Given F1 = -5i+3j-4k F2 = 5i-5j-2k The position vectors of the forces F1 and F2 are r1 and r2 respectively From figure we can write r1 = 4 i -0.15 j + 0.25 k r2 = 4 i +0.15 j + 0.25 k Now the equivalent force at O is FR = F1 + F2 = -5i+3j-4k+ 5i-5j-2k FR = -2 j -6 k Therefore FRx, FRy,FRz = (0, -2, -6) Now the resultant moment about O is MR,O = r1XF1 + r2X F2 MR,O =(4 i -0.15 j + 0.25 k)X(-5i+3j-4k)+(4 i +0.15 j + 0.25 k)X( 5i-5j-2k) MR,O =12 k +16 j -0.75 k+0.6 i -1.25 j -0.75 i -20 k+8 j -0.75 k -0.3 i +1.25 j + 1.25 i MR,O = 0.8 i + 24 j -9.5 k so (MR,Ox,MR,Oy, MR,Oz) = (0.8, 24, -9.5) Solution Given F1 = -5i+3j-4k F2 = 5i-5j-2k The position vectors of the forces F1 and F2 are r1 and r2 respectively From figure we can write r1 = 4 i -0.15 j + 0.25 k r2 = 4 i +0.15 j + 0.25 k Now the equivalent force at O is FR = F1 + F2 = -5i+3j-4k+ 5i-5j-2k FR = -2 j -6 k Therefore FRx, FRy,FRz = (0, -2, -6) Now the resultant moment about O is MR,O = r1XF1 + r2X F2 MR,O =(4 i -0.15 j + 0.25 k)X(-5i+3j-4k)+(4 i +0.15 j + 0.25 k)X( 5i-5j-2k) MR,O =12 k +16 j -0.75 k+0.6 i -1.25 j -0.75 i -20 k+8 j -0.75 k -0.3 i +1.25 j + 1.25 i MR,O = 0.8 i + 24 j -9.5 k so (MR,Ox,MR,Oy, MR,Oz) = (0.8, 24, -9.5).