Question

1. A water trough of semicircular cross section of radius 0.6 m consists of two symmetric parts hinged to each other at the bottom, as shown in Fig. 1. The two parts are held together by a cable and turnbuckle placed every 3 m along the length of the trough. Calculate the tension in each cable when the trough is filled to the rim. Draw a sketch to illustrate the problem solution and show all your work (write formulae, substitutions with units, detailed calculations). (8 points) 1.2 m Cable Hinge Fieure 1

please help...add sketch

Answer 1

Answer consider half part as shown TE I'! 'Fr W=Fv - Horizontal force on vertical Surface is F = eghe A = 29) = 1000 (9.81)0-6x3) 5297.4 N vertical for on the horisontal surface is Ev=kl= egy Eeg(wx moy = 1000 (9.81)(3x (06) = 8316.YIN Resultant force FR is 2 FR = F E 5297.4 +8316.91 = 9860.7 N e = tanic 8316.91 52.97.4 -) = 57.5°

Take moment about point a EMANO [Fre sin (90-57.5)) R - TR = 0 9860.7 sín (32.5)(0.6) - T (0.6) = 0 T=5297.39 Navy Good Luck

Answer 2

I realize that Fh is the only other horizontal force acting on the system apart from the tension (T).

Therefore Fh = T as well.