# laplace test Aug 24, 2020

## First Book: On the General Laws of Equilibrium and Motion.

It is my intention to give in this book the general principles of the equilibrium and motion of bodies, and to solve those problems of mechanics which are indispensable in the theory of the system of the world.

### Chapter I: On the Equilibrium and Composition of Forces Which Act on a Material Point.

#### Section 1: On motion and force, also on the composition and resolution of forces

§ 1.] A body appears to us to be in motion, when it changes its relative situation with respect to a system of bodies supposed to be at rest; but as all bodies, even those which appear in the most perfect repose, may be in motion; a space is conceived of, without bounds, immoveable, and penetrable by the particles of matter; and we refer in our minds the position of bodies to the parts of this real, or ideal space, supposing the bodies to be in motion, when they correspond, in successive moments, to different parts of this space.

The nature of that singular modification, by means of which a body is transported from one place to another, is now, and always will be, unknown; it is denoted by the name of Force. We can only ascertain its effects, and the laws of its action. The effect of a force acting upon a material point, or particle, is to put it in motion, if no obstacle is opposed; the direction of the force is the right line which it tends to make the point describe. It is evident, that if two forces act in the same direction, the resultant is the sum of the two forces; but if they act in contrary directions, the point is affected by the difference of the forces. If their directions form an angle with each other, the force which results will have an intermediate direction between the two proposed forces. We shall now investigate the quantity and direction of this resulting force.

For this purpose, let us consider two forces, $$x$$ and $$y$$, acting at the same moment upon a material point $$M$$, in directions forming a right angle with each other. Let $$z$$ be their resultant, and $$\theta$$ the angle which it makes with the direction of the force $$x$$. The two forces $$x$$ and $$y$$ being given, the angle $$\theta$$ and the quantity $$z$$ must have determinate values, so that there will exist, between the three quantities $$x$$, $$z$$ and $$\theta$$, a relation which is to be investigated.

Suppose in the first place that the two forces $$x$$ and $$y$$ are infinitely small, and equal to the differentials $$dx$$, $$dy$$. Then suppose that $$x$$ becomes successively $$dx$$, $$2\,dx$$, $$3\,dx$$, &c., and $$y$$ becomes $$dy$$, $$2\,dy$$, $$3\,dy$$, &c., it is evident that the angle $$\theta$$ will remain constant, and the resultant $$z$$ will become successively $$dz$$, $$2\,dz$$, $$3\,dz$$, &c., and in the successive increments of the three forces $$x$$, $$y$$ and $$z$$, the ratio of $$x$$ to $$z$$ will be constant, and may be expressed by a function of $$\theta$$, which we shall denote by $$\phi(\theta)$$1; we shall therefore have $$x=z\cdot\phi(\theta)$$, in which equation we may change $$x$$ into $$y$$, provided we also change the angle $$\theta$$ to $$\frac{\pi}{2}-\theta$$, $$\pi$$ being the semi-circumference of a circle whose radius is unity.

Now we may consider the force $$x$$ as the resultant of two forces $$x'$$ and $$x''$$, of which the first $$x'$$ is directed along the resultant $$z$$, and the second $$x''$$ is perpendicular to it.2 The force $$x$$, which results from these two new forces, forms the angle $$\theta$$ with the force $$x'$$, and the angle $$\frac{\pi}{2}-\theta$$ with the force $$x''$$; we shall therefore have

$$x'=x\cdot\phi(\theta)=\frac{x^2}{z}\mbox{;}\qquad x''=x\cdot\phi(\frac{\pi}{2}-\theta)=\frac{xy}{z}\mbox{;}$$

and we may substitute those two forces instead of the force $$x$$. We may likewise substitute for the force $$y$$ two new forces, $$y'$$ and $$y''$$ , of which the first is equal to $$\frac{y^2}{z}$$ in the direction $$z$$, and the second equal to $$\frac{xy}{z}$$ perpendicular to $$z$$; we shall thus have, instead of the two forces $$x$$ and $$y$$, the four following:

$$\frac{x^2}{z}\mbox{, }\frac{y^2}{z}\mbox{, }\frac{xy}{z}\mbox{, }\frac{xy}{z}\mbox{;}$$

the two last, acting in contrary directions, destroy each other;3 the two first, acting in the same direction, are to be added together, and produce the resultant $$z$$; we shall therefore have4

$$x^2 + y^2 = z^2\mbox{;}$$

whence it follows, that the resultant of the two forces $$x$$ and $$y$$ is represented in magnitude, by the diagonal of the rectangle whose sides represent those forces.

