Engineering mechanics is that branch of science which deals with the behaviour of a body when the body is at rest or in motion. The engineering mechanics may be divided into statics and dynamics. The branch of science, which deals with the study of a body when the body is at rest, is known as statics while the branch of science which deals with the study of a body when the body is in motion, is known as dynamics. Dynamics is further divided into kinematics and kinetics. The study of a body in motion, when the forces which cause the motion are not considered, is called kinematics and if the forces are also considered for the body in motion, that branch of science is called kinetics. The classification of Engineering Mechanics are shown in Fig. 1.1 below.

**Note**. Statics deals with equilibrium of bodies at rest, whereas dynamics deals with the motion of bodies and the forces that cause them.

The following terms are generally used in Mechanics :

1. Vector quantity, 2. Scalar quantity, 3. Particle, 4. Law of parallelogram of forces,

5. Triangle law and 6. Lame’s theorem

**1.1.1. Vector Quantity**. A quantity which is completely specified by magnitude anddirection, is known as a vector quantity. Some examples of vector quantities are : velocity,acceleration, force and momentum. A vector quantity is represented by means of a straight line with an arrow as shown in Fig. 1.2. The length of the straight line (i.e., AB) represents the magnitude and arrow represents the direction of the vector.

** Fig. 1.2. Vector quantity**

The symbol AB also represents this vector, which means it is acting from A to B .

**1.1.2. Scalar Quantity**. A quantity, which is completely specified by magnitude only, is known as a scalar quantity. Some examples of scalar quantity are : mass, length, time and temperature.

**1.1.3. A Particle**. A particle is a body of infinitely small volume (or a particle is a body of” negligible dimensions) and the mass of the particle is considered to be concentrated at a point. Hence a particle is assumed to a point and the mass of the particle is concentrated at this point.

**1.1.4. Law of Parallelogram of Forces**. The law of parallelogram of forces is used to determine the resultant* of two forces acting at a point in a plane. It states, “If two forces, acting at a point be represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point.”

Let two forces P and Q act at a point 0 as shown in Fig. 1.3. The force P is represented in magnitude and direction by OA whereas the force Q is presented in magnitude and direction by OB. Let the angle between the two forces be ‘a’. The resultant of these two forces will be obtained in magnitude and direction by the diagonal (passing through O) of the parallelogram of which OA and OB are two adjacent sides. Hence draw the parallelogram with OA and OB as adjacent sides as shown in Fig. 1.4. The resultant R is represented by OC in magnitude and direction.

**Magnitude of Resultant (R)**

From C draw CD perpendicular to OA produced.

Let a = Angle between two forces P and Q = LAOB

Now < DAC = < LAOB (Corresponding angles)

In parallelogram OACB, AC is parallel and equal to OB .

AC=Q.

In triangle ACD,

AD = AC cos a = Q cos a

and CD =AC sin a= Q sin a.

In triangle OCD,

OC^{2} = OD^{2} + DC^{2}.

But OC = R, OD = OA +AD = P + Q cos a

and DC =Q sin a

R ^{2} = (P + Q cos a)^{2} + (Q sin a)^{2} = p^{2} + Q^{2} cos^{2} a+ 2PQ cos < X+ Q^{2} sin^{2} a

= p^{2 }+ Q^{2} (cos^{2} a+ sin^{2} a)+ 2PQ cos a

= P^{2 }+ Q^{2} + 2PQ cos a

R = √ p^{2} + Q^{2} + 2PQ cos a

Equation (1.1) gives the magnitude of resultant force R.

** **

**Direction of Resultant**

Let θ = Angle made by resultant with OA.

Then from triangle OCD,

tan θ = CD / OD = Q sin a / P + Q cos a

θ = tan ^{-1} ( Q sin a / P + Q cos a)

Equation (1.2) gives the direction of resultant (R). The direction of resultant can also be obtained by using sine rule [In triangle OAC, OA = P, AC = Q, OC = R, angle OAC = (180 – a), angle ACO = 180- [θ + 180 – a] = (a- θ)]

sin θ / AC = sin (180 –a) OC = sin (a – θ) / OA

sin θ / Q = sin (180 – a) / R = sin (a – θ) / P

Two cases are important.

**1st Case**. If the two forces P and Q act at right angles, then

a = 90°

From equation (1.1), we get the magnitude of resultant as

From equation (1.2), the direction of resultant is obtained as

θ = tan^{ -1} ( Q sin a / P + Q sin a)

= tan^{ -1} ( Q sin 90^{0} / P + Q cos 90^{0 }) = tan ^{-1} Q /P

** **

**2nd Case.** The two forces P and Q are equal and are acting at an angle a between them.

Then the magnitude and direction of resultant is given as

It is not necessary that one of two forces, should be along x-axis. The forces P and Q may be in any direction as shown in Fig. 1.5. If the angle between the two forces is ‘a’, then their resultant will be given by equation (1.1). The direction of the resultant would be obtained from equation (1.2). But angle e will be the angle made by resultant with the direction of P.

**1.1.5. Law of Triangle of Forces**. It states that, “if three forces acting at a point be represented in magnitude and direction by the three sides of a triangle, taken in order, they will be in equilibrium.”

**1.1.6. Lame’s Theorem**. It states that, “If there forces acting at a point are in equilibrium, each force will be proportional to the sine of the angle between the other two forces.”

Suppose the three forces P, Q and Rare acting at a point 0 and they are in equilibrium as

shown in Fig. 1.6.

Let a = Angle between force P and Q.

β =Angle between force Q and R.

y = Angle between force R and P.

Then according to Lame’s theorem, P is proportional sine of angle between Q and R a sin β.

P / sin β = constant

Similarly Q / sin γ = constant R / sin a = constant

P / sin β = Q / sin γ = R / sin a

**Proof of Lame’s Theorem**. The three forces acting on a point, are in equilibrium and hence they can be represented by the three sides of the triangle taken in the same order. Now draw the force triangle as shown in Fig. 1.6 (a ).

Now applying sine rule, we get

P / sin (180 – β) = Q / sin (180 – γ) = R / sin (180 –a)

This can also be written

P / sin β = Q / sin γ = R / sin a

This is same equation as equation (1.5).

**Note**. All the three forces should be acting either towards the point or away from the point.