HW_3_solutions

# HW_3_solutions - C r€6 7of Hw J,1 rI 2.20 The hydraulic...

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Unformatted text preview: C r€6 7of Hw+: J-,1- rI) 2.20 The hydraulic jack in Fig. P2.20 is filled rvith oil at 56 lbf/ft'. Neglecting piston u"eights. what force F on the handle is required to support the 2000-lbf n'eight shorvn? Fig. P2.20 Solution: First sum moments clockwise about the hinge A of the handle: I M ^ = 0 = F ( 1 s + 1 ) - P ( 1 ) , or: F - P/16, rvhere P is the force in the small (1 in) piston. Meanwhile figure the pressure in the oil from the weight on the large piston: w 2000 lbf A,-,n (.r,-lX3i 12 ft)2 = 40744 pst, Poil z , I \ ) Hence P = o^.,A-_-,, -t40744\: -l =222 lbf , 4 ( r ' ) / Therefore the handle force required is F=P/16 =222116 = lllbf Ans. i i n /l 2.35 Water flows upu'ard in a pipe slanted at 30'. as in Fig. P2.35. The mercury manometer reads h = 12 cm. What is the pressure diff-erence between points (l) and (2) in the pipe? Solution: The vertical distance between points I and 2 equals (2.0 m)tan 30'or 1.155 m. Go aror.rnd the U{ube h.vdro- statically from point I to point 2: p, +9790h...
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HW_3_solutions - C r€6 7of Hw J,1 rI 2.20 The hydraulic...

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