Tiêu đề Statics Engineering Practice Sheet Solution.pdf ✅

w ̄’wZwe` ̈v  Engineering Practice Sheet Solution 1 08 w ̄’wZwe` ̈v Statics WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. f~wgi Dci Lvovfv‡e `Ûvqgvb GKwU Mv‡Qi mv‡_ 20 wgUvi `xN© GKwU k3 `woi GKcÖvšÍ euvav Av‡Q Ges Aci cÖvšÍ a‡i GKRb †jvK wbw`©ó ej cÖ‡qv‡M Uvb‡Q| MvQwUi †Kvb ̄’v‡b `wo evua‡j Zv †jvKwUi c‡ÿ Dwë‡q †djv m¤¢e n‡e? [BUET 23-24] mgvavb: awi, ‘x’ •`‡N© ̈ `wo evua‡Z n‡e| AB = x, OA = 20 m B we›`yi mv‡c‡ÿ F e‡ji †gv‡g›U, = F  BC = F  BC OB  OB = F  sin  20cos = F  10  2sincos = 1Fsin2 O B  x A C F GLv‡b, BC OB = sin OB 20 = cos  OB = 20cos †gv‡g›U m‡ev©”P n‡e hw`, sin2 = 1 nq|  2 = 90   = 45 GLb, AB OA = sin  x 20 = sin45  x = 10 2 m (Ans.) 2. 2x `xN© Ges W IR‡bi GKwU mylg Z3v `yBwU LyuwUi Dci Avbyf~wgKfv‡e Aew ̄’Z hv‡`i g‡a ̈ `~iZ¡ y| Z3vwU bv Dwë‡q Gi `yB cÖv‡šÍ ch©vqμ‡g me©vwaK W1 I W2 IRb Szjv‡bv hvq| Zvn‡j,     x y †K wewfbœ IR‡bi gva ̈‡g cÖKvk Ki| [BUET 22-23] mgvavb: hLb ïay W1 Szjv‡bv, W1(x – d) = Wd  d = W1x W + W1 hLb ïay W2 Szjv‡bv, W2 (x – d) = Wd  d = W2x W + W2 x – d d d x – d x x W1 W W2 A B y  d + d = W1x W + W1 + W2x W + W2 [d + d = y]  x y = 1 W1 W + W1 + W2 W + W2 (Ans.) 3. c P P + W W d Q Q + W W cÖgvY Ki †h, W = cP – dQ d – c [BUET 21-22] mgvavb: P  c = W(l – c)  P W = l – c c  P + W W = l c ....... (i) c P P + W W l – c l A B E Q  d = W(l – d)  Q W = l – d d  Q + W W = l d ....... (ii) d Q Q + W W l – d l A B F (i)  (ii) K‡i, P + W Q +W = d c  cP + cW = dQ + dW  (c – d)W = dQ – cP  W = cP – dQ d – c (Proved) 4. 5 N 4 N w GKwU †njv‡bv mgZ‡ji •`N© ̈ I f‚wgi mgvšÍiv‡j wμqvkxj ej h_vμ‡g 4 N I 5 N| ejØq cÖ‡Z ̈‡K GKKfv‡e W IR‡bi †Kv‡bv e ̄‘‡K Z‡ji Dci w ̄’i ivL‡Z cv‡i| W Gi gvb wbY©q Ki| [BUET 21-22]
2  Higher Math 2nd Paper Chapter-8 mgvavb: R 4 N W C R1 D  A B  5 N wP‡Î, R1, W I 4 N G wZbwU ej D we›`y‡Z mvg ̈ve ̄’v m„wó Ki‡e| W sin90 = 4 sin  cosec = W 4 .......... (i) Avevi, R, W I 5 N G wZbwU ej D we›`y‡Z mvg ̈ve ̄’v m„wó Ki‡e| 5 sin(180 – ) = W sin(90 + )  5 sin = W cos  cot = W 5 ............... (ii) GLb, cosec2 – cot2 = 1  W 2 16 – W 