Question

Four masses A, B, C and 0 as shown below are to be completely balanced.

	A	B	C	D
Mass (kg) Radius(mm)	- 180	30 240	50 120	40 150

The planes containing masses Band Care 300 mm apart. The angle between planes containing Band C is 90°. Band C make angles of 210° and 120° respectively with 0 in the same sense. 

Find: 

1. The magnitude and the angular position of mass A ; and 

2. The position of planes A and D

Answer 1

Given: r_A = 180 mm = 0.18 m ; m_B = 30 kg; r_B = 240 mm = 0.24 m ; m_C = 50 kg; r_C = 120 mm = 0.12 m, m_D = 40 kg; r₀ = 150 mm = 0.15 m ; BOC = 900 ; BOD = 2100 ; COD = 1200

1. The magnitude and the angular position of mass A Let

m_A = Magnitude of Mass A,

x = Distance between the planes Band D, and

y = Distance between the planes A and B.

The position of the planes and the angular position of the masses is shown in Fig. (a) and (b) respectively.

Assuming the plane B as the reference plane (R.P.) and the mass B (m_B) along the horizontal line as shown in Fig. (b), the data may be tabulated as below: 

Plane(1)	Mass(m) kg (2)	Radius(r) m (3)	Cent.force/ω2(m.r)kg-m(4)	Distance from plane B(l) m (5)	Couple/ω2(m.r.l) kg-m2(6)
A B(R.P) C D	mA 30 30 50 40	0.18 0.24 0.12 0.15	0.08 mA 7.2 6 6	-y 0 0.3 x	-0.18 mAy 0 1.8 6x

The magnitude and angular position of mass A may be determined by drawing the force polygon from the data given in Table (Column 4), as shown in Fig.(c), to some suitable scale. Since the masses are to be completely balanced, therefore the force polygon must be a closed figure. The dosing side (i.e. vector do) is proportional to 0.18 m A. By measurement, 0.18 m A = Vector do = 3.6 kg-m or m A = 20 kg .

In order to find the angular position of mass A, draw O_A in Fig (b) parallel to vector do. By measurement, we find that the angular position of mass A from mass B in the anti dock wise direction is AOB = 2360

2. Position of planes A and 0.

The position of planes A and D may be obtained by drawing the couple polygon, as shown in Fig (d), from the data given in Table (column 6). The couple polygon is drawn as discussed below:

1. Draw vector 0 c parallel to OC and equal to 1.8 kg-m₂, to some suitable scale.

2. From points can do, draw lines parallel to Ol) and OA respectively, such that they intersect at point d . By measurement, we find that 6 x = vector cd = 2.3 kg-m² or x = 0.383 m We see from the couple polygon that the direction of vector c dis opposite to the direction of mass D.

Therefore the plane of mass D is 0.383 m or 383 mm towards left of plane Band not towards right of plane B as already assumed.

 Fig. (a) Position of planes

All dimensions in mm.

Fig. (b)  Angular position of masses.

Fig. (c) Force polygon

Fig. (d) Couple polygon

Again by measurement from couple polygon,

- 0.18 m A.y = vector od = 3.6 kg-m²

- 0.18 x 20 y = 3.6 or y = -1 m 

The negative sign indicates that the plane A is not towards left of B as assumed but it is 1 m

or 1000 mm towards right of plane B.