第二章 2-1解: 设加水后左边水银面下降Δh ①设容器空着时,左边水注为h, 则有 ?h??水银h水银 h=13.6×0.6×sin30°=4.08(m) ②容器充满水后 ???3.0?4.08??h???水银??0.6??h?sin30???h? 7.08+Δh=13.6 (0.3+1.5Δh) Δh=0.155(m) 读数l =0.6+0.155+0.155/sin30°=1.065(m) 2-2解: pB?pA??(h1?h2)??s2h2??s1h1?98?9.8(0.30?0.15)?0.8?9.8?0.15?0.9?0.8?0.3 ?97.412kpa2-3解:
由?0.22?13.6?9.8?(25?20)?0.8?9.8?0.8?9.8h1得h1?5?0.22?13.6?1.26(m)?126(cm)0.8
由0.8?9.8?1.26?9.8?(20?15)?9.8h2h2?0.8?1.26?5?6.008(m)?600.8(cm)由9.8?(6.008?15?10)?13.6?9.8hh?0.809(m)?80.9(cm)
2-4解: ?'??1???'?h??2h2 h1?2h2 ?'?11??h1?h22-5解:
设大气压强为10m水柱
相对压强pA??2?9800??19600(pa) pB?2.5?9800?24500(pa) po??3?9800??29400(pa) 绝对压强pA绝?8?9800?78400(pa)
pB绝?12.5?9800?122500(pa)
po绝?7?9800?63600(pa) 2-6解:
g(?水h2??煤气H)?(?空气H??水h1)g
y?3m
?煤气??空气??水2-7解:
h2?h10.115?0.1?1.28?1000??0.53(kg/m3) H20p?pa??水银(h4?h3?h2?h1)??水(h2?h3?H?h1) ?98?13.6?9.8(1.1?1.1)?9.8(1.3?1.6)?98?9.8?(13.6?2.2?2.9)?362.796(kpa)
2-8解:
?d2设A杯水位下降ΔH,则B杯水位上升ΔH,ΔH=??d?1?h?5??h?h? ???50100???22?1gH1??2gH2
p1??1g(H1??H?h)?p2??2g(H2??H?h)
?p?p1?p2?(?1??2)g?H?(?1??2)gh?156.6(pa) 2-9解:
(1)zA1?h2?h1?3?2?5(m) zB1?h2?3(m)
pA1?pB1 ?(0.7?1)?10??3(m水柱) ??3?2??1(m水柱)?
PA????z?2m水柱) A???5?3?(???1PB???z?2m水柱) ?B????3?1?(??1
(2)zA2?h1?2(m) zB1?0 pA1
?pB1 ??3(m水柱) ??1(m水柱)?PA?PB???????z??z? AB??????1(m水柱)????2??2图略。 2-10解:
101325+1.03×9800(y+0.35)=0.84×13.6×9800+0.70×13.6×9800 y=9.95(m) 2-11解:
小活塞上受力F?F2b?a75?15?147??882(N) a15
活塞下部压强p?4F24F ?2?d1?d22F24905?5?11.79(cm) F882 d2?d2-12解: 取图示隔离体分析其受力,有: 支承力F?G水?G容器 ??g(D2a?d2b)?9804????9800??0.82?0.4?0.32?1.5??0.1? ?4??3989(N)容器水? 2-13解: 取半球内的水为隔离体,其受力如图 拉力为Fn Fn?P?G水cos??22?R3????g??R?cos30???3???2???g?R2?1?Rcos30???3?1?21??9800?3.1416??1??cos30??4?32??9919(N) P=ρgHπR 水? 2πρ=3切力为Fτ 2?R32?11F??G水sin???gsin30??9800????1283(N) 33822-14解: 将坐标原点取在自由表面中心点,则自由液面方程为 ax+gz=0
由题意 x=-5 时,z=1 将其代入上式得 a??gz9.8??1.96(m/s2) x5a3.5?x??10 g9.8当车被密封时,前后等压面铅直高度差为?z???p??g?z?0.92?9.8?3.5?10?32.2(Kpa) (后部大于前部) 9.82-15解:
自由液面方程 ax+gy=0 将x??l z?2(h1?h2) 代入得 32?al?g?(h1?h2)?0
3
a?2g(h1?h2) 3l2-16解:
a?4.9(m/s2) 时, p底???g?a??h?4.9?2?9.8(kPa) a?9.8(m/s2) 时, 底部相对压强为零
2-17解:
1?D2?D2?a?b??b a=b 水刚好不溢出时,有 ?2442-18解:
以中心处管子底部为坐标原点,则自由液面方程为 z?由边界条件x=1m, z=2m. 得 z??2x22g ?22g,??2gzx?19.6?2?6.62(1/s)
自由液面方程z?2x2 可求得 x=0.5m时, z=0.5m.
由于是小管,每个管内不考虑液面高度差,于是由液体体积守恒,得 5?r2h?2?r2?0.5?2? h?1(m)
2-19解:
2??75?1?x?x????0.352?0.12??0.354(m) ?z????2g?60?19.62221?222-20解:
以旋转轴处假想的自由液面位置为坐标原点,则自由液面方程为z?处比A点的自由液面高度低
H?8.0??1?sin30???2g22?2x22g,由此可得旋转轴
64?14?8 2gg?8??pB??g?1?cos30?????487(N) pC??g?2cos30??16974(N) g??2-21解:
绘图 略 2-22解:
闸门开启时,水压力绕O轴的力矩和恰好为零。于是有
1H11?1h?1h???h???x????h??x?? 2sin??3sin??2sin??3sin??x?H2?h2??1?H3?h3? 3sin?
H3?h323?0.43x???0.80(m) 3sin?(H2?h2)3?sin60???22?0.42?2-23解:
d??d?d?d?P??g??10?10???100?g?980??769.692(kN)
2?444?2222-24解:
??d2d2??d?3?11?1???g?d3?F??g??????g??23??10.263(kN) ????423?2??81224????2-25 绘图 略
2-26解:
BC?R2??H?h??R2?H2?7.52?(5.8?4.8)2?7.52?5.82?2.678(m)
2??cos?1H?hH15.8?cos?1?cos?1?cos?1?43? RR7.57.5??压力体体积V???R2
?360?121?Rsin??BC?h??6.4 22?4311??????7.52??7.52sin43???2.678?4.8??6.4?53.463(m3)
36022??Pz??V?9.8?53.463?523.94(kN)9.8Px?b??4.82?6.4?722.53(kN)22 P?px2?pz2?892.5(kN)?h2?? 作用线与铅直线的夹角为 ??tg?1Px722.53?tg?1?54? Pz523.94作用线通过O点 作用点距底部的距离?H?Rcos??1.40(m) 2-27解: ?D02??0.22P??gH?sin45???g?0.5??cos45??108.78(N) 水平方向x44