Engineering Mechanics

(^16) „„„„„ A Textbook of Engineering Mechanics
Note. It the angle (α) which the resultant force makes with the other force F 2 ,
then^1
21
sin
tan
cos
F
FF
θ
α=
+θ
Cor.

1. If θ = 0 i.e., when the forces act along the same line, then
   R=F 1 + F 2 ...(Since cos 0° = 1)


2. If θ = 90° i.e., when the forces act at right angle, then
   22
   θ= =R FF 12 + ...(Since cos 90° = 0)


3. If θ = 180° i.e., when the forces act along the same straight line but in opposite directions,
   then R = F 1 – F 2 ...(Since cos 180° = – 1)
   In this case, the resultant force will act in the direction of the greater force.


4. If the two forces are equal i.e., when F 1 = F 2 = F then

RFF F=++θ=+θ22 2 2 cos 2 F^2 (1 cos )

22cos^22

2

F

⎛⎞θ

=×⎜⎟

⎝⎠

... 1 cos 2 cos^2

2

⎡ ⎛⎞θ ⎤

⎢⎥+θ= ⎜⎟

⎣⎦⎝⎠

Q

4 22 cos 2 cos

22

FF

⎛⎞ ⎛⎞θθ

==⎜⎟ ⎜⎟

⎝⎠ ⎝⎠

Example 2.1. Two forces of 100 N and 150 N are acting simultaneously at a point. What is

the resultant of these two forces, if the angle between them is 45°?

Solution. Given : First force (F 1 ) = 100 N; Second force (F 2 ) = 150 N and angle between

F 1 and F 2 (θ) = 45°.

We know that the resultant force,

22

RFF FF=++12 122cosθ

=++×× °(100)^22 (150) 2 100 150 cos 45 N

=++×10 000 22 500 (30 000 0.707) N

= 232 N Ans.

Example 2.2. Two forces act at an angle of 120°. The bigger force is of 40 N and the

resultant is perpendicular to the smaller one. Find the smaller force.

Solution. Given : Angle between the forces ∠=°AOC 120 , Bigger force (F 1 ) = 40 N and

angle between the resultant and FBOC 2 ()90∠=°;

Let F 2 = Smaller force in N

From the geometry of the figure, we find that ∠AOB,

α = 120° – 90° = 30°

We know that

2

12

sin

tan

cos

F

FF

θ

α=

+θ

22

22

sin120 sin 60

tan 30

40 cos120 40 (– cos 60 )

FF

FF

°°

°= =

+°+ °

Fig. 2.1.