) V Belts

V belts have trapezoidal cross-section. If $\alpha$ is the angle made by the imaginary triangle at its vertex (see the figure), then we have

${\frac {T_1}{T_2} = e^{\big( \frac {\mu \beta}{sin \ \alpha} \big)}}$

$\alpha$ is also known as the angle of groove.

• ## Brakes (Band Brake)

Friction can be used as a power-transmitting agent (in belts-pulleys) as well as a power-absorbing agent (in band brakes). The idea is, the band resists the motion of rotating wheel through friction. The braking torque is obtained by the relations

${\frac {T_1}{T_2} = e^{\mu \beta}}$

and

${Torque \ = \ (T_1 - T_2) \times \ radius \ of \ wheel}$

In this case, the values of $T_1$ and $T_2$ depend on the applied force $P$ as well. Drawing the free body diagram of lever and pulley is necessary.

• ## Wedge and Block Friction

A wedge is a triangular tool and an inclined plane, used for

I) producing large forces or

II) giving small displacements

When a force applied to its blunt end, the force gets divided into forces, normal to its inclined surfaces. (See the figure below)

While analyzing the wedge action, the weight of wedge is generally neglected.

## Statics : Trusses, Cables and Frames

The analysis of trusses, cables and frames is based on the static equilibrium of these structures. So, the equations of equilibrium are extensively used. There are certain differences as far as the definitions are concerned, but the analysis is similar.

• # Trusses

• ## Structure of Howrah Bridge

The Howrah bridge is a cantilever truss bridge, built in 1943. Typical structure of a cantilever bridge is shown below, with specific details mentioned.

The bridge contains an engineering structure, known as truss.

• ## What is a Truss?

A truss is a structure which contains only two-force members. A two-force member is the one, where force is applied at only two points. The members of a truss are pinned to each other. The members are generally straight and their ends are connected at joints. These joints are referred to as joints. In general, 5 (or more) triangular units are constructed and their ends are pinned at nodes.

The forces are considered to act only on the nodes. No torque acts on any member (An assumption, generally holds well). The forces can be tensile or compressive.

Trusses are used in construction of bridges, towers, roofs etc.

• ## Terms Associated with Trusses

I) Top Chords – The top beams in a truss, generally are in compression

II) Bottom Chords – The bottom beams in a truss, generally are in tension

III) Webs– The interior beams, the areas inside the webs are called panels

IV) Plane Truss– All members are in a plane

V) Space Truss – Members form a 3D structure

VI) Perfect Truss– Let $n$ be the number of members, $j$ be the number of joints and $r$ be the number of total reactions. Then a truss is perfect, when

$n = 2j-r$

VII) Redundant Truss$n > 2j-r$

VIII) Deficient Truss$n < 2j - r$

IX) Cantilever Truss– It is fixed on one side and free on the other. Howrah bridge contains 2 cantilever arms, as shown.

• ## Analysis of Truss – Method of Joints

We will be dealing with perfect planar trusses.

The deformation due to loads is neglected and the friction between the members is assumed to be negligible, so that no torque acts on the member. So, the external forces are in complete balance with the reactions. This proves that although a truss contains multiple members, as a whole, it behaves as a single unit. The general problem is to find the members with highest tension or compression. (Recall that the forces can be tensile or compressive, but they cannot be shear).

In the method of joints, we assume that all forces are tensile. We draw the free body diagram at each joint. The starting point is generally the load application point/ the supports. The unknowns are obtained by

$\sum F_x = 0, \sum F_y = 0, \sum M = 0$

Sometimes, Lami’s theorem is useful.

• ## Analysis of Truss – Method of Sections

The method is useful, when we are asked to find the forces in a limited number of members (generally three). We choose an imaginary plane, cutting those members and apply the conditions of equilibrium on either of the 2 sections. The unknown forces are obtained by solving a system of simultaneous equations.

