Question

Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that P= 1400 N, determine the value of Q for which cable AD is just taut. 220 mm 960 mm 320 mm 380 mm 240 mm 960 mm The value of Q is

Answer 1

FBD at Ponnit A TAD TAC p=1200N As the magnitude of force a increases the magnitude of the force TAD decreases. As the magnitude of force Q goes on increasing at some point the magnitude of TAD becomes equals to zero Q = aman TAO=0 AB = € 960 mm)i + (-240 mm) , +(380 mm k | AB) = √(960)2+(240)2 +880)2 = 1060 mm AC = -960 mm + (-240 mm + (-320mm & IĀC) = NE960)2 + (-240)? H(3203 = 1040 mm

DELTA Pg No Unit vector along AB, DAB = AB TABI TAB = TAB LAB TAB) TAB = TAB [(-960 mm) + (-240mm)+380 mm 1060 mm = (- 9067 MABê + (-0.226 TAB) ↑ + (0.358 TAB K Unit vector along AC = XAC = AC - 0 TACT TAC = T C TAC = TAC AC ACT TAC = Toc [(-960 mm) î + (-24 0 mm)i + (-320 mm) kJ 1040 mo = 60.923 TAC) î + 60-230 TA c) ĵ + (-308 TA c R P = 0200N) e ③ Q = Qman ĵ - -@ Ef=0 TABI TACITAD+P+B=0 60.906 TAB]Í + (-0.226 TAB) +(0.358 TAB)k + (-0.923TA CH" +(-0-23). Tad +(-0-308 TAC Kt of 1200N) i t aman Î = 0 G0.906 TAB- 0.923 TAC +1200N) î+ (-0.226 TAB-0-231 TAR+Qman) 1 + (0.358 TAB - 0.308 TAC) R = 0 - 9 - 6 -0.906 TAB - 0.923 TAC +1200 = 0 -0.226 TAB 0.231 TAC+ Qman = 0 1.358 TAB -.308 TAC = 0 - 1 from TAB = 0.860 TAG.

Date DELTA PO NO Put in 6 -.90 6 TAB - 0.923 Tac + 1200 = 0 -0.906 C0-860 TAC) - 923 TAC +1200=0 | TAC = 705N bom (TAB = 606.50 Put in Eq -22 G TAB-0.231 To t Q man = 0 (Qmax = 299.50 The calde AD will remain saut far the following gange os Q <299-5N)