Tiêu đề Statics Engineering Practice Sheet Solution.pdf ✅

w ̄’wZwe` ̈v  Engineering Practice Sheet Solution 3 8. P Ges Q `ywU mgvšÍivj I m`„k ej| P e‡ji wμqv †iLv‡K Gi mgvšÍivj eivei Q e‡ji w`‡K x `~i‡Z¡ miv‡bv n‡j G‡`i jwä d `~i‡Z¡ m‡i hvq| cÖgvY Ki †h, d = Px P + Q [BUET 16-17; BUTex 05-06] mgvavb: P A d P P + Q (P + Q) Q C D E B x GLv‡b, (P, x (P + Q, d)  Px = (P + Q)d  d = Px P + Q (Proved) 9. f‚wgi mv‡_  †Kv‡Y †njv‡bv GKwU mgZ‡ji Dci GKwU 20 kg IR‡bi e ̄‘‡K Zj I f‚wgi mgvšÍiv‡j 10 kg-wt gv‡bi `yBwU mgvb ej cÖ‡qv‡M w ̄’i Ae ̄’vq ivLv n‡q‡Q| Z‡ji Dci wμqviZ e‡ji cwigvY wbY©q Ki| [BUET 14-15] mgvavb: 10 + 10cos R 10  20cos + 10sin 20sin   20 GLb, R = 20cos + 10sin ........... (i) Avevi, 20sin = 10 + 10cos  2sin = 1 + cos = 2cos2 2  2  2sin 2 cos  2 = 2cos2 2  cos  2     cos  2 – 2sin 2 = 0 nq, cos  2 = 0   2 =  2   =  [hv Am¤¢e] A_ev, cos  2 – 2sin 2 = 0  tan  2 = 1 2   = 53.13  (i) n‡Z, R = 20 kg-wt (Ans.) 10. †Kv‡bv we›`y‡Z wμqviZ wZbwU ej P, Q Ges R fvimvg ̈ m„wó K‡i| P I Q ci ̄úi j¤^ Ges Q I R Gi ga ̈eZ©x †KvY 120 n‡j, Q I R Gi AbycvZ KZ? [BUET 13-14; MIST 22-23] mgvavb: Q 90 150 120 P R jvwgi Dccv` ̈ Abyhvqx, Q sin150 = R sin90  Q R = 1 2  Q : R = 1 : 2 (Ans.) 11. GKwU †mvRv mylg i‡Wi GK cÖv‡šÍ 10 kg IR‡bi GKwU e ̄‘ Syjv‡bv n‡j, H cÖvšÍ n‡Z 1 m `~‡i GKwU LuywUi Dci Avbyf‚wgKfv‡e w ̄’i _v‡K| LuywUi Dci Pv‡ci cwigvY 30 kg-wt n‡j, iWwUi •`N© ̈ I IRb wbY©q Ki| [BUET 12-13] mgvavb: W l m R = 30 kg-wt P = 10 kg-wt l m ( ) l 2 –1 m l 2 m O C A awi, LuywUi Dci Pvc = R = 30 kg-wt wPÎ n‡Z, R = P + W  30 = 10 + W  W = 20 kg-wt (Ans.) C we›`yi mv‡c‡ÿ †gv‡g›U wb‡q cvB, W    l 2 – 1 = P × 1  l 2 – 1 = 10 20  l 2 – 1 = 0.5  l = 3 m (Ans.) 12. ABC GKwU mgwØevû mg‡KvYx wÎfzR| mgvb evû AB Ges AC cÖ‡Z ̈KwUi •`N© ̈ 4 wgUvi| A, B Ges C we›`y‡Z GKwU e‡ji åvgK h_vμ‡g 8, 8 Ges 16 kg-m; ejwUi gvb I MwZc_ wbY©q Ki| [BUET 12-13] mgvavb: 4 2 B H  A F x 4 C 4 G
