Question

A shaft carries four masses A, B, C and 0 of magnitude 200 kg, 300 kg, 400 kg and 200 kg respectively and revolving at radii 80 mrn, 70 mm, 60 mm and 80 mm in planes measured from A at 300 rnrn, 400 mm and 700 mm. The angles between the cranks measured anticlockwise are A to B 45°, B to C 70° and C to 0 120°. The balancing masses are to be placed in planes X and Y. The distance between the planes A and X is 100 mm, between X and Y is 400 mm and between Y and 0 is 200 mm. If the balancing masses revolve at a radius of 100 mm, find their magnitudes and angular positions. 

Answer 1

Given: m_A = 200 kg ; m_B =300 kg; m_C = 400 kg ; m_D = 200 kg ; r_A = 80 mm = 0.08 m ; r_B = 70 mm = 0.07 m; r_C = 60 mm = 0.06 m ; r_D = 80 mm = 0.08 m ; r_X = r_Y = 100 mm = 0.1 m

Let m_X = Balancing mass placed in plane X, and m Y = Balancing mass placed in plane Y. Assume the plane X as the reference plane (R.P.). The distances of the planes to the right of plane X are taken as + ve while the distances of the planes to the left of plane X are taken as - ve.

The data may be tabulated as shown in Table

Plane(1)	Mass(m)(2)	Radius(r) m(3)	cent.force/ω2(m.r) kg-m(4)	Distance from Plane x(l) m(5)	Couple/ ω2(m.r.l)kg-m2(6)
A X(R.P.) B C Y D	200 mx 300 400 my 200	0.08 0.1 0.07 0.06 0.1 0.08	16 0.1 mx 21 24 0.1 my 16	-0.1 0 0.2 0.3 0.4 0.6	-1.6 0 4.2 7.2 0.04 my 9.6

The balancing masses m_X and m_V and their angular positions may be determined graphically as discussed below:

1. First of all, draw the couple polygon from the data given in Table  (column 6) as shown in Fig. (c) to some suitable scale. The vector d₀ represents the balanced couple. Since the balanced couple is proportional to 0.04 m_V, therefore by measurement, 0.44 m_y = vector do =.7.3 kg-m² m_V = 182.5 kg .

The angular position of the mass m_V is obtained by drawing Om_V in Fig.(b), parallel to vector do. By measurement, the angular position of m_V is 1J_Y = 120 in the clockwise direction from mass m_A (200)

fig. (a) Position of planes.

fig. (b)  Angular position of masses

2. Now draw the force polygon from the data given in Table (column 4) as shown in Fig.(d). The vector represents the balanced force. Since the balanced force is proportional to 0.1 m_X, therefore by measurement, O.lm_X = vector eo = 35.5 kg-m or m_X = 355 kg 

The angular position of the mass m_X is obtained by drawing Om_X in Fig.(b), parallel to vector eo. By measurement, the angular position of m_X is 1J_X = 145°in the clockwise direction from mass m_A (i.e.200 kg)