TRIANGLE 121 



10. Three beads of weights TF a , TF&, W c are placed on a circular wire, and 

 when the beads are at the points A, B, C on the circle, the center of gravity 

 of the whole is found to coincide with 0, the center of the circle. Show that 



Wg = W b W C 



sinBOC sinCOA sin A OB 



CENTER OF GRAVITY OF A LAMINA 



91. It is often of importance to be able to find the position 

 of the center of gravity of a lamina, i.e. of a thin, plane shell of 

 uniform thickness and density, such, for instance, as is obtained 

 by cutting a figure out of a sheet of cardboard. 



92. Center of gravity of a triangle. Let ABC represent a 

 triangular lamina of which it is required to find the position of 

 the center of gravity. Let us imagine the triangle divided by 

 lines parallel to the base B C into a very 



great number of infinitely narrow strips. 

 Let pq be any single strip. Since, by 

 hypothesis, we may regard this strip as 

 of vanishingly small width and thick- 

 ness, we may treat it as a thin uniform 

 rod. The center of gravity of a thin B 

 uniform rod is at its middle point, so 

 that the weight of the strip pq may be supposed to act at r, the 

 middle point of pq. 



The weights of the other strips may be treated in the same 

 way, so that the weight of the whole triangle may be replaced by 

 the weights of a system of particles situated at the middle points 

 of these strips. 



Now if D is the middle point of the base BC, the middle points 

 of all the strips lie in the line AD. Thus the weight of the tri- 

 angle is replaced by the weights of a number of particles, all of 

 which are situated in the line AD. It follows that the center of 

 gravity of the whole triangle must lie in the line AD. 



We might equally well have supposed the triangle divided into 

 strips parallel to the side AC. We should then have found that 



