Mechanical Engineering Principles

FORCE, MASS AND ACCELERATION 145

Thus, forceαacceleration, or force=a constant×
acceleration, this constant of proportionality being
the mass of the object, i.e.

force=mass×acceleration

The unit of force is the newton (N) and is defined
in terms of mass and acceleration. One newton is
the force required to give a mass of 1 kilogram an
acceleration of 1 metre per second squared. Thus

F=ma

whereFis the force in newtons (N),mis the mass
in kilograms (kg) andais the acceleration in metres

per second squared (m/s^2 ), i.e. 1 N=

1kgm
s^2
It follows that 1 m/s^2 =1 N/kg. Hence a gravi-
tational acceleration of 9.8 m/s^2 is the same as a
gravitational field of 9.8 N/kg.

Newton’s third law of motionmay be stated as:

For every force, there is an equal and opposite reacting

force

Thus, an object on, say, a table, exerts a downward
force on the table and the table exerts an equal
upward force on the object, known as areaction
forceor just areaction.

Problem 1. Calculate the force needed to

accelerate a boat of mass 20 tonne uniformly

from rest to a speed of 21.6 km/h in

10 minutes.

The mass of the boat,m, is 20 t, that is 20000 kg.

The law of motion,v = u+at can be used to
determine the accelerationa.
The initial velocity,u, is zero,

the final velocity,v= 21 .6km/h=

21. 6

3. 6

=6m/s,

and the time,t=10 min=600 s.
Thusv=u+at, i.e. 6= 0 +a×600, from which,

a=

6

600

= 0 .01 m/s^2

From Newton’s second law,F=ma

i.e. force= 20000 × 0 .01 N=200 N

Problem 2. The moving head of a machine

tool requires a force of 1.2 N to bring it to

rest in 0.8 s from a cutting speed of

30 m/min. Find the mass of the moving head.

From Newton’s second law,F=ma, thusm=

F

a

,

where force is given as 1.2 N. The law of motion

v=u+at can be used to find accelerationa,where

v=0,u=30 m/min=

30

60

m/s= 0 .5m/s,and

t= 0 .8s.

Thus, 0 = 0. 5 +a× 0. 8

from which, a=−

0. 5

0. 8

=− 0 .625 m/s^2 or

a retardation of 0.625 m/s^2.

Thus themass,m=

F

a

=

1. 2

0. 625

= 1 .92 kg

Problem 3. A lorry of mass 1350 kg

accelerates uniformly from 9 km/h to reach a

velocity of 45 km/h in 18 s. Determine

(a) the acceleration of the lorry, (b) the

uniform force needed to accelerate the lorry.

(a) The law of motionv=u+at can be used to

determine the acceleration, where final velocity

v=

45

3. 6

m/s, initial velocityu=

9

3. 6

m/s and

timet=18 s.

Thus

45

3. 6

=

9

3. 6

+a× 18

from which,

a=

1

18

(

45

3. 6

−

9

3. 6

)

=

1

18

(

36

3. 6

)

=

10

18

=

5

9

m/s^2 or 0 .556 m/s^2

(b) From Newton’s second law of motion,

force,F=ma= 1350 ×

5

9

=750 N

Problem 4. Find the weight of an object of

mass 1.6 kg at a point on the earth’s surface

where the gravitational field is 9.81 N/kg (or

9.81 m/s^2 ).