Mechanical Engineering Principles

34 MECHANICAL ENGINEERING PRINCIPLES

15 kN

37 °

28 °

Figure 3.22

3. Four coplanar forces acting on a body
   are such that it is in equilibrium. The
   vector diagram for the forces is such that
   the 60 N force acts vertically upwards,
   the 40 N force acts at 65°to the 60 N
   force, the 100 N force acts from the nose
   of the 60 N force and the 90 N force
   acts from the nose of the 100 N force.
   Determine the direction of the 100 N and
   90 N forces relative to the 60 N force.

[100 N force at 263° to the 60 N force,

90 N force at 132° to the 60 N force]

3.10 Resolution of forces

A vector quantity may be expressed in terms of its
horizontalandvertical components. For example,
a vector representing a force of 10 N at an angle
of 60° to the horizontal is shown in Figure 3.23.
If the horizontal lineoa and the vertical lineab
are constructed as shown, thenoa is called the
horizontal component of the 10 N force, andabthe
vertical component of the 10 N force.
By trigonometry,

cos 60°=

oa

ob

,

hence the horizontal component,

oa=10 cos 60°

Also, sin 60°=

ab

ob

,

hence the vertical component,ab=10 sin 60°

10 N

b

60 °

0

a

Figure 3.23

This process is calledfinding the horizontal and

vertical components of a vectororthe resolution

of a vector, and can be used as an alternative to

graphical methods for calculating the resultant of

two or more coplanar forces acting at a point.

10 N b

60 °

30 °a

0

d

20 N

c

Figure 3.24

For example, to calculate the resultant of a 10 N

force acting at 60°to the horizontal and a 20 N force

acting at− 30 °to the horizontal (see Figure 3.24) the

procedure is as follows:

(i) Determine the horizontal and vertical compo-

nents of the 10 N force, i.e.

horizontal component,oa=10 cos 60°

= 5 .0N,and

vertical component,ab=10 sin 60°

= 8 .66 N