. A text book of physics, for the use of students of science and engineering . sting on two inclined planes. _P_=Q V2 2 P QV3 R —r- + t> =6 V2 2 P P A/o fi- • P- 6V^ • •(4).(5) P . • ? li 6 V2 1+V3Inserting this value in (3), we have 18 .(6) whence =6+2x,1+V3 x = 0-294 foot. Example 2.—In Fig. 159 {a) AB and AC are two uniform bars eachweighing 10 lb. and 6 feet long. The bars are smoothly jointed at A,and rest at B and C on a smooth horizontal surface. B and C are con-nected by an inextensible cord -1 feet long. A load of 40 isattached to D. Find, in terms of BD. the tension in
. A text book of physics, for the use of students of science and engineering . sting on two inclined planes. _P_=Q V2 2 P QV3 R —r- + t> =6 V2 2 P P A/o fi- • P- 6V^ • •(4).(5) P . • ? li 6 V2 1+V3Inserting this value in (3), we have 18 .(6) whence =6+2x,1+V3 x = 0-294 foot. Example 2.—In Fig. 159 {a) AB and AC are two uniform bars eachweighing 10 lb. and 6 feet long. The bars are smoothly jointed at A,and rest at B and C on a smooth horizontal surface. B and C are con-nected by an inextensible cord -1 feet long. A load of 40 isattached to D. Find, in terms of BD. the tension in the cord and thereactions communicated across the joint at A. First consider ABC to be a rigid body, acted upon by vertical forces ofl, tO and 10 together with the reactions P and Q. Then P+Q-10-40-10=0; ,\ P+Q = 60 (1) EQUILIBRIUM OF UNIPLANAR FORCES 133 Taking moments about C gives (P x 4) - (10 x 3) - (40 x CE) - (10 x 1) =0 ;/. 4P = 30 + 10+, P = 10 + 10CE From(l), Q=60-P=60-10-10CE = 50-10CE Also, CE = BC - BE = 4 - BE. .(2)•(3). Fig. 159. —Equilibrium of two rods. Draw AF perpendicular to BC, then, in the similar triangles BED, BFA,we have BE BF 2 bdba^B .\ BE = iBD;.. CE=4-?;, from (2) and (3), P = 10 +10(4 •(4) £BD) = 50 -^ BD And Q=50-10(4-?.BD) = 10 + \PBD (5) Now consider the bar AC separately (Fig. 159 (?>)). The forces appliedto it are its weight, acting vertically through G2, the vertical reaction Q, atC, the horizontal pull T of the cord BC, and a reaction S at A. S is exertedby the other bar AB, and its direction is guessed in Fig. 159 (b) ; theprecise direction will be determined in the following calculation. Resolve 134 DYNAMICS chap. S into horizontal and vertical components Sx and S^, and apply the con-ditions of equilibrium. T-S,=0; .*. T = SZ (G) Q-10-S„=0; .-. Q = 10 + S„ (7) Take moments about A, first calculating the length of AF : AF =VAC2-CF - VW^l - V32.(Qx2)-(10x1)-(Txa/32)=0; .*. 2Q = 10+TV32 ., (8
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