Notes on Chapter 3 examples

B15 agg w vg disp liq b b 2 ag g 100 g cm3 b expt

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Unformatted text preview: Buoyant force up = Weight of block down Wdisplaced liquid oil 2 soln B 0.75 bSGg 15 g / cm3 . B soln A w 15 . 3.9 Let Volume displaced: Vd 1 hs 1 WA + WB h hb1 for b p 1 hb 1 Vw Ap h p1 bg subst. 3 for h p 1 in b 2 g, solve for h b1 hb1 i d Subst. (1) for Vd 1 , solve for h p1 hb1 i (2) Ap h p1 Vd 1 WA WB pw g Vw Ap bW A h p1 WB pw g bi g Vw Ap g LM 1 NM A p 3-3 Vw Ap h p1 WA WB pw gAp 1 Ab OP QP (1) WA WB weight of displaced water Volume of pond water: Vw subst. 2 d (object sinks) Ab hp1 hb1 wVd 1 g WA WB pw gAb h p1 hb1 Ab hsi Archimedes 1 Before object is jettisoned bg density of water. Note: w d Ab h p1 hb1 (3) (4) i...
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