vi PARALLEL FORCES AND CENTRE OF GRAVITY 81 



A uniform beam, 12 feet long, is supported at its two ends, 

 A and B. The beam weighs 4 cwb., and hanging from it at a 

 distance of 3 feet from A is a weight of 2 cwt. What is the 

 pressure on each support ? 



As the rod is uniform, its weight, as we shall see later (p. 82), 

 may be considered to be concentrated at the middle point. 

 Then taking moments about A, and denoting by S the force 

 exerted by B, we have the following equation : 



S x BA = (2 x AD) + (4 x AC) 

 Or S x 12 = (2 x 3) + (4 x 6) 

 Hence 12S = 30 



30 



Therefore S = = 2i cwt* 

 12 



In a similar way, by taking moments about the point B, we 

 have, when S 1 denotes the force exerted by A : 



S 1 x AB = (2 x DB) + (4 x CD) 

 Or S 1 x 12 = (2 x 9) + (4 x 6) 

 Hence 12S 1 = 42 



And S 1 = 4 -? - 3J cwt. 



It will be noticed that S + S 1 = 6 cwt., which is the total 

 weight borne. From this it follows that if the force exerted by 

 one support is found, the force exerted by the other support can 

 be determined 'by subtracting the first force from the total 

 weight borne. 



Centre of Gravity. Consider a large number of weights, 

 some heavier than others, suspended from a horizontal rod 

 arranged as in Expt. 61. A certain position could be found at 

 which the spring balance would have to be attached in order 

 to keep the rod in equilibrium. When the rod is hung from 

 ' this point the tendency to turn in one direction is counteracted 

 by the tendency to turn in the other, so the rod remains 

 horizontal. The weights may be regarded as parallel forces, and 

 the spring balance as equal to their resultant. Now consider a 

 stone, or any other object, suspended by a string. Every par- 

 ticle of the stone is being pulled downwards by the force of 

 gravity, as indicated in Fig. 33. The resultant of these parallel 



