154 MECHANICAL ENGINEERING PRINCIPLES
500
Force / N
020
Extension / mm
Figure 14.2
=^12 ×( 20 × 10 −^3 )m×500 N
=5J
It is shown in Chapter 13 that force = mass×
acceleration, and that if an object is dropped from
a height it has a constant acceleration of around
9.81 m/s^2. Thus if a mass of 8 kg is lifted vertically
4 m, the work done is given by:
work done=force×distance
=(mass×acceleration)×distance
=( 8 × 9. 81 )× 4 = 313 .92 J
The work done by a variable force may be found by
determining the area enclosed by the force/distance
graph using an approximate method such as themid-
ordinate rule.
To determine the area ABCD of Figure 14.3
using the mid-ordinate rule:
(i) Divide baseADinto any number of equal
intervals, each of width d (the greater the
number of intervals, the greater the accuracy)
(ii) Erect ordinates in the middle of each interval
(shown by broken lines in Figure 14.3)
(iii) Accurately measure ordinatesy 1 ,y 2 ,y 3 ,etc.
(iv) AreaABCD=d(y 1 +y 2 +y 3 +y 4 +y 5 +y 6 )
A
dddddd
B
C
D
y 1 y 2 y 3 y 4 y 5 y 6
Figure 14.3
In general, the mid-ordinate rule states:
Area=
(
width of
interval
)(
sum of
mid-ordinates
)
Problem 1. Calculate the work done when
a force of 40 N pushes an object a distance
of 500 m in the same direction as the force.
Work done=force×distance moved in the
direction of the force
=40 N×500 m
=20000 J(since 1 J=1Nm)
i.e. work done=20 kJ
Problem 2. Calculate the work done when
a mass is lifted vertically by a crane to a
height of 5 m, the force required to lift the
mass being 98 N.
When work is done in lifting then:
work done=(weight of the body)
×(vertical distance moved)
Weight is the downward force due to the mass of an
object. Hence
workdone=98 N×5m=490 J
Problem 3. A motor supplies a constant
force of 1 kN which is used to move a load
a distance of 5 m. The force is then changed
to a constant 500 N and the load is moved a
further 15 m. Draw the force/distance graph
for the operation and from the graph
determine the work done by the motor.
The force/distance graph or work diagram is shown
in Figure 14.4. Between points A andB a con-
stant force of 1000 N moves the load 5 m; between