MECHANICAL PHILOSOPHYSTATICS. 



[RESULTANT* or TOKOS. 



P.Y.. P.Y., and 1', Y,. perpendicular to Ay, 



x,, P, x 3> ... dart 



, and 



1 Jl. 



i i 



Thon, Pro;. Ill . A X,. AY,, X, and Y,, are the 

 resolved parts of l' l in the direction of the axes Ax 

 and Ay. 



and AY. - X. and Y, those of P,, AX, and 



AY S -X 3 and Y 3 those of P 3 and- AX 4 and 



AY. X , and Y 4 those of P 4 along the same axes. 

 Let K, be the resultant of the forces P. and P,. 



R t the resultant of the forces R, and P,, or of P,, 

 P,, and P. v 



e resultant of the forces R 3 and P,, or of 

 PM P t , P s , and P 4 . 



; ie resolved part of R.along the 



axis of AT = X, -(- X,, that of R, - X, + X, + X 3> 

 and that of R s - X. + .\ X 4 . 



Similarly the resolved part of R. along the axis Ay 



- Y.. that of R, - Y, + Y, + Y,, and that of R 3 

 -Y, + Y. Y 5 Y 4 . 



X - X, + X. + X, X 4 , the algebraical sum 



of the resolved parts of the four forces along the axis 



Y, + Y, Y, Y,. the algebraical 



! the resolved parts of these forces along the axis 

 Ay. 



Take AX - X along the axis AX, and AY - Y 

 Jong the axis A Y. Fig. S3. 



iuli X draw 

 :i2) pcr- 

 'ular to AX, 

 a id through Y, 

 i i : , perpendicular 

 to A Y meeting in 

 thepoint R s . Join A I: 



III . A i: will represent the resultant 

 of the forces P,, P x , P,, and P 4 , in magnitude and 

 direct i 



This single force A R,, acting alone, will produce on 

 A the same effect as the forces P,, P., P., and P 4 . 

 Produce .\ 3, make A 8 - A R,. 



Then. II., A force represented in magnitude 



and direction by A 8 will exactly counteract the effect of 

 the force represented by A I : , 



Consequently, a force represented in magnitude and 

 direction by A 8, will keep the point A in equilibrium 

 when acted on l>y the forces P,. P.., P.., and l' t . 



From 8 let fall 8X perpendicular to AX, and 8Y 

 perpendicular to A Y. 



1 1 1 , A X and A Y are the resolved parts 

 of A 8 along the rectangular axes. Now in the triangles 

 AX H, and AS X, the angle X A R 3 - angle SAX, 

 and the angles at X and X are right angles, and also 

 AR, -AS. 



BMntev AX AX i \. a,,dsx - xi:,. 



A X R, Y and AX 8 Y arc parallelograms. 



Consequently, 8 X - A Y, and X R, - A Y~. 



Therefore AY- AY- 



Tho resolved narU of A S are therefore X and Y . 



i, aud-Y-Y.+Y, 

 Y, - 



Therefore X, + X . + X 3 X 4 X - 0, and Y. + 

 Y, Y. Y, -f-Y-0. 



"or if live forces represented in m <nd direc- 



tion by A P,, A P,, A P s , A P 4 , and A S keen a point 

 in e,|m!iliriuin, the algebraical sum of the resolved i 

 of these forces along the rectangular axes passing through 

 this point, will each be equal to zero. 



The same demonstration may be applied to any num- 

 ber of forces. It may also be observed i li.it tin- position 

 of the rectingular axes is perfectly arbitrary, pro-. 

 only that they are in the same plane in which the forces 

 are supposed to act 



The conditions of equilibrium when all the forces are 

 not in the same plane, involve the di-.-u--i..n of olid 

 geometry. We shall reserve the consideration of this 

 subject to a more advanced portion of the section. 



The student will observe that all the conditions of 

 equilibrium, for a material point which we have con- 

 sidered, are geometrical deductions from the parallelo- 

 gram of forces, and involve no new mechanical principles; 

 and that the parallelogram of forces depends upon one 

 mechanical principle, that of the transmission of a force 

 from any one point to another rigidly connected with it, 

 and in the direction of its action, without altering its 

 effect 



We subjoin two or three examples, to show the method 

 of applying the principles we have determine! Me- 

 chanical problems may be solved in two ways graphi- 

 cally, i.e., by accurately drawing the geometria] li_ 

 by mathematical instruments and scales, and obtain 

 result, which can be measured by these instruments ; or 

 trigonometrically, i.e., by computing the relations of 

 the figures according to the principles of tri-. m,.m 

 and thus arriving by means of trigonometrical taUes at 

 a more accurate result than the former will afford. 



Vie would recommend the student to use both of these 

 methods, as he will by this means obtain a clearer view 

 of the subject than by accustoming himself only to one 

 of them. 



PBOBI.KM I. Tioo forces, represented by 12H). ., 



are inclined to each other at an angle of 60 ; required 

 the magnitude of the resultant ; and its inclination to 

 the greater. 



Fig. S3. 



1st, Graphical solution. 



Draw a line A B. 



Make AB equal to 15 parts of any convenient scale, 

 as the 8th or the 10th of an inch. 



Draw A C inclined to A It at an angle of 60. 



Take A C equal to 12 parts of the same scale, that A B 

 is equal to 15 parts. 



Through C draw C E parallel to AB, and through B, 

 BE parallel to A C, meeting in the point E. 



Then, Prop. I., A E will represent the resultant of the 

 forces represented by AC and A B in magnitude and 

 n : and if A E be measured by the scale used in 

 drawing A B and A C, it will be found to be 23,^ parts 

 of that scale, and the angle E A B will measure very 

 nearly 26 2<X. 



The resulting force will therefore bo represented by 

 23 A Ibs. , and its inclination to the greater force will be 

 26* W nearly. 



The process might have been shortened by the use of 

 Prop. I V., in whu-h case we should have drawn A B 

 !." HOti <.f the scile. 1! K niakui- an an-le ISO --(Hi or 

 l-ii with A I!, and i: K - 12 parts of the icale, and 

 kwtly joined A and K. 



2nd. Trigonometrical solution. 



