36 MECHANICAL ENGINEERING PRINCIPLES
By Pythagoras’ theorem,
r=
√
1. 5352 + 9. 5302
= 9. 653 ,
and by trigonometry, angle
φ=tan−^1
9. 530
1. 535
= 80. 85 °
Hence the resultant of the two forces shown in
Figure 3.26 is a force of 9.653 N acting at 80.85°
to the horizontal.
Problems 9 and 10 demonstrate the use of resolution
of forces for calculating the resultant of two coplanar
forces acting at a point. However the method may
be used for more than two forces acting at a point,
as shown in Problem 11.
Problem 11. Determine by resolution of
forces the resultant of the following three
coplanar forces acting at a point: 200 N
acting at 20°to the horizontal; 400 N acting
at 165°to the horizontal; 500 N acting at
250 °to the horizontal.
A tabular approach using a calculator may be made
as shown below:
Horizontal component
Force 1 200 cos 20° = 187.94
Force 2 400 cos 165° = −386.37
Force 3 500 cos 250° = −171.01
Total horizontal component = −369.44
Vertical component
Force 1 200 sin 20° = 68.40
Force 2 400 sin 165° = 103.53
Force 3 500 sin 250° = −469.85
Total vertical component = − 297. 92
The total horizontal and vertical components are
shown in Figure 3.28.
Resultant r=
√
369. 442 + 297. 922
= 474. 60 ,
and angle φ=tan−^1
297. 92
369. 44
= 38. 88 °,
from which, α= 180 °− 38. 88 °= 141. 12 °
–369.44
–297.92
φ
r
α
Figure 3.28
Thus the resultant of the three forces given
is 474.6 N acting at an angle of − 141. 12 ° (or
+ 218. 88 °) to the horizontal.
Now try the following exercise
Exercise 17 Further problems on resolu-
tion of forces
- Resolve a force of 23.0 N at an angle
of 64° into its horizontal and vertical
components. [10.08 N, 20.67 N] - Forces of 5 N at 21° and 9 N at 126°
act at a point. By resolving these forces
into horizontal and vertical components,
determine their resultant.
[9.09 N at 93.92°]
In questions 3 and 4, determine the mag-
nitude and direction of the resultant of the
coplanar forces given, which are acting at
a point, by resolution of forces.
- ForceA, 12 N acting horizontally to the
right, force B, 20 N acting at 140° to
force A,forceC, 16 N acting 290° to
forceA.[3.1Nat− 45 °to forceA] - Force 1, 23 kN acting at 80°to the hor-
izontal, force 2, 30 kN acting at 37° to
force 1, force 3, 15 kN acting at 70°to
force 2.
[53.5 kN at 37°to force 1
(i.e. 117°to the horizontal)] - Determine, by resolution of forces, the
resultant of the following three coplanar
forces acting at a point: 10 kN acting at
32 °to the horizontal, 15 kN acting at 170°
to the horizontal; 20 kN acting at 240°to
the horizontal.
[18.82 kN at 210.03°to the horizontal]