Engineering Mechanics

(^594) „„„„„ A Textbook of Engineering Mechanics

1. First of all, draw the space diagram, with the given masses and their positions as shown in
   Fig. 29·5 (a).


2. Take some convenient point a and draw ab equal to m 1 × r 1 (10 × 100 = 1000) to some
   scale and parallel to the 10 kg mass.


3. Through b draw bc equal to m 2 × r 2 (9 × 125 = 1125) to scale and parallel to the 9 kg mass.


4. Similarly, through c draw cd equal to m 3 × r 3 (16 × 50 = 800) to scale and parallel to the
   16 kg mass.


5. Join da, which represents the magnitude and direction of the assumed resultant force.
   Now the assumed balancing force will be given by ad to the scale.


6. Measuring ad to scale, we find that ad = 1835 kg-mm.


7. Therefore m × 150 = 1840

or

1835

12·2 kg

150

m== Ans.

8. Now measuring the inclination of da with respect to the body A, we find that
   θ = 237.1° Ans.
   Example 29·5. A circular disc, rotating around a vertical spindle, has the following masses
   placed on it.
   Position of load
   Load Magnitude
   θ with respect to Y-Y Distance from centre
   A 0 degree 250 mm 2·5 kg
   B 60 degree 300 mm 3·5 kg
   C 150 degree 225 mm 5·0 kg

Determine the unbalanced force on the spindle, when the disc is rotating at 240 r.p.m. Also

determine the magnitude and angular position of a mass, that should be placed 262·5 mm, to give

balance when rotating.

Solution. Given : No. of revolution (N) = 240 r.p.m and radius of balancing mass

(r) = 262·5 mm = 0·2625 m

The example may be solved analytically or graphically. But we shall solve this problem

graphically as discussed below :

Fig. 29.6.