4  Higher Math 2nd Paper Chapter-8 wPÎ n‡Z, AG = BH = x (awi)  Fx = 8 GLb, F(x + 4) = 16 [ CG  GH]  Fx + 4F = 16  8 + 4F = 16  F = 2 kg-wt (Ans.) Avevi, Fx = 8  2x = 8 x = 4 m  e‡ji wμqv‡iLv AB Gi mgvšÍiv‡j I AB n‡Z 4 m `~i‡Z¡ A n‡Z B Gi w`‡K wμqvkxj| (Ans.) 13. †Kv‡bv we›`y‡Z wμqviZ P I Q (P > Q) gv‡bi `ywU e‡ji jwä P e‡ji w`‡Ki mv‡_ 60 †KvY Drcbœ K‡i| P ejwU‡K wØ ̧Y Ki‡j D3 †KvY 30 nq| ej `yÕwUi AšÍf©y3 †KvY wbY©q Ki| [BUET 11-12; KUET 06-07] mgvavb: Q P P 30  60 cÖ_g †ÿ‡Î, tan60 = Qsin P + Qcos wØZxq †ÿ‡Î, tan30 = Qsin 2P + Qcos  tan60 tan30 = 2P + Qcos P + Qcos  3 = 2P + Qcos P + Qcos  2P + Qcos = 3P + 3Qcos  P = – 2Qcos  tan60 = Qsin – 2Qcos + Qcos  3 = Qsin – Qcos  tan = – 3 14. f‚wgZ‡ji mgvšÍivj GKB †iLv ̄’ `yÕwU gm„Y †c‡iK P I Q Gi Dci 8 wgUvi `xN© GKwU euv‡ki cÖvšÍØq Ae ̄’vb Ki‡Q| euvkwUi Dci ̄’ R we›`y‡Z GKwU fvix †evSv Szjv‡bv n‡jv, hw` PR = 3RQ nq Ges Q we›`y‡Z Pvc P we›`y‡Z Pvc A‡cÿv 325 MÖvg IRb †ewk nq, Z‡e †evSvwUi IRb wbY©q Ki| [BUET 09-10] mgvavb: P x W x + 325 6 R 2 Q awi, P †Z Pvc = x Q †Z Pvc = x + 325 Ges PR = 3RQ  PR RQ = 3  PR : RQ = 3 : 1  PR = 8 × 3 4 = 6  RQ = 2 wPÎ n‡Z cvB, 6  x = 2 × (x + 325)  6x = 2x + 650  4x = 650  x = 162.5 GLb, W = x + x + 325 = 162.5 + 162.5 + 325  W = 650 kg-wt (Ans.) 15. GKwU fvix Mvwoi PvKvi IRb W Ges e ̈vmva© 20 BwÂ| PvKvi †K›`awe›`y‡Z b~ ̈bZg wK cwigvY ej f‚wgi mgvšÍiv‡j cÖ‡qvM Ki‡j PvKvwU 10 Bw D”PZvwewkó Lvov cÖwZeÜK cvi n‡Z cvi‡e? [BUET 05-06] mgvavb: W C B D O F E 20 A 10 GLv‡b, F.CD  W.CE CD = r – h CE2 = OC2 – OE2 = r2 – (r – h)2 CE = 2rh – h 2  F  W 2rh – h 2 r – h  Fmin = W 2rh – h 2 r – h = W 2 × 20 × 10 – 102 20 – 10 W 3 (Ans.) 16. e„ËPvc AvKv‡ii GKwU nvjKv Zv‡ii `yB cÖvšÍ Gi †K‡›`ai mv‡_  †KvY Drcbœ K‡i| ZviwUi `yB cÖvšÍ n‡Z P I Q IR‡bi `ywU e ̄‘ Szj‡Q Ges ZviwUi DËj w`K wb‡Pi w`K †_‡K w ̄’Zve ̄’vq Av‡Q| †K›`a w`‡q AwZμvšÍ Djø¤^ †iLv, P IR‡bi w`‡Ki e ̈vmv‡a©i mv‡_ †h †KvY Drcbœ K‡i, Zv wbY©q K‡iv| [BUET 04-05] mgvavb: P  r W B  r F A D Q + MD = 0