2 25 = 1  W 2 = 400 9  W = 6.67 N (Ans.) 5. W1 W2 60 20 45 T3 T2 T1 wP‡Îi g‡Zv `ywU Dj¤^ †`qvj †_‡K myZvi mvnv‡h ̈ 5 kg f‡ii 2 wU †`vjbv Av‡Q| cÖ_g †`vjbvq evj‡Ki fi 40 kg| e ̈e ̄’vwU mvg ̈e ̄’vq _vK‡j 2q †`vjbvwU‡Z _vKv evj‡Ki fi KZ? [BUET 20-21] mgvavb: cÖ_g †`vjbvq, W1 = (40 + 5) kg-wt = 45 kg-wt = (45 × 9.8) N = 441 N W1 T2 T1 45 20 jvwgi Dccv` ̈ n‡Z, T2 sin135 = 441 sin115  T2 = 441 sin115 × sin135  T2 = 344 N wØZxq †`vjbvq, jvwgi Dccv` ̈ n‡Z, W2 T3 T2 30 20 T2 sin120 = W2 sin170  W2 = T2 sin120 × sin170  W2 = 344 sin120 × sin170  W2 = 68.98 N  W2 = 7.04 kg-wt  evj‡Ki fi = (7.04 – 5) kg = 2.04 kg (Ans.) 6. 10 m j¤^v Ges 20 N IR‡bi GKwU mylg `Ð AB `‡Ði Dcwiw ̄’Z C we›`y‡Z gy3fv‡e Szjv‡bv Av‡Q| Gi B cÖv‡šÍ 12 N IRb ivL‡j `ÐwU Avbyf‚wg‡Ki mv‡_ mgvšÍivj _v‡K| C we›`yi Ae ̄’vb wbY©q Ki| `ÐwUi A I B cÖv‡šÍ AviI 5 N K‡i IRb †hvM Ki‡j `ÐwU Avbyf‚wgK Ae ̄’vq ivLvi Rb ̈ C we›`y‡K g~j Ae ̄’vb †_‡K KZUv miv‡Z n‡e Zv wbY©q Ki| [BUET 19-20] mgvavb: C 20 N 12 N (5 – x) B O A x wPÎ n‡Z, AB = 10 m, OB = 5 m, IRb ga ̈we›`y O †Z wμqviZ| C we›`yi mv‡c‡ÿ †gv‡g›U wb‡q cvB, 12x = 20 × (5 – x)  12x = 100 – 20 x  x = 100 32 m = 3.125 m  C we›`yi Ae ̄’vb B n‡Z 3.125 m †fZ‡i| (Ans.) C 20 N 17 N A B y m 5 N 5 5 – y O Avevi, A I B cÖv‡šÍ 5 N †hvM Kivi ci C we›`yi mv‡c‡ÿ †gv‡g›U wb‡q cvB, 17y = 20 × (5 – y) + 5 × (10 – y)  17y = 100 – 20y + 50 – 5y  42y = 150  y = 150 42 = 3.571 m  C we›`y‡K g~j we›`y †_‡K A Gi w`‡K (3.571 – 3.125) m ev 0.446 m miv‡Z n‡e| (Ans.) 7. 20 cm •`‡N© ̈i nvjKv AB `ÐwU 10 cm e ̈eav‡b `yBwU †c‡i‡Ki Dci Avbyf~wgKfv‡e Aew ̄’Z| A I B we›`y‡Z h_vμ‡g 2W I 3W IRb Szjv‡bv n‡j, †c‡iK `yBwUi †Kvb Ae ̄’v‡bi Rb ̈ G‡`i Dci Pvc mgvb n‡e? [BUET 17-18; CUET 04-05] mgvavb: AB = 20 cm awi, mgvb Pvc = p  p + p = 2W + 3W  2p = 5W  p = 5W 2 ........ (i) 2W 3W p A C B x 10 D p A we›`yi mv‡c‡ÿ åvgK wb‡q, 2W × 0 + 3W × AB = p × AC + p × AD 0 + 3W × 20 = 5W 2 x + 5W 2 (x + 10) [(i) n‡Z]  3W × 20 = 5W 2 (2x + 10)  2x + 10 = 24  2x = 14  x = 7  †c‡iK `ywU A cÖvšÍ †_‡K 7 cm I 17 cm `~‡i ̄’vcb Ki‡Z n‡e| (Ans.)
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