Four masses A, B, C and D, as shown below are to be completely balanced:
| D | C | B | A | |
| 40 | 50 | 30 | – | Mass (kg) |
| 150 | 120 | 240 | 180 | Radius (mm) |
The planes containing masses B and C are 300 mm apart. The angle between the planes containing B and C is 90°. B and C make angles of 210° and 120° respectively with D in the same sense. Find (i) magnitude and angular position of mass A (ii) the positions of planes A and D.
\text { Given: } \angle B O C=90^{\circ}, \angle B O D=210^{\circ}, \angle C O D=120^{\circ} .
Reference plane B (Fig. 4.40)
\text { Let } M_{A}=\text { mass at } A .
a = distance of plane A from B
b = distance of plane D from B
Refer to Fig.12.51.
| Table 12.33 | |||||
| Couple Mrl (kg.m² ) |
Distance from RP, l (m) |
Centrifugal force, Mr (kg.m) |
Radius, r (m) |
Mass, M (kg) |
Plane |
| –0.18M _{A} a | -a | 0.18M _{A} | 0.18 | M _{A} | A |
| 0 | 0 | 7.2 | 0.24 | 30 | B |
| 1.8 | 0.3 | 6.0 | 0.12 | 50 | C |
| 6 b | b | 6.0 | 0.15 | 40 | D |
Force Polygon: Draw the force polygon with the data from column 4 as shown in Fig.12.51(c).
1. Draw ob = 7.2 units parallel to OB.
2. Draw bc = 6 units parallel to OC.
3. Draw cd = 6 units parallel to OD.
4. Join od and measure it
o d=3.6=0.18 M_{A} .
M_{A}=20 kg .
To locate the angular position of A, draw OA from O in Fig. 4.40 (b) parallel to od. \angle A O B=235^{\circ} \text { and } \angle A O D=25^{\circ} .
Couple polygon: Form the data in column 6 draw the couple polygon as shown in Fig.12.51(d).
\text { 1. Draw } o^{\prime} c^{\prime} \text { parallel to } O C \text { and equal to } 1.8 \text { units upwards. }\text { 2. Draw a line from } c^{\prime} \text { parallel to } O D \text { and another line from o' parallel to } O A \text { to intersect at } d^{\prime}.
\text { 3. } o^{\prime} d^{\prime}=3.8 \text { units }=-0.18 M_{A} .
a=\frac{-3.8}{0.18 \times 20}=-1.05 m .
Negative sign indicates that plane A is towards right of B instead of the left as assumed.
c^{\prime} d^{\prime}=2.4 \text { units }=6 b .
b = 0.4 m.
We observe that the direction of c’d’ is opposite to the direction of mass D. Therefore, the plane of mass D is 0.4 m towards left of plane B and not towards right of plane B as assumed.