• # Cables

The famous Vandre-Worli sea link is a bridge supported by cables. Such bridges are known as cable stayed bridges. (See figure below). Another application of cables is in the transmission lines.

• ## Analysis of Cables

The analysis of cables is done by considering the usual equilibrium equations, i.e.

$\sum F_x = 0, \sum F_y = 0, \sum M = 0$

The loads are generally point loads, thus the flexible and inextensible cable will sag or hog only at the load points. Weight of cable is neglected. The forces are assumed to act along the cable.

The analysis is similar to the analysis of trusses by the method of joints.

• ## Frames

A frame is a structure, made up of members pinned (hinged) together. The members may be straight or bent into a shape. The difference between a frame and a truss is that, there exists at least one multi-force member in a frame. Rest of the analysis is similar, because frame is in static equilibrium. The equations

$\sum F_x = 0, \sum F_y = 0, \sum M = 0$

hold true for frames as well.

The load analysis of machine frames plays an important role in the design and theory of machines.

• ## Topics Covered

Dependent Motion and d’Alembert’s Principle

• ## Dependent Motion and Related Problems

When motion of one body depends on the motion of other body (or bodies), it is known as dependent motion. Generally, these bodies are connected by a continuous, inelastic string.

An inelastic string does not change in length under the load, hence the total length of string always remains constant.

The objective of solving problems on dependent motion is to find the resulting motion of dependent bodies, when one of them is moved.

• ### How to approach such problems?

We assume that all motions have at the most constant acceleration, i.e. $\frac {da}{dt}=0$

1) The very first decision one has to take is about the datum or a reference point. If the motions are vertical, a horizontal datum is chosen and if the motions are horizontal, a vertical datum is chosen. Note that for a system with both vertical and horizontal motions, 2 datums will be there.

If the motion is at angle, it is to be resolved vectorially in horizontal and vertical directions (with reference to the datum).

A sign convention is selected for each datum.

2) Now we make use of the fact that string is inextensible. We note the lengths of string connecting each member of the system and form an equation of the kind:

$\sum_{i=1}^{n} x_i = L$

Here, $L$ denotes the total length of the string and $x_i$ denotes the length of $i$th section of the string.

3) On differentiating above equation, we get the velocities and on further differentiating, we get the accelerations of each member.

Thus,

$\sum_{i=1}^{n} v_i = 0$

and

$\sum_{i=1}^{n} a_i = 0$

4) If there $k$ different strings, there will be $k$ equations for lengths.

• ## Special Case – 2 Particles Connected by Single String

Suppose there are 2 particles $A$ and $B$ connected by a string. The arrangement is such that there are $n_A$ parts connected to $A$ and $n_B$ parts are connected to $B$. Then,

$\frac {n_A}{n_B}= \frac {x_A}{x_B} = \frac {v_A}{v_B}= \frac {a_A}{a_B}$

• ## D’Alembert’s Principle

The principle is named after the French mathematician Jean le Rond d’Alembert (1717-1783), who discovered it. It is a modified statement of Newton’s 2nd law of motion.

Consider an accelerating rigid body. This is a dynamic system. D’Alembert showed that the system can be converted into an equivalent static system by adding inertial force and inertial torque.

• ### How is it used?

Consider a body of mass $m$, being pulled by a steel wire having tension $T$ in it. If the block accelerates upwards with an acceleration $a$, we can write

$\frac {T-mg}{m} = a$

This is Newton’s 2nd law.

On rewriting,

$T-mg=ma, \ i.e.\ T-mg + (-ma)=0$

If a force equal to $ma$ is added in opposite direction, the body appears to be under static equilibrium. This force is a fictitious force, called inertial force.

• ### Application

It is easier to determine unknown forces on bodies, which are under static equilibrium than bodies which are moving. The analysis of machines and mechanisms gets tremendously simplified, when they are analyzed as static bodies, using d’Alembert’s principle.