PHYS 1441 – Section 501 Lecture #19Problem #35Problems 5, 25 and 29Problems 36 and 42Problems 56 and 57Monday, Aug. 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu1PHYS 1441 – Section 501Lecture #19Monday, Aug. 9, 2004Dr. Jaehoon Yu•Problem #35•And many othersMonday, Aug. 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu2HdyyhProblem #35Water is filled to a height H (=32m) behind a dam of width w (=1200m). Determine the resultant force exerted by the water on the dam.Since the water pressure varies as a function of depth, we will have to do some calculus to figure out the total force. Therefore the total force exerted by the water on the dam isP =The pressure at the depth h isThe infinitesimal force dF exerting on a small strip of dam dy isdF =F =( )0y Hyg H y wdyr==- =�2012y Hygw Hy yr==� �- =� �� �212gwHr =( )23 9110 9.8 1200 32 6.0 102N= � � � � = �PdA =( )g H y wdyr -ghr =( )g H yr -Monday, Aug. 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu3Bad problem since it does not provide all necessary information!!Problems 5, 25 and 29Problem 5: Rotational kinematicsa =f itv v-=D47 472.0- -=247 /rad s-Problem 25: Elastic properties of matter100 9.8 980W N= � =Weight of the personTotal force to be supported issup4 3920portF W N= =Thus, from the definition of tensile strengthA =sup portFTensileStength=5 2 2839201.96 10 0.22.00 10Nm cm-= � =�Problem 29: Conditions for equilibriumsin 500 0 xF T Nq= - =�cos 0yF N T q= - =�Where is the angle between the support cable and the poleThus, solving Fx equation for T, we obtain3500 5002.5 10sin25 1NTq= = = �+Monday, Aug. 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu4Problems 36 and 42Problem 36:Pressure measurements1.0 16 14.7 16Air gaugeP P P atm psi psi psi= + = + = +Problem 42: Flow rate and equation of continuity problem( )20.05 1.4 / 0.011 /Air Air AirMA v m s kg str r p rD= ��= �� � = �DFlow rate through the vent5 530.730.7 1.01 10 2.1 1014.7psi N= = � � = �Total mass of air in the room4.0 3.5 2.5 35air room Air Air AirM V kgr r r-= = � � � = �Thus the time needed to replace the entire air in the room is353182 53min0.011 /air room AirAirM kgsFlow rate kg srr-�= = =- �Monday, Aug. 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu5Problems 56 and 57Problem 56: Simple Harmonic Motion( )22 2 21; Thus, 4 4 10 1.68 1114 /2kf k m f N mmp pp= = � � � = � � � =Problem 57: Equation of Motion of a SHM( )( ) ( ) ( ) ( )22 2sin ; sin 3.33 4.22 sin 4.22 1.15 58.7 /x A ta A t m svv v=-=- =- � � =( ) ( )2 2331 11140.662 ; Thus, 645002 2 0.662A AAkf Hz m kgfp p= = = = =��
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