Let us now determine the angle $$\theta$$. If we increase the force $$x$$ by the differential $$dx$$, without varying the force $$y$$, that angle will be diminished by the infinitely small quantity $$d\theta$$;5 now we may conceive the force $$dx$$ to be resolved into two other forces, the one $$dx'$$ in the direction $$z$$, and the other $$dx''$$ perpendicular to $$z$$; the point $$M$$ will then be acted upon by the two forces $$z+dx'$$ and $$dx''$$, perpendicular to each other, and the resultant of these two forces, which we shall call $$z'$$, will make with $$dx''$$ the angle $$\frac{\pi}{2}-d\theta$$;6 we shall thus have, by what precedes,

$$dx'' = z'\cdot\phi(\frac{\pi}{2}-d\theta)$$

consequently the function \$φ(\frac{\pi}{2}-dθ) is infinitely small, and of the form $$-kd\theta$$, $$k$$ being a constant quantity, independent of the angle $$\theta$$;7 we shall therefore

## Footnotes:

1

A quantity $$z$$ is said to be a function of another quantity $$x$$, when it depends on it in any manner. Thus, if $$z$$, $$y$$ be variable, $$a$$, $$b$$, $$c$$, &c. be constant, and we have either of the following expressions, $$z=ax+b$$, $$z=ax^2+bx+c$$; $$z=a^x$$, $$z=\sin(ax)$$ &c., $$z$$ will be a function of $$x$$; and if the precise form of the function is known, as in these examples, it is called an explicit function. If the form is not known, but must be found by some algebraical process, it is called an implicit function.

2

For

illustration, suppose the forces $$x$$ and $$y$$ to act at at the point $$A$$, in the directions $$AX$$, $$AY$$, respectively, and that the resultant $$z$$ is in the direction $$AZ$$, forming with $$AX$$, $$AY$$, the angles $$ZAX=\theta$$, $$ZAY=\frac{\pi}{2}-\theta$$. Then as above, we have $$x=z\cdot\phi(\theta)$$, $$y = z\cdot\phi\left(\frac{\pi}{2}-\theta\right)$$. Draw $$EAF$$ perpendicular to $$AZ$$, and suppose the force $$x$$ in the direction $$AX$$ to be resolved into two forces, $$x'$$, $$x''$$, in the directions $$AZ$$, $$AE$$, respectively, so that the angle $$ZAX=\theta$$, and $$XAE=\frac{\pi}{2}-\theta$$. Then, in the same manner in which the above values of $$x$$, $$y$$, are obtained from $$z$$, we may get $$x'=x\cdot\phi(\theta)$$; $$x''=x\cdot\phi(\frac{\pi}{2}-\theta)$$. If in these we substitute the values $$\phi(\theta)=\frac{x}{z}$$; $$\phi(\frac{\pi}{2}-\theta)=\frac{y}{z}$$, deduced from the above equations, we obtain $$x'=\frac{x^2}{z}$$; $$x''=\frac{xy}{x}$$. In like manner, if the force $$y$$, in the direction $$AY$$, be resolved into the two forces $$y'$$, $$y''$$, in the directions $$AZ$$, $$AF$$, making the angle $$YAZ=\frac{\pi}{2}-\theta$$, $$YAF=\theta$$, we shall have $$y'=y\cdot\phi(\frac{\pi}{2}-\theta)$$; $$y''=y\cdot\phi(\theta)$$; which, by substituting the above values of $$\phi(\frac{\pi}{2}-\theta)$$, $$\phi(\theta)$$, become $$y'=\frac{y^2}{z}$$, $$y''=\frac{xy}{z}$$, as above.

3

For, by the preceding note, the force $$x''=\frac{xy}{z}$$, is in the direction $$AE$$, and the force $$y'' = \frac{xy}{z}$$, is in the opposite direction $$AF$$, and as they are equal they must destroy each other.

4

The sum of the two forces $$x'=\frac{x^2}{z}$$, $$y'=\frac{y^2}{z}$$, in the direction $$AZ$$, being put equal to the resultant $$z$$, gives $$\frac{x^2}{z}+\frac{y^2}{z} = z$$, which multiplied by $$z$$ becomes $$x^2+y^2=z^2$$.

5

The

resultant of the forces $$x$$, $$y$$, is, by hypothesis, in the direction $$AZ$$, and, by increasing the force $$x$$ by $$dx$$, the forces become equal to $$z$$ in the direction $$AZ$$, and $$d x$$ in the direction $$AX$$, and the resulting force $$z'$$, must evidently fall between $$AZ$$, $$AX$$, on a line as $$AG$$, forming with $$AZ$$ an infinitely small angle $$ZAG$$, represented by $$d \theta$$. Then the force $$d x$$, in the direction $$AX$$, may be resolved into two forces, the one $$d x'$$ in the direction $$AZ$$, the other $$d x''$$ in the direction $$AE$$, and as this last force is inclined to $$AX$$ by the angle $$XAE=\frac{\pi}{2}-\theta$$, we shall have as above $$d x'' = d x\cdot\phi(\frac{\pi}{2}-\theta)$$; or by substituting the preceding value of $$\phi(\frac{\pi}{2}-\theta)=\frac{y}{z}$$, $$dx''=\frac{y\,dx}{z}$$.

6

This angle is equal to $$GAE=\frac{\pi}{2}-d\theta$$; and if the force $$z'$$ in the direction $$AG$$ is resolved into two forces in the directions $$AZ$$, $$AE$$, the last will (by the nature of the function φ) be represented by $$z'\cdot\phi(\frac{\pi}{2}-d\theta$$).

7

Because $$\phi(\frac{\pi}{2}-d\theta)$$ contains only the quantities $$\frac{\pi}{2}$$, $$d\theta$$, but does not explicitly contain $$\theta$$. Moreover, the function $$\phi(\frac{\pi}{2}-d\theta)$$ being developed in the usual manner, according to the powers of $$d\theta$$, by Taylor’s theorem [xxx], or by any other way, will be of the form