Question

2.81/ F= 17.3 KN and yr= 0.938m need to see how to work it out

36. 2-80. The tank truck is filled to its top with water. Determine the magnitude of the resultant force on the elliptical back plate of the tank, and the location of the center of pressure measured from the top of the tank. Solve the problem using the formula method. 37. 2-81. Solve Prob. 2-80 using the integration method. (0.75 m) 9.75 m Im Probs. 2-80/81 2.3-96 Eull Alternative Text

Answer 1

(a) ya given, (0.752 + x² Ah 0.75 for ellipse az Ilm b=0.75m 16 Resultant force on the plade F = lgh A here, h = 0.75 m A= ñaba 5x0.75 m² 30, F = 9.81x10²x0.750 0.75 AN [f= 17.3 kr and, "Center of pressure, G= 5+ I L = 0.15+ IT ab? 7 Tab 0:15 Yp = ons + b 2 2015+ 0.952 uxor Jp = 0.9875m?

14 Gratally - (b) By Integration, dy I furtat Force on the small arla da dA=2&dy df = fg (ons-y) dA = J fg (ons-y) 2 sedy F = f 289 (0.1511- ( 4 3 2 1 1 1 2 2) dy F=288[p3s. (youguye+oque surt) - 0.75 10.75 20.75 to.rs Lon5 F = 2X981*18* et xlsxx ]-2x98 1x16[o] TE= 17.3 kn gy For centre of pressure (Jp) ge' . To En taking the sum of moments of all a Tox pressure forces about se axis to the moment of Resultant force. ((0.15-y) dF = FR. J. » (0.15-y) fg (0.75-9) dA = FR Jp Slo.ns-yj² gg 2 x dy = FR Jp Fayp = 28g Slons-yeſ 1-10 y dy

0.75 OL FRYp=2fg | (0.152/1-14 2 + y2/1-14 J _2x0.75g, H -0.75 *T-203 ore so ay per -0.75 -0.75 o 73 0.75 0.75 7 [0.15x+ / 9² / 1- yz dy |-- -OT '0-752 solving on 0752 I = J y ²/1-42 dy let sino = u5 Cosodo= dy Is 0.75?sin?0 Ni-sin’O x0.75 Cosodo -0.75 m2 I= 07 sino costo do - - I- Sons (1-сove) do e corp ? [o- sipuram I - 0,75x2+] = 0.1656 putting in eqli) the value of I - FRJp = 289 [ 0.75 + 0.1656] => 17.3 Yp=249.81 X 0.8282 Typ = 0.939 3m