Engineering Mechanics

Chapter 3 : Moments and Their Applications „„„„„ 37

Let α = Angle which the resultant force makes with the horizontal.

∴

–15.656

tan – 1.6214

9.656

V

H

Σ

α= = =

Σ

or α = 58.3° Ans.

Since ΣH is positive and ΣV is negative, therefore resultant lies between 270° and 360°. Thus

actual angle of the resultant force = 360° – 58.3° = 301.7° Ans.

Magnitude and sense of two forces along AJ and JH

Let P 1 = Force in units along AH, and
P 2 = Force in units along AJ.
The system of forces along with the resultant (R) is shown in Fig. 3.15. From the geometry of
the figure, we find that

(^12)
0.5
tan tan 0.5
1
θ= θ = =
or θ 1 = θ 2 = 26.6°
Resolving the forces P 1 and P 2 horizontally, and equating it
with the horizontal component of the resultant force,
P 1 cos θ 1 + P 2 sin θ 2 = 9.656
P 1 cos 26.6° + P 2 sin 26.6° = 9.656
P 1 × 0.8944 + P 2 × 0.4472 = 9.656
∴ 2 P 1 + P 2 = 21.59 ...(i)
and now resolving the forces P 1 and P 2 vertically and equating them with the vertical component of
the resultant force,
P 1 sin θ 1 + P 2 cos θ 2 = 15.656
P 1 sin 26.6° + P 2 cos 26.6° = 15.656
P 1 × 0.4472 + P 2 × 0.8944 = 15.656
∴ P 1 + 2P 2 = 35.01
or 0.5 P 1 + P 2 = 17.505 ...(ii)
Subtracting equation (ii) from equation (i),
1.5 P 1 = 4.085
∴ 1
4.085
2.72 units
1.5
P==^ Ans.
Now substituting the value of P 1 in equation (i),
2 × 2.72 + P 2 = 21.59
∴ P 2 = 21.59 – 5.44 = 16.15 units Ans.
Note. Since the two forces P 1 and P 2 together will balance the resultant force, therefore their
directions will be in the opposite directions as assumed i.e. along HA and JA.
Fig. 3.15.