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two forces f1=2i-5j-6k and f2=-i+2j-k are acting on a boby at points (1,1,0) and (0,1,2) respectively. Find torque acting on body about point (-1,0,1)
Let point A = (1,1,0) and B = (0,1,2) and k = (-1,0,1).Position of A with respect to K is {1-(-1),1-0,0-1} = (2,1,-1)Similarly position of B w.r.t k is (1,1,1)Now, Torque by F1 = cross product of A*F1=(-11i , 10j , -12k)Torque by F2 = cross product of B*F2=(-3i , 3k)Resultant of all torque = T(a) + T(b) =
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