Mechanical Engineering Principles

60 MECHANICAL ENGINEERING PRINCIPLES

i.e. force,F=

1340

20

=67 N

(b) The clockwise moment is now due to a force
of 23 N acting at a distance of, say,dfrom the
fulcrum. Since the value ofFis decreased to
21 N, the anticlockwise moment is( 21 × 20 )+
( 12 × 80 )Nmm.

Applying the principle of moments,

23 ×d=( 21 × 20 )+( 12 × 80 )

i.e. distance,d=

420 + 960

23

=

1380

23

=60 mm

Problem 5. For the centrally supported

uniform beam shown in Figure 5.6,

determine the values of forcesF 1 andF 2

when the beam is in equilibrium.

F 1 F^2

R = 5 kN

3 m 7 m

Figure 5.6

At equilibrium:

(i) R=F 1 +F 2 i.e. 5=F 1 +F 2 ( 1 )

and

(ii) F 1 × 3 =F 2 × 7 ( 2 )

From equation (1),F 2 = 5 −F 1. Substituting forF 2

in equation (2) gives:

F 1 × 3 =( 5 −F 1 )× 7

i.e. 3 F 1 = 35 − 7 F 1

10 F 1 = 35

from which, F 1 = 3 .5kN

SinceF 2 = 5 −F 1 , F 2 = 1 .5kN

Thus at equilibrium, forceF 1 = 3 .5 kN and force

F 2 = 1 .5kN

Now try the following exercise

Exercise 26 Further problems on equilib-

rium and the principle of

moments

1. Determine distancedand the force acting
   at the support A for the force system
   shown in Figure 5.7, when the system is
   in equilibrium. [50 mm, 3.8 kN]

1 kN 2.8 kN

140 mmd

A

RA

Figure 5.7

2. If the 1 kN force shown in Figure 5.7 is
   replaced by a forceF at a distance of
   250 mm to the left ofRA, find the value
   ofFfor the system to be in equilibrium.
   [560 N]


3. Determine the values of the forces acting
   atAandBfor the force system shown in
   Figure 5.8. [RA=RB=25N]

20 mm

76 mm

50 mm

20 N 30 N

AB

RA RB

Figure 5.8

4. The forces acting on a beam are as shown
   in Figure 5.9. Neglecting the mass of the
   beam, find the value ofRAand distance
   dwhen the beam is in equilibrium.
   [5 N, 25 mm]