Lecture 20. ACT 2 Hydraulics Consider the systems shown to the right In each case, a block of mass m is placed on the piston of the large cylinder, resulting in a difference d in A1 A1 the liquid levels >If A2=2 A1, compare da and dB A)da=(1/2)dB b)da= db c)da=2d A ))If A10=2A20, compare da and dc A)da=(1/2)dc B)da= dc c)dA=2do A Physics 121: Lecture 20, Pg 11
Physics 121: Lecture 20, Pg 11 A1 A10 A2 A10 A1 A20 M M dC M dB dA A) dA = (1/2)dB B) dA = dB C) dA = 2dB »If A10 = 2´A20, compare dA and dC. A) dA = (1/2)dC B) dA = dC C) dA = 2dC Lecture 20, ACT 2 Hydraulics Consider the systems shown to the right. In each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference dI in the liquid levels. »If A2 = 2´A1 , compare dA and dB
Example problems(1) At what depth is the water pressure two atmospheres? It is one atmosphere at the surface. What is the pressure at the bottom of the deepest oceanic trench (about 104 meters)? Solution: P2=P+ ogd is the depth 2.02×105Pa=1.01×105Pa The pressure increases one +103kgm3*9.8m/s2*d atmosphere for every 10m d=10.3 Ford=10 This assumes that water is P2=1.01×105Pa+103kg/m3*98m/s2*104m incompressible =9.81×107Pa=971Atm Physics 121: Lecture 20, Pg 12
Physics 121: Lecture 20, Pg 12 At what depth is the water pressure two atmospheres? It is one atmosphere at the surface. What is the pressure at the bottom of the deepest oceanic trench (about 104 meters)? P2 = P1 + gd 2.02105 Pa = 1.01105 Pa + 103 kg/m3 *9.8m/s2 *d d = 10.3 m P2 = 1.01105 Pa + 103 kg/m3*9.8m/s2*104 m = 9.81107 Pa = 971 Atm Solution: d is the depth. The pressure increases one atmosphere for every 10m. This assumes that water is incompressible. For d = 104 m: Example Problems (1)