STATICS. 



[190. 



The construction is somewhat simplified when < = <' since then the 

 intersections of the total reactions lie on the circle described about 

 ABC (Fig. 58). 



W 



Fig. 58. 



190. As another example consider the ordinary jack intended to raise 

 an eccentric load JF acting vertically downwards through A (Fig. 59) by 



a force P passing vertically upwards through 

 the pitch-line B of the rack. Near C and D 

 the rack is pressed against the casing. The 

 directions of the total reactions C, D at 

 these points are found by applying the 

 friction angle to the normals. 



The four forces W, P, C, D can be in 

 equilibrium only if the resultant of W and 

 D is equal and opposite to the resultant 

 of P and C ; hence, if E be the intersec- 

 tion of W and D, F that of .P and C, each 

 of these resultants must act along EF. 



If the load W be known, the other 

 forces can now be found by constructing 

 the force polygon. Draw i 2 = W in 

 position (i.e. through A) draw 2 3 paral- 

 Fig> 59< lei to C; 41 parallel to D] and through 



the intersection 4 of 4 i with EF draw the vertical 3 4 to the intersec- 

 tion 3 with 2 3. 



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