108 



STATICS OF EIGID BODIES 



ILLUSTRATIVE EXAMPLES 



1. Two equal uniform planks are hinged at one end, and stand with their free 

 ends on a smooth horizontal plane, being prevented from slipping by a rope which 

 is tied to each plank at the same height up. Find the tension in this rope and 

 the action at the hinge. 



In the figure let AB, AC represent the two planks, hinged at A, and let 

 PQ be the rope. The forces acting on the plank AB will consist of 

 A (a) the action at the hinge A 



(b) the tension of the rope acting along PQ; 



(c) the reaction at the foot B ; 



(d) the weight. 



Of these four forces, (a) and (b) are the forces which 

 it is required to find. Force (c) also is at present 

 unknown. Force (d), as explained in 77, can be 

 regarded as a single force W, the total weight of the 

 plank, and since we are told that the plank is uniform, 

 this must be supposed to act through its middle point. 



r> 



FIG. 60 



There is a simple way of finding force (c), the reaction at B. Since we are 

 told that the contact at B is smooth, the direction of the reaction must be 

 vertically upwards. Let its amount be R. From symmetry, there must be 

 an exactly similar reaction at the foot C of the second plank. Now consider 

 the equilibrium of the whole system which consists of the two planks and the 

 rope. The only external forces which act on this system consist of 



(a) the weight ; 



(b) the two reactions at B and C. 



If we resolve vertically, we obtain, since this 

 system is in equilibrium, 



2 W - 2 E = 0, 



so that E = W ; each reaction is Just equal to the 

 weight of one plank, as we might have anticipated. 



Of the four forces acting on the plank AB, 

 the last two are now known, while the first two 

 remain unknown. If we take moments about A, 

 we shall get an equation between forces (&), (c), 

 and (d), and this will enable us to find the un- 

 known force (b), the tension. 



If we denote the tension by T, and the angle 

 BA C by 2 6, ^ the equation obtained on taking 

 moments about A is / 



FIG. 61 



R AB sin 6 - W - \ AB sin d - T- AP cos 6 = 0, 

 so that, remembering that R = W, we have 



ZAP 



