PROBLEMS.] MECHANICAL PHILOSOPHY". STATICS. 



699 



AE 2 =AB 2 -f BE 2 -2AB-BEcos. 120. 

 = AB 2 + BE 2 + 2 ABBE cos. 6U. 

 = 12 2 -f 15-' + 2 X 12 X 15 X '5. 

 = 144 -f 225 + 180 = 549. 



Hence, AE = ^549 = 23 43. 



sin^EAB _ EB sin. EAB _ 12_ 

 EA' or sin. 120~ ~^3-43 



Hence, sin. EAB = 



12 



; sin. 120=-^ sin. 60. 



And Log. sin. E A B = Log. 12 + Log. sin. 60 

 Log. 23-43. 



= 1-0791812 + 99375306 1-3697723 

 = 9-6469395. 



= Log. sin. 26 20' nearly. 

 Angle E A B = 26 2ff. 



When the resultant of more than two forces is to be 

 found, it will be more 

 convenient to resolve 

 the forces as in Prop. 

 DL We shall there- 

 fore work out the 



above example by a P>=' 



that method. 



Draw two straight 

 lines, A z and A y at 



Fig. SI. 



angles to each 

 other (Fig. 34). 



Take A P, = 15 = 

 force P, along A x. 



Draw AP, = 12 = 

 with A P,. 



force P a , making an angle 60 

 j/a perpendicular to Ay, and P 2 



Through P, draw P 4 

 x t perpendicular to Ax. 



Then Ax, Ay,, are the resolved parts of P 3 along Ar, 

 and Ay ; A P, is the resolved part of P, along the axis 

 Ax, and its resolved part along Ay is zero. 



Produce A P! to x, and make P, x" = Az 2 . 



Then Ax" = A P. + Ax 2 is the resolved part of the 

 resultant of P , and P 2 along Ax and Ay a is the resolved 

 part along Ay. 



Through a? draw x 1 R perpendicular to Ax, and pro- 

 duce y 2 Pj to meet x 1 R iu R. Join A R. 



Then A R represents the resultant of P, and P a in 

 magnitude and direction, Ay y = A P 2 cos. 60 = 12 x '5 

 - 6, and Ay a = P. X a = AP t sin. 60 12x '866 = 

 10-392. 



But A R2 = Ay 2 -f Ri/ 2 2 = Ay a + Az,* = (10-392) 

 -f (15 + 6) = (10 : 392) 2 + (21). = 107-993664 + 441 = 

 548.993664. _ 



Therefore A R - ^548. 993664 = 23-430. 



Again, cos. 



And therefore angle R Ax 1 = 26 20 7 . 



~ = 23-743 = '89628 = cog. 26 



nearly. And therefore angle R Ax 1 = 26 20 7 . And 

 this is the inclination of the resultant to the greater 

 force. 



Tig. 3J. 



PROBLEM II. Three forces which are to each other as 3, 

 4, and 5, act upon a point, and keep it at rest ; rerjuired 

 the angles at which these forces are inclined to each other. 

 1st. Graphical solution. 

 Draw a line A B = 4 parts of some scale, with A as a 



centre and radius A C = 3 parts of this scale, describe 



an arc. 



Also, with B as centre, and radius B C = 5 parts 



of the scale, describe another arc. 



Let C be the point where the two arcs intersect. 



Join A C and B C. Through C draw C E parallel to 

 A B and through B, B E parallel to A C, meeting in E. 

 Join A E and produce A E to F, making A F = A E. 

 Then the forces which are to each other as 3, 4 and 5, 

 acting upon the point A, and keeping it at rest, will be 

 represented in magnitude and direction by the lines A C, 

 A B, and A F ; and the angles being measured by a pro- 

 tractor, or any other means used for measuring angles, 

 it will be found that the angle B A C = 90, the angle 

 B A F = 143 8', and the angle C A F = 126 52', 

 nearly. 



2nd . Trigonometrical solution. 



Since 3 2 + 4 2 = 9 + 16 = 25 = 5 s . 



The forces repre- 

 sented by 3 and 4 

 must be at right 

 angles to each other, j 

 and their resultant i 

 5, will be the dia- j 

 gonal of the rec- j 

 tangular parallel- j 

 ogram whose sides 

 are as 3 and 4. 



Let a be the angle 

 between tho result- 

 ant and the force 

 represented by 3. 

 4 



Then tan. a = -.,- 

 o 



Fig. 35. 



1-33333 = tan. 53 8', nearly. 



Hence, by a reference to the diagram it will bo readily 

 seen that the angle between the forces 3 and 5 is 180 

 a= 180 53 8' = 126 52', while that between tho 

 forces 4 and 5 is 90 + a = 90 + 53 8' = 143 8'. 



PROBLEM III. Three forces represented by the numbers 3, 

 5, and 9, cannot under circumstances produce equi- 

 librium on a point. 



This is evidently true, since by Prop. IV. , three forces 

 can only produce equilibrium on a point, when a triangle 

 can be described whose sides are respectively proportional 

 to the magnitudes of the forces. Now, the sides of a 

 triangle never can bear to each other the proportion of 

 the numbers 3, 5, and 9, since in every triangle the sum 

 of any two sides must always be greater than the third, 

 and 3 + 5 = 8, a number less than 9. 



If we attempt to solve the problem graphically, wo 

 shall soon perceive its impossibility. 



Fig. 37. 



Thus, if we draw a line A B = 9 parts of any scale, 

 and with A as a centre and radius A C => 5 parts of the 

 scale, and with B as a centre and radius B D = 3 parts 

 of the scale, we describe two circles ; they evidently will 

 not cut each other in any point, and consequently we 

 cannot construct a triangle whose sides are to each other 

 as the numbers 3, 5, and 9. 



Wo add a few more problems for the practice of tho 

 student, taken from the Cambridge Examination Papers 

 and other sources. 



Three forces acting in the same plane keep a point at 

 rest ; the angles between the directions of the forces 

 are 135, 120 , and 105 ; compare their magnitudes. 

 Four forces represented by 1, 2, 3 and 4, act in the samo 

 plane on a point. Tho directions of the first and 

 third are at right angles to each other ; and so are tho 

 directions of the second and fourth ; and the second is 

 inclined at an angle of 60 to the first. Find the mag- 

 nitude and direction of the resultant. 



