(^70) A Textbook of Engineering Mechanics
Fig. 5.26.
From the geometry of the triangle BDF, we find that
8
sin 0.4132 or 24.4
19.36
DF
BD
β== = β=°
and BF==(19.36) – (8)^22 17.63 m
From the geometry of the triangle ABF, we also find that
17.63
cos 0.7052
25
BF
ABF
AB
∠== = or ∠ ABF = 45.1°
∴θ = 45.1° – 24.4° = 20.7°
We know that resultant of the forces in members BC and BE (acting along BD)
R = 2P cos α = 2P cos 14.5°
= 2P × 0.9680 = 1.936 P
The system of forces acting at B is shown in Fig 5.26.
Applying Lami’s equation at B,
1.936 250
sin (180 – 24.4 ) sin 45.1 sin (180 – 20.7 )
TP
°° ° °°
1.936 250
sin 24.4 sin 45.1 sin 20.7
TP
°°°
∴
250 sin 24.4 250 0.4131
291.5 kN.
sin 20.7 0.3543
T
×°×
°
Ans
and
250 sin 45.1 250 0.7090
258.4 kN.
1.936 sin 20.7 1.936 0.3543
P
×° ×
×° ×
Ans
5.6. GRAPHICAL METHOD FOR THE EQUILIBRIUM OF COPLANAR FORCES
We have studied in Art 5.5 the equilibrium of forces by analytical method. Sometimes, the
analytical method is too tedious and complicated. The equilibrium of such forces may also be studied,
graphically, by drawing the vector diagram. This may also be done by studying the
- Converse of the Law of Triangle of Forces.
- Converse of the Law of Polygon of Forces.