Proposition 100

Table of Contents

PROPOSITION 100. PROBLEM

If a body is acted on by three forces, of which the directions Mf, Mg and MQ (Fig.

are parallel to the three coordinate axes AP, PQ and QM, to determine the motion of the body and the orbit in which it moves.

SOLUTION

Since Mf and Mg are parallel to AP and PQ, the plane fMg is parallel to the plane APQ. In this plane, Mi is drawn parallel to the element Qq; and this line Mi is also placed in the [vertical] plane Mq. A perpendicular Qd is sent from Q to the element mM produced; and from f and g the perpendiculars fi and gk are sent to Mi. Then from i and k the perpendiculars ib and kc fall on Md. Moreover fi and gk are perpendicular to the plane Mq, since the plane fMg is normal to the plane Mq. Now with AP = x, PQ = y and QM = z remaining as before, let the force pulling the body along Mf be equal to P, the force which pulls along the body along Mg is equal to Q, and the force pulling along MQ is equal to R. Therefore these forces, in order that their effect can become known, must be resolved into forces acting along the tangent Mm, placed along the normal to Mm in the plane Mq, and normal to the plane Mq.

Since the angle Mfi =Qqp it is the case that [p. 349] and Pdy ( dx 2 + dy 2 ) expresses the force arising from P acting along if , and if P alone acts, then by (802) : Following this, the force pulling along Mi is equal to :

Again this force can be resolved into a force pulling along bi equal to : and a force pulling along Mb equal to : Therefore from P we have the contributions

In a like manner the force Q, the direction of which is Mg, can be resolved into the force acting along kg equal to : Qdx ( dx 2 + dy 2 ) and the force along Mk equal to : Qdy ( dx 2 + dy 2 )

This is finally resolved into the force along ck equal to : and the force along Mc equal to : Whereby, if this force alone acts, we have :

Finally the force R, having the direction MQ, is resolved into the force along Md equal to : and the force along dQ equal to :

Therefore from the force R there becomes :

Therefore from these three forces P, Q, and R acting at the same time, the tangential force arising from all : the normal force in the plane Mq which is in place : and the second normal force

…

With these values T, N and M substituted in place in the equations (809), the three following equations are produced : and

Which equations determine the equations of motion of the body. Q.E.I

Corollary 1

These two last equations give that ratio: From which it is found :

Corollary 2

Therefore the plane Mmμ (Fig. 76), in which the elements Mm and mμ are present, can be defined in this manner. Let −Qdz + Rdy and the tangent of the angle POR = Pdz − Rdx and the tangent of the angle, which the plane Mmμ makes with the plane APQ,

Corollary 3

If the force P vanishes, there is found : from the two equations found above. And from the third there is produced : the integral of which is

Corollary 4

Therefore from this hypothesis the time, or

From which it is understood that the motion of the body progressing along parallel to the axis AP is uniform.

Corollary 5

Again from the same hypothesis, we have : since ddy : ddz = Q : R (824). From which equations the curve itself described by the body can be determined.

Scholium

From the resolution of forces into three forces that we have considered in this proposition, it is easily seen how all forces which can be devised can generally be reduced. Whereby, with these forces given, it is not difficult to find the curve described by the body, also for whatever cases are proposed this theorem has the maximum usefulness.