Higher Engineering Mathematics

228 VECTOR GEOMETRY

Figure 21.9

H=7 cos 0◦+4 cos 45◦= 7 + 2. 828 = 9 .828 N

Vertical component of force,

V=7 sin 0◦+4 sin 45◦= 0 + 2. 828 = 2 .828 N

The magnitude of the resultant of vector addition

=

√

(H^2 +V^2 )=

√

(9. 8282 + 2. 8282 )

=

√

(104.59)= 10 .23 N

The direction of the resultant of vector addition

=tan−^1

(

V

H

)

=tan−^1

(

2. 828

9. 828

)

= 16. 05 ◦

Thus, the resultant of the two forces is a single
vector of 10.23 N at 16.05◦to the 7 N vector.

Problem 5. Calculate the resultant velocity of

the three velocities given in Problem 2.

With reference to Fig. 21.5:
Horizontal component of the velocity,
H=10 cos 20◦+15 cos 90◦+7 cos 190◦
= 9. 397 + 0 +(− 6 .894)= 2 .503 m/s

Vertical component of the velocity,

V=10 sin 20◦+15 sin 90◦+7 sin 190◦

= 3. 420 + 15 +(− 1 .216)= 17 .204 m/s

Magnitude of the resultant of vector addition

=

√

(H^2 +V^2 )=

√

(2. 5032 + 17. 2042 )

=

√

302. 24 = 17 .39 m/s

Direction of the resultant of vector addition

=tan−^1

(

V

H

)

=tan−^1

(

17. 204

2. 503

)

=tan−^16. 8734 = 81. 72 ◦

Thus, the resultant of the three velocities is a sin-

gle vector of 17.39 m/s at 81.72◦to the horizontal.

Now try the following exercise.

Exercise 93 Further problems on vector

addition and resolution

1. Forces of 23 N and 41 N act at a point and
   are inclined at 90◦to each other. Find, by
   drawing, the resultant force and its direction
   relative to the 41 N force. [47 N at 29◦]


2. ForcesA,BandCare coplanar and act at
   a point. ForceAis 12 kN at 90◦,Bis 5 kN
   at 180◦andCis 13 kN at 293◦. Determine
   graphically the resultant force. [Zero]


3. Calculate the magnitude and direction of
   velocities of 3 m/s at 18◦and 7 m/s at 115◦
   when acting simultaneously on a point.
   [7.27 m/s at 90.8◦]


4. Three forces of 2 N, 3 N and 4 N act as shown
   in Fig. 21.10. Calculate the magnitude of the
   resultant force and its direction relative to the
   2 N force. [6.24 N at 76.10◦]

Figure 21.10