Applied Mechanics for Engineers by J. Duncan

By J. Duncan

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A bent lever (Fig. 43) has its arms at = 1 5 inches, BC = 6 pivoted at C. AC A force P of 35 Ib. is applied at A inches. at 1 5 to the horizontal, and another Q is applied Find the magnitude at B at 20 to the vertical. of Q and the magnitude and direction of the reaction at C required to balance P and Q. 5. 90 and ^p is A body weighing 24 Ib. is kept at rest incline which makes 40 with the horizontal by a force P which is parallel to the plane (Fig. 44). Assume that the reaction of the plane is at 90 to its surface, and find P.

Fig. 55 (a)). their resultant. R=2P =2+5+7+3 =^7 of sense downward. , Taking moments about A, we have =92, feet. i 3/&1. lh---- j ---x f-2-x ----- --; '->[ fR ; -s'-- i ' u --------- 6 ----- ^J - 7' FIG. 55. EXAMPLE 2. Parallel forces act on a body as shown in Fig. 5 5 (b). Find their resultant. , of sense downward. convenient to take moments about a point O on the line of the 3 lb. force. (3 xo)+(4 x iJ)-(8 x 4) + (2 x 6J), 3=-j[7 The feet. negative sign indicates that R falls on the left side of O.

51). Find the relation of sense applied at Q A P and Q if the rod is balanced. AC = #, BC = & Let Taking moments about C, Q * Again we may say that each force is proportional to the distance of the other force from the pivot, and that the equilibrant of P and Q acts through C. EXAMPLE ~*~ ~ A 3. ,. horizontal rod the weight of which AB, W may be glected, has a weight applied at C and B, the reactions P supported at being vertical (Fig. 52). Find P and Q. A . B AB = Let is Q ,, BC = /-. then Taking moments about ne- and and B, Px/=W(/-) + (Q*o), P=(^)W (i) Taking moments about A, a = (Qx/) + (Pxo), W (2) PARALLEL FORCES of interest to find the It is above ; sum thus of P and Q, using (l-a\ = 43 their values as found a W-ji-~+^ =w.

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