Title Statics & Dynamics Sub.pdf ✅

Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159 TRIGONOMETRICAL RATIO 2 Sol. B C O A R   P Q D  Q In  CAD ; cos  = Q AD AD = Q cos  R cos  = Q = OA + AD Q = P + Q cos   cos  = Q Q − P Q (1 – cos ) = P Q (1 – 1 + 2 sin2/2) = P sin /2 = P / 2Q  = 2 sin–1 (P / 2Q) proved R 2 = P2 + Q2 + 2PQ cos  R 2 = P2 + Q2 + 2PQ         − Q Q P R 2 = P2 + Q2 + 2PQ – 2P2 R = Q P 2PQ 2 2 − + Q.21 A weight w is lifted with the help of a string upto a height h from rest to rest. The string can safely bear a tension upto nw. Prove that the minimum time in which the weight can be lifted is 1/ 2 (n 1)g 2nh       − . Sol. For minimum time T – W = mf nw – w = mf w = mg  nw – w = g w f f = (n – 1) g... (1) t1 t2 time V V velocity 2 1 Vt1 = h1, 2 1 Vt2 = h2 V = ft1, V = gt2 h1 + h2 = 2 1 V (t1 + t2) h = 2 1 Vt... (2) t1 + t2 = V         + g 1 f 1 t = V         + g 1 f 1 ... (3) divide (2) by (3) t h =         + g 1 f 1 2 t  t 2 = 2h         + g 1 f 1 t 2 = 2h         + − g 1 (n 1)g 1  t 2 = 2h         − + − (n 1)g 1 n 1 t = (n 1)g 2nh − Q.24 Three forces P, Q and R act along the sides BC, CA and AB of a triangle ABC, taken in order. Show that if their resultant passes through - (a) the circumcentre if P cos A + Q cos B + R cos C = 0 (b) the incentre if P + Q + R = 0 (c) circumcentre & incentre both then prove cosB cosC P − = cosC cosA Q − = cosA cosB R − Sol. (i) If resultant passes through incentre P(IP) + Q (IE) + R (IF) = 0 P.r + Q. r + R.r = 0 P + Q + R = 0 (ii) cos A =  OD OD =  cos A OE =  cos B OF =  cos C B A C F E D R P Q A O    Resultant passes through circumcentre taking moment about O h2 B h1 O A A Q F R r D B C P I r E