Basic Engineering Mathematics, Fifth Edition

268 Basic Engineering Mathematics

0

c

F 1

F 2

a

b

Figure 29.5

Problem 1. A force of 5N is inclined at an angle

of 45◦to a second force of 8N, both forces acting at

a point. Find the magnitude of the resultant of these

two forces and the direction of the resultant with

respect to the 8N force by (a) the nose-to-tail

method and (b) the parallelogram method

The two forces are shown in Figure 29.6. (Althoughthe

8N force is shown horizontal, it could have been drawn

in any direction.)

45 

5N

8N

Figure 29.6

(a) Nose-to-tail method

(i) The 8N force is drawn horizontally 8 units

long, shown as 0 ain Figure 29.7.

(ii) From the nose of the 8N force, the 5N force

is drawn 5 units long at an angle of 45◦to the

horizontal, shown asab.

(iii) The resultant force is given by length 0band

is measured as12Nand angleθis measured

as 17 ◦.

^458

5N

8N a

0

b

Figure 29.7

(b) Parallelogram method

(i) In Figure 29.8, a line is constructed which

is parallel to and equal in length to the 8N

force.

(ii) A line is constructed which is parallel to and

equal in length to the 5N force.



45 

5N

8N

b

0

Figure 29.8

(iii) The resultant force is given by the diagonal

of the parallelogram, i.e. length 0b,andis

measured as12Nand angleθis measured

as 17 ◦.

Thus, the resultant of the two force vectors in

Figure 29.6 is 12N at 17◦to the 8N force.

Problem 2. Forces of 15 and 10N are at an angle

of 90◦to each other as shown in Figure 29.9. Find,

by drawing, the magnitude of the resultant of these

two forces and the direction of the resultant with

respect to the 15N force

15N

10N

Figure 29.9

Using the nose-to-tail method,

(i) The 15N force is drawn horizontally 15 units

long, as shown in Figure 29.10.

R 10N

15N



Figure 29.10

(ii) From the nose of the 15N force, the 10N force

is drawn 10 units long at an angle of 90◦to the

horizontal as shown.