Mechanical Engineering Principles

FORCES ACTING AT A POINT 35

(ii) Determine the horizontal and vertical compo-

nents of the 20 N force, i.e.

horizontal component,od=20 cos(− 30 °)

= 17 .32 N,and

vertical component,cd=20 sin(− 30 °)

=− 10 .0N

(iii) Determine the total horizontal component, i.e.

oa+od= 5. 0 + 17. 32 = 22 .32 N

(iv) Determine the total vertical component, i.e.

ab+cd= 8. 66 +(− 10. 0 )=− 1 .34 N

Total horizontal component = 22.32Total vertical

Resultant component=–1.34

0

r

f

Figure 3.25

(v) Sketch the total horizontal and vertical com-

ponents as shown in Figure 3.25. The resultant

of the two components is given by lengthor

and, by Pythagoras’ theorem,

or=

√

22. 322 + 1. 342

= 22 .36 N

and using trigonometry, angle

φ=tan−^1

1. 34

22. 32

= 3. 44 °

Hence the resultant of the 10 N and 20 N forces
shown in Figure 3.24 is22.36 N at an angle of
− 3. 44 °to the horizontal.

Problem 10. Forces of 5.0 N at 25°and

8.0 N at 112°act at a point. By resolving

these forces into horizontal and vertical

components, determine their resultant.

The space diagram is shown in Figure 3.26.

(i) The horizontal component of the 5.0 N force,

oa= 5 .0 cos 25°= 4. 532 ,

and the vertical component of the 5.0 N force,

ab= 5 .0sin25°= 2. 113

xx

c 0 a

b

5.0 N

112 °

25 °

d 8.0 N

y

y

Figure 3.26

(ii) The horizontal component of the 8.0 N force,

oc= 8 .0 cos 112°=− 2. 997

The vertical component of the 8.0 N force,

cd= 8 .0 sin 112°= 7. 417

(iii) Total horizontal component

=oa+oc= 4. 532 +(− 2. 997 )

=+ 1. 535

(iv) Total vertical component

=ab+cd= 2. 113 + 7. 417

=+ 9. 530

9.530

1.535

φ

r

Figure 3.27

(v) The components are shown sketched in

Figure 3.27.