A rectangular lock gate hinged at the bottom that is 100\' wide and 25\' tall ha
ID: 2996706 • Letter: A
Question
A rectangular lock gate hinged at the bottom that is 100' wide and 25' tall has 20' of water on the upstream side. It has 10' of water on the downstream side. What force must be applied to the top of the gate to keep it closed? While shooting Class III rapids, a canoeist hits a sharp rock and pokes a 1" diameter hole in the bottom of the canoe. She places a rubber sheet over the hole and weights it with a flat rock. What is the minimum rock weight to keep the hole plugged? The water line on the canoe is 8" above the bottom of the canoe.Explanation / Answer
Volume of the water coinciding with the gate in upstream side= 100*25*20=50000 ft3
Volume of the water coinciding with the gate in downstream side= 100*25*10= 25000 ft3
Volumetric gradient= 50000-25000= 25000 ft3
Since the force due to potential = m*g= v*? *g
25000*(1/3.28)3*1000*9.8= 6942.95 KN
DIA OF HOLE= 1/12 FT= .025 m
DIA OF WATER LINE CANOE= 8/12 FT= .203m
AGAIN VOLUME DIFFERENCE= 4/3 ? (.2033-.0253)
=.0351m3
Mass of rock = mass of the hole material= v*?*g = .0351*1000*9.8=344.04 N