2.1 解： A
f =τ 而 dy d μμτ= )(0565.015.01199.02m dl A =⨯⨯==ππ )(1000005.012
1199.012.0--=-=s dy d μ ∴ )(725.36065.010*******.0N dy
d A f =⨯⨯==μμ
2.3 解： 取等压面0-0，建立静压力平衡方程 gh gx P x H g P B A 汞水水ρρρ+-=-+)( ])([gx x H g gh P P B A 水水汞ρρρ+--=- )(H h g P P B A 水汞ρρ-=-
)1100.136.01036.1(8.934⨯⨯-⨯⨯⨯= )(108220.34Pa ⨯=
2.4 解： )(10210
2500500256111m N A W P ⨯=⨯==- )(1031050001500266222m N A W P ⨯=⨯==
- 21P P <
∴ 缸1先运动，缸2后运动
min)/(410250010106
311m A q V =⨯⨯==-- min)/(210
500010106322m A q V =⨯⨯==-- )(01.500)0(111N V q W F =-+=ρ )(005.15000)0(222N V q W F =-+=ρ
2.10 解： ① 2
244d F d F A F p ππ===∆ 032
12v d p l d q δπμδπ-∆= （因柱塞移动方向与压差方向相反，所以取“-”） 而 0204v d Av q π=
=, mm d D 05.02=-=δ ∴ 023022
4124v d d F l d v d δππμδππ-= 2
32403)(1uld F v d πδδ⨯+= 23333205.042002
.007.010503)1005.0(10)(1⨯⨯⨯⨯⨯⨯⨯⨯+=----π )/(10885.13s m -⨯= )(5310
885.11.030s v h t =⨯==- ② 2244d
F d F A F p ππ===∆ 0232)5.11(12v d p l d q δπεμδπ-+∆= 而 0204v d Av q π=
= mm d D 05.02=-=δ 0223022
)5.11(4124v d d F l d v d δπεπμδππ-+= )5.11(3)(1223240επδδ+⨯+=uld
F v d 偏心距 )(05.0mm r R e =-= )(05.02
mm d D =-=δ 偏心率 105
.005.0===δεe ∴ )/(107125.4)5.11(3)(1322
3240s m uld F v d -⨯=+⨯+=επδδ )(21107125.41.030s v h t =⨯==-
