024
THEORY AND PRACTICE OF ENGINEERING.
Book II.
Fig. 1233.
the plane, and they will no longer be in equilibrio: thus the equilibrium of a body whichrests on a plane by a single point is submitted to two conditions; first, the result of theforces applied to bodies passes by the point of support, and, secondly, its direction is per-pendicular to the plane.
Equilibrium of an inclined Plane. —Let us represent any plane cut perpendicularly toits horizontal, and following the line of its greatest inclination, A B. Although this planebe supposed to be indefinitely prolonged, its in-clination is sufficient to define its base A C and itsheight B C. Let us imagine a heavy body placedon this plane of which G shall be the centre ofgravity : the equilibrium of this body in the planewill require, first, that the resultant of its weightshould pass by one of the points of the base whichsupports the body; secondly, that that resultant beperpendicular to the plane: this latter conditioncannot be generally satisfied with regard to any plane whatever, because the action ofgravity is always vertical ; hence this action will be decomposed into two others, the oneperpendicular to the plane, which will be destroyed by its resistance, and the other parallelto the plane, which will cause the body to move or descend according to its length. Wemust, however, remark that if the vertical of the centre of gravity meets the interior of thebase of support of the body, this latter will only slide in the direction of the plane; hut itwill roll if the vertical falls without the base. This will happen with a base, becausethe vertical always passes outside its point of support on the plane.
Let us suppose that a force, p, parallel to theplane, is opposed to the descent of the ball: callF the weight of the latter; we know that thebody ought to remain in equilibrio under theaction of the two forces p and P; consequently theirresultant, G M, will he perpendicular to A B, andwill pass by the point of support m, and then let GMbe the intensity of this resultant; it will be suchthat the sides of the parallelogram G G' M N willrepresent, the one G G' equal M N the vertical forceP, and the other G N the force p, parallel to A B.
The triangle G M N is similar to the triangleA B C, for G M and M N are perpendicular toA B and A C ; moreover they are rectangles, theone in G, the other in C : we shall have the pro-portion
Or p : P : : B C :
Then the force parallel to the plane which holds the body, is to the weight of this body, asthe height of the plane is to its length : and we have
A B,A B.
9 = P x
B_C
Air
Suppose, instead of drawing in the direction of the plane, the force /?, which retains thebody on the plane, was horizontal and acted in the direction of the power, which prevents anut from descending along the screw; G M, theresultant of the weight of the body and of this newforce,will not be the less perpendicular to A B; andthe triangle G 31 N, representing the two forceswhich draw the body as well as their resultant,will be similar to the triangle ABC: but as G Nis perpendicular to B C, and the right angle of thefirst triangle is in N, we shall have the proportion,as G N : MN:: BC: AC;
Or, p : P : : B C : AC,
which shows us that the horizontal force retainingthe body in the plane is to the weight of the bodyas the height of the plane is to its base : or, that
Fig. 1236.
T> BC
f=Px rc
It will be remarked, that in the value of p the numerator is the same for two cases in pro-portion to the height of the plane; but the denominator is only the side of the planeparallel to the force applied.