By Thomas R. Kane

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**Extra info for Analytical Elements of Mechanics**

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P' = —p = 4ni — 3n2 — 2n3 ft Problem (b): Show that the distance d from a point P to a line L is given by d = |n X p| where n is a unit vector parallel to L, and p is the position vector of P relative to any point 0 on L. FIG. 1b Solution (See Fig. 2 The relationship between the position vector p and the coordinates x, y, z, of a point P Notation (See Fig. (x,,y,z,) FIG. 2a FIG. 2) are respectively equal to x,y,z: P = xni + 2/112 + zn 3 This relationship is the link between vector analysis and scalar analytic geometry.

2b): 0 the midpoint of line P1P2. 5n ft Result: See Fig. 2c. 4 47 Problem (b): The strengths of the points Pi, . , P 4 shown in Fig. 2d are Ni = 2, N2 = - 3 , Nz = 5, ΛΓ4 = 3. Locate the centroid of this set of points. \ ^ % s k y t, 3 P z 8' <£ J_ *■rz^ FIG. 2d FIG. 2. 3 Position vector First moment ni n2 n3 ni n2 Π3 0 2 0 -2 0 0 -8 -8 0 3 3 3 0 -6 0 6 0 0 -40 24 0 -9 15 -9 0 -16 -3 p*, the position vector of the centroid P* relative to Pi, is given by P* = I6112 — 3n3 1 = — 16n2 — 3ιΐ3 ft Result: See Fig.

CENTROIDS AND MASS CENTERS; SECTION 2 . 6 If the points of a set are arranged in such a way that corresponding to every point on one side of a certain plane there exists a point of equal strength on the other side, the two points being equidistant from the plane, but not necessarily lying on the same normal to it, then the centroid of the set lies in this plane. Such a plane is called a plane of symmetry. Proof: Let the plane of symmetry be the X-Y plane of a rec tangular cartesian coordinate system.