A rod of length L and mass M is acted on by two unequal force F1 and F2 (< F1) as shown in fig.The tension in the rod at a distance x from end B will be

Net force on the rod = F1 - F2∴ Acceleration of the rod=F1−F2/MNow consider t}re motion of part BC. Mass of this part = (M / L) x. ThenF1−