Rectilinear motion is termed as the motion of particles of a rigid body such that each of them travels in a straight line and parallel to each other. Moreover, they travel the same distance as compare to one-another in the same direction.

Suppose the mass of one particle is m travels in x direction. The force applied in the particle is F in the same direction then,

According to Newton’s 2^{nd} law of motion, distance travelled by the particle due to the action of external force is given by,

_{x} = m. a_{x} (This is a scalar product in x-direction)

Where,

a_{x} = acceleration in x-direction

Considering the force in y-direction and z-direction,

_{y} = 0

_{z} = 0

Let us consider the particle has three components of acceleration such that a_{x}, a_{y} and a_{z}. In this case, force can be written as,

_{x} = m. a_{x}

_{y} = m. a_{y}

_{z} = m. a_{z}

To calculate the value of acceleration, we determine,

= a_{x}+ a_{y}+ a_{z}

||= a = (a_{x}^{2} + a_{y}^{2} + a_{z}^{2})

Similarly,

Force, = _{x} +_{y}+ _{z}

||=F = (F_{x}^{2} + F_{y}^{2} + F_{z}^{2})

The S.I. Unit of Force is

1 Newton (N) = 1 kg x 1 m/ s^{2}

= 1 kgm/ s^{2}

**Example 5.1:**

Suppose a body was in rest when an external force is acted on it. The force on the body produces 6 m/s^{2} of acceleration to give it some motion. If the mass of the object is 15 kg, then calculate the magnitude of force.

**Solution:**

**Given Data:**

Acceleration, a = 6 m/s^{2}

Mass, m = 15 kg

We know that,

Force, F = m . a

= 15 . 6

= 90 N

**Links of Previous Main Topic:-**

- Introduction to statics
- Introduction to vector algebra
- Two dimensional force systems
- Introduction concept of equilibrium of rigid body
- Friction introduction
- Introduction about distributed forces
- Area moments of inertia in rectangular and polar coordinates
- Mass moment of inertia introduction
- Work done by force
- Kinematics of particles
- Position vector velocity and acceleration
- Plane kinematics of rigid bodies introduction
- Combined motion of translation and rotation

**Links of Next Mechanical Engineering Topics:-**