# Applied Mechanics for Engineers by J. Duncan

By J. Duncan

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31 and hence the be represented in Fig. 3 1 (b} by FORCES ACTING AT A POINT R2 and T their resultant R AD. being the only forces remaining in Fig. 31(0), be found from the triangle of forces ADA' will E which gives their equilibrant represented by A' A, 3 reversal gives R. It will be noticed that, had the given forces been in equilibrium, (Fig. 3i(^)), , and on E 3 would have been This case is zero, and A' would have coincided with A. shown ABCD, in Fig. 32, giving a closed polygon the sides of which, taken in order, represent respectively the given forces.

Is In Fig. 41 Forces acting at a point but not in the same plane. shown in outline a pair of sheer legs such as is used for moving heavy loads. Two legs AB and and are hinged at the ground rotating as a whole about the BC are jointed together at the top B, at A and C so as to be capable of The legs are line AC in the plan. supported in any given position by means of a back leg DB, which is FORCES ACTING AT A POINT 35 D capable of BD in the plan. jointed to the other legs at B, and has its end moved horizontally in the direction of the line W hung from B and produces forces T, Q, Q in the It will be noted that T and these are shown acting at B.

Check these by calculation as shown below, and plot P and x. Since P, and T are respectively horizontal, vertical and along Hence is the triangle of forces for them. AB, it follows that W ABC -, /I Wtana. (0 MATERIALS AND STRUCTURES TAB/ Also, Wseca FIG. 33. Measure P and T /, also by Tabulate thus x and inserting : Experiment on a pendulum. h, for the The curve it is required Weight of bob Length of straight each position of the bob, and calculate AB in quantities =W= inches = /= in (i) and (2).