Chapter 29 : Balancing of Rotating Masses 593
Let m = Mass of the balancing body D, and
θ = Angle, which the balancing body makes with A.
The example may be solved analytically or graphically. But we shall solve by both the meth-
ods, one by one.
Analytical method
Resolving all the assumed forces horizontally,
∑H = m 1 r 1 cos θ 1 + m 2 r 2 cos θ 2 + m 3 r 3 cos θ 3
= (10 × 100 cos 0°) + (9 × 125 cos 60°) + (16 × 50 cos 135°)
= (1000 × 1·0) + (1125 × 0·5) + 800 (– 0·707) = 997 ...(i)and now resolving the assumed forces vertically,
∑V = m 1 r 1 sin θ 1 + m 2 r 2 sin θ 2 + m 3 r 3 sin θ 3
= (10 × 100 sin 0°) + (9 × 125 sin 60°) + (16 × 50 sin 135°)
= (1000 × 0) + (1125 × 0·866) + (800 × 0·707) = 1540 ...(ii)
∴ Resultant assumed force,(^) RH V=Σ +Σ =( )^22 ( ) (997)2 2+(1540) = 1835
We know that m × r = 1835
∴ (^18351835) 12·2 kg
150
m
r
=== Ans.
and
1540
tan 1·5446
997
V
H
Σ
θ= = =
Σ
or θ = 57°.1°
Since ∑H and ∑V are both positive, therefore the resultant of these forces lies in the first
quadrant. It is thus obvious, that the balancing force must act in its opposite direction. Therefore
actual angle of the balancing body,
θ = 180° + 57.1° = 237.1° Ans.
Graphical Method
Fig. 29.5.