Vectors 273
the horizontal component of the 20m/s velocity is
20cos90◦=0m/s
and the horizontal component of the 15m/s velocity is
15cos195◦=− 14 .489m/s.
The total horizontal component of the three velocities,
H= 8. 660 + 0 − 14. 489 =− 5 .829m/sThe vertical component of the 10m/s velocity
=10sin30◦=5m/s,
the vertical component of the 20m/s velocity is
20sin90◦=20m/s
and the vertical component of the 15m/s velocity is
15sin195◦=− 3 .882m/s.
The total vertical component of the three forces,
V= 5 + 20 − 3. 882 = 21 .118m/s5.82921.118R Figure 29.27
From Figure 29.27, magnitude of resultant vector,
R=√
H^2 +V^2 =√
5. 8292 + 21. 1182 = 21 .91m/sThe direction of the resultant vector,
α=tan−^1(
V
H)
=tan−^1(
21. 118
5. 829)
= 74. 57 ◦Measuring from the horizontal,
θ= 180 ◦− 74. 57 ◦= 105. 43 ◦.
Thus,the resultant of the three velocities is a single
vector of 21.91m/s at 105. 43 ◦to the horizontal.
Now try the following Practice Exercise
PracticeExercise 114 Addition of vectors
by calculation (answers on page 352)- A force of 7N is inclined at an angle of 50◦to
a second force of 12N, both forces acting at
a point. Calculate the magnitude of the resul-
tant of the two forces and the direction of the
resultant with respect to the 12N force. - Velocities of 5m/s and 12m/s act at a point
at 90◦to each other. Calculate the resultant
velocityanditsdirectionrelativetothe12m/s
velocity.
3. Calculate the magnitude and direction of the
resultant of the two force vectors shown in
Figure 29.28.
10N13N
Figure 29.28- Calculate the magnitude and direction of the
resultant of the two force vectors shown in
Figure 29.29.
22N18N
Figure 29.29- A displacement vectors 1 is 30m at 0◦.A
second displacement vectors 2 is 12m at 90◦.
Calculate the magnitude and direction of the
resultant vectors 1 +s 2 - Three forces of 5N, 8N and 13N act as
shown in Figure 29.30. Calculate the mag-
nitude and direction of the resultant force.
5N13N8N708608Figure 29.30