By D. G. Gorman and W. Kennedy (Auth.)
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Additional resources for Applied Solid Dynamics
2) Component 2 = r$(j cos φ — i sin φ). 5a, is termed the tangential component of the acceleration of Q relative to P, aPQ, of magnitude νφ. The line of action of this component is perpendicular to the line joining P and Q, and its direction is obtained by rotating the direction of the vector SPO through 90° in the direction of rotation of Q about P if φ is positive. If φ is negative, rotate the direction of the vector SPQ through 90° in the opposite direction to that of the rotation of Q about P.
Also find the angular acceleration of link AC. 1. 8c respectively. Similarly, as before, the reader should also be in a position to locate the points g' and b' on the velocity vector diagram, and g" and b" on the acceleration vector diagram, noting that A, G, B and C lie on the same solid link. 8b). e. vBD = vBD (since length BD is constant). e. perpendicular to the line BD, which is drawn in on the velocity vector diagram passing through point B'. e. horizontal. e. 35 m/s -►. 8c). 4 and Once again, the line of action of aBD is parallel to BD and acts in a direction towards B as shown.
Kinematics of general plane motion 37 Consider now vAB. Once again, since the length BA is constant, vAB = vAB. e. perpendicular to AB, we do not at this stage know either its magnitude or direction. Consider, however, vCB = ν^Β (since the length BC is constant). Once again, we know the line of action of VcB, perpendicular to BC, but neither its magnitude nor its direction. e. b'. 7b. e. e. 7c). Once again starting at pole point (o",c"), consider the acceleration of A relative to O. 4, f = f = (j) = 0.
Applied Solid Dynamics by D. G. Gorman and W. Kennedy (Auth.)