Question:

Physics Torquehelp?

Physics Torquehelp?

Answer:

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 6.12210³ N and T1 7.8910³ N

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