|QUIZ 1- CH 12 - FLUIDS|
|READ CHAPTER EXAMPLES AS NEEDED. A QUICK SURVEY OF FLUIDS ~ DENSITY: 1 ; WATER PRESSURE: 9; BUOYANT FORCE: 25, 26, 27 ; BERNOULLI'S LAW: 41, 44..|
SOMETIMES THE HINTS (INDICATED BY "HINT POSTED" TAG) ARE
ALREADY POSTED AT nvaphysics.com with the LECTURE NOTES,
Exercise 12.2 is for 2 point(s)
Exercise 12.1 is for 2 point(s)
Exercise 12.9 is for 2 point(s)
Exercise 12.21 is for 2 point(s)
Exercise 12.25 is for 2 point(s) HINT POSTED
Exercise 12.26 is for 2 point(s) HINT POSTED
Exercise 12.27 is for 2 point(s) HINT POSTED
Exercise 12.30 is for 2 point(s)
Exercise 12.31 is for 2 point(s)
Exercise 12.42 is for 2 point(s)
Exercise 12.41 is for 2 point(s)
Exercise 12.44 is for 2 point(s)
Exercise 12.48 is for 2 point(s)
Problem 12.86 is for 2 point(s)
HERE'S A HINT TO #86 (a). YOU WANT TO GET dP = differential change in pressure IN THE RADIAL direction (along r) and then integrate. The inward force along r on a differential ring of mass dm is dF = dP*2*pi*r*dh, where r is the inner ring radius , pi = 3.14...., and dh is the differential height of the ring. Note dF = dP*d(area), where the differential area = d(area) = 2*pi*r*dh. But dF = dm*v2/r = centripetal force on differential ring of radius r. NOTE: dm = (density)*2*pi*r*dh*dr, where density = density of fluid. Note: v2 means speed v squared. THUS, dP*2*pi*r*dh = (density)*2*pi*r*dh*dr*v2/r. THUS dP/dr = (density)*v2/r. FINISH PROOF WITH THE SUBSTITUTION: w*r = v.
Problem 12.54 is for 2 point(s)
For the torque we have an integral above and also below the axis shown by dot:
IT SHOULD BE CLEAR THE AVERAGE FORCE ON BOTTOM ( BELOW AXIS) IS HIGHER THAN THAT ON TOP (ABOVE AXIS) DUE TO THE HIGHER (ON AVERAGE) PRESSURE BELOW AXIS OF ROTATION. LET US CONTINUE WITH THE COMPUTATION.
NOW LET'S WORK BELOW THE AXIS:
dTorque = differential torque = Differential force*y = dF*y, where y is the positive vertical downward directed coordinate whose origin is at the central gate axis shown . Clearly dF = (density)*g*y*W*dy, where W is the gate width (not shown in the side view figure above.); W = 4.00 m.
dTorque = (density)*g*y*W*dy*y. Please integrate this from y = 1.00 m to y = 2.00 m
Also apply this method to the top: dTorque = (density)*g*y*W*dy*y. Please integrate this from y = 0.00 m to y = 1.00 m, where y is now positive directed upward.
SUBTRACT THE TWO RESULTS TO GET THE NET TORQUE SINCE THE FORCES ABOVE AND BELOW THE AXIS TEND TO TWIST THE GATE IN OPPOSITE ROTATIONAL DIRECTIONS .
Problem 12.94 is for 2 point(s)
Problem 12.74 is for 2 point(s)
The mass measured on scale is due to the mass of (A) beaker, (B) the liquid in beaker and (C) the force from the.
pressure exerted by the block on the fluid below it. 7.50 kg = Mb +ML + (fluid density* V, where V = 0.0038 m3 . THE LAST TERM COMES FROM THE BUOYANT FORCE: SINCE THE BUOYANT FORCE IS THE FORCE OF WATER ON BLOCK, THE WATER PUSHES ON BLOCK WITH A FORCE OF THIS SAME MAGNITUDE, WHICH IN TURN PROVIDES THE ADDITIONAL PRESSURE DOWNWARD BY BLOCK ON BOTTOM MASS SCALE. Find the fluid density and all other requested quantities from this information.