A Course of Mathematics for Engineers and Scientists. Volume by Brian H. Chirgwin, Charles Plumpton

By Brian H. Chirgwin, Charles Plumpton

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If {X1 0 Z1}, {X2 0 Z2 } are the reactions at P,Q respectively, and Y is the stop reaction, show that Xi = X2 = (Wb/2h) sing, Z1 + Z2 = Wb(11b — Y = (Wbd/hp), + cosa/h), Z2 - The forces are : at G, {b sin a. 0 b cos a}, the weight {0 0 —W} ; at P, {0 c 0}, the action at the hinge {X, 0 Zi}; at Q, (0 — c 0), the action at the hinge {X2 0 Z2 }; at D, {0 — a 0}, the action of the stop {0 Y 0} ; at K, {p sin a d p cos a), the tension {T1T2 T3}. = (Wbd/cp). 54 A COURSE OF MATHEMATICS From Fig. 18 we see (by resolution) that the magnitude T and the resolutes {T, T2 T3 } of the tension in the string are related by T1 — p sinoc T2 T3 T —d h — p cosoc HK • Moments about H.

The necessary and sufficient condition for equilibrium is that F= F, = O. 2) where {X, Y1 Z,} are the resolutes of the force FL in an arbitrary frame of reference. The conditions for equilibrium of a rigid body must include the turning effects of all the forces acting on the body. In plane systems the turning effect of a force P about a point A was measured by its moment P . AN, where AN is the perpendicular distance of the line of action of P from A, and P = I P 1. If r is the position vector of a point Q (see Fig.

It is also necessary to choose suitable directions for resolution and for taking moments. For the first part of this problem the forces T1, T2 , T3and the weights W1, W2 , W3 are absent (see Fig. 17). Each rod is subject to the normal reaction from the floor, R, the weight W, the couple Wk0, whose axis is perpendicular to the appropriate plane, and the action from the smooth joint at 0. Since the arrangement is symmetrical we consider only one rod. We do not need to know the action at 0 and so we take moments for the rod OA, about an axis through 0 perpendicular to the plane AOG.

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