Question

Hi can you do this proble. There is solutions on cheg but it is not the same solutoin. please specify your solution. Thank You!

Prob3 The gate in is 5 ft wide, is hinged at point B, and rests against a smooth wall at point A. Compute (a) the force on the gate due to seawater pressure, (b) the horizontal force P exerted by the wall at point A, and (e) the reactions at the hinge B. Wall P. Seawater 64 In 15 61

Answer 1

F and P known the reactions Bri the ca to CP is given by The distancu l from since Pa is neglected l= -cp + Inn sino hea A (417 6+9) G) (12 F 1) (506+2) l=0.417 ft 4,58367 thus The distance from point B to force F lord - 5 Summing moments Counterclockwise about B gives PL sino - F(5-1) = P(64+) - [(38400 lbf) (4.583f+)] P=29300 lbf and are with forces ed By found by summing on the Gate Efx = 0 = Bx + Fsino -p = Bu +(38400) (06)-29300 or Br=6300 lbf {fy=0 = By + F(oso By -(38400) (0.8) B 11 as 30700 lbf CS Scanned with CamScanner

Solution Pa Wall Seawater 64 467/63 15 Ft A Pa -> 6ft &ft Shinge A P -5Ft 'cp Bx - loft By is 15-3 = 12 ft and its centroid is halfway to fd from Geometry the Gate AB long between or at Elevation 3 fit above point B. The depth hea is thus The Gate area is 5 (10) = 50 ft2 Neglect Pa as acting on both sides of the gate The hydrostatic force on the gate is Vhoa A= F = 38400 lbf calculation of center of pressure of E bu Scanned Ith = (5 ftJx (10 fty? CS 12 CamScanner 12 - = 417 444