(^36) A Textbook of Engineering Mechanics
Fig. 3.14.
and now resolving all the forces vertically,
ΣV = 7 + 8 cos α – 5 cos 45° – 9 cos 45° + 6 cos α
= 7 + (8 × 0.447) – (5 × 0.707) – (9 × 0.707) + (6 × 0.447) kN
= 3.36 kN ...(ii)
We know that magnitude of the resultant force,
RH V=Σ +Σ =( )^22 ( ) (10.96)2 2+(3.36) =11.46 kN Ans.
Direction of the resultant force
Let θ = Angle, which the resultant force makes with BC i.e., with the horizontal.
∴
3.36
tan 0.3066
10.96
V
H
Σ
θ= = =
Σ
or θ = 17.05° Ans.
Note : Since both the values of ΣH and ΣV are + ve, therefore resultant actual angle of the
resultant force lies between 0° and 90°.
Position of the resultant force
Let x = Perpendicular distance between the point E and the line of action of the
resultant force.
Taking moments about *E and equating the same,
11.46 x = (7 × 0) + (8 × 0) + (12 × 10) + (5 × 0.707) + (9 × 0.707) + (6 × 0)
= 129.9
∴
129.9
11.33 cm
11.46
x== Ans.
Example 3.8. ABCD is a square. Forces of 10, 8 and 4 units act at A in the directions AD,
AC and AB respectively. Using the analytical method, determine
(i) resultant force in magnitude and direction ;
(ii) magnitude and sense of two forces along the directions AJ and AH, where J and H are
the mid-points of CD and BC respectively, which together will balance the above resultant.
Solution. The system of forces is shown in Fig. 3.14
(i) resultant force in magnitude and direction
Resolving the forces horizontally,
ΣH = 4 + 8 cos 45° = 4 + (8 × 0.707) units
= 9.656 units
and now resolving the forces vertically,
ΣV = – 10 + (– 8 cos 45°) = – 10 – (8 × 0.707)
= – 15.656 units
We know that magnitude of the resultant force,
(^) RH V=Σ +Σ =( )^22 ( ) (9.656)^2 +(–15.656)^2 = 18.39 units Ans.
- The point E has been intentionally selected as the moments of these forces (i.e. 7 kN,
8 kN and 6 kN) about this point are zero, because these forces pass through E.