A crane lifts a 4.7m long 820 kg reinforced concrete pipe using a 5.6m long steel cableThe pipe's centre of mass is 1.45m long one endThe ends of cable are attached to each end of pipe remains horizontal when the crane hook is attached at some point between the ends of cableWhat is a) the angle made by each part of cable to the pipe?b) the tension in each part of cable? (Answer given: a) 23.9°, 44.8° , b) 6.12 x 10?N, 7.88 x 10?N)Thanks in advance! :)
Let T1 and T2 be tensions and let theta be te angle that T1 makes with vertical and phi be the angle that T2 makes with verticalWe have T1cos theta + T2cos phi 8209.81 .(1) also T1sin theta T2sin phi -(2) Again if h is the perpendicular distance of center of mass of the pipe from the hook, we have sq rt{1.45^2 +h^2} [5.6 - sq rt{3.25^2 +h^2}] squaring both sides we get {1.45^2 +h^2} 5.6^2 + {3.25^2 +h^2} - [25.6sq rt{3.25^2 +h^2}] or [25.6sq rt{3.25^2 +h^2}] 5.6^2 + 3.25^2 - 1.45^2 39.82 or squaring again we get (11.2^2){3.25^2 +h^2} 1585.6324 or h^2 12.6405 - 10.5625 2.078 or h 1.44 theta arc tan (1.45/1.44) 45.17 degree angle made with the pipe 90 -45.17 44.83 degree and phi arc tan(3.25/1.44) 66.10 and the angle made with pipe 23.90 degree T1 sin 45.17 T2sin 66.10 or T1 (0.9142/0.7092)T2 or T1 1.289T2, substituting in (1) we get {(1.289cos theta) + cos phi}T2 8209.81 or T2 6.12210? N and T1 7.8910? N
