Vectors 271

total horizontal component of the two forces,

H=F 1 cosθ 1 +F 2 cosθ 2

The vertical component of forceF 1 isF 1 sinθ 1 and the
vertical component of forceF 2 isF 2 sinθ 2. The total
vertical component of the two forces,

V=F 1 sinθ 1 +F 2 sinθ 2

Since we haveHandV, the resultant ofF 1 andF 2
is obtained by using the theorem of Pythagoras. From
Figure 29.19,

0 b^2 =H^2 +V^2

i.e. resultant=

√

H^2 +V^2 at an angle

given byθ=tan−^1

(

V

H

)

V

b

Resultant

(^0) H a

Figure 29.19
Problem 7. A force of 5N is inclined at an angle
of 45◦to a second force of 8N, both forces acting at
a point. Calculate the magnitude of the resultant of
these two forces and the direction of the resultant
with respect to the 8N force
The two forces are shown in Figure 29.20.
458
8N
5N
Figure 29.20
The horizontal component of the 8N force is 8cos0◦
and the horizontal component of the 5N force is
5cos45◦. The total horizontal component of the two
forces,
H=8cos0◦+5cos45◦= 8 + 3. 5355 = 11. 5355
The vertical component of the 8N force is 8sin0◦and
the vertical component of the 5N force is 5sin45◦.The
total vertical component of the two forces,
V=8sin0◦+5sin45◦= 0 + 3. 5355 = 3. 5355

Resultant
H11.5355 N
V3.5355 N
Figure 29.21
From Figure 29.21, magnitude of resultant vector

√
H^2 +V^2

√
11. 53552 + 3. 53552 = 12 .07N
The direction of the resultant vector,
θ=tan−^1
(
V
H
)
=tan−^1
(
3. 5355
11. 5355
)
=tan−^10. 30648866 ...= 17. 04 ◦
Thus,the resultant of the two forces is a single vector
of 12.07N at 17. 04 ◦to the 8N vector.
Problem 8. Forces of 15N and 10N are at an
angle of 90◦to each other as shown in Figure 29.22.
Calculate the magnitude of the resultant of these
two forces and its direction with respect to the
15N force
10N
15N
Figure 29.22
The horizontal component of the 15N force is 15cos0◦
and the horizontal component of the 10N force is
10cos90◦. The total horizontal component of the two
velocities,
H=15cos0◦+10cos90◦= 15 + 0 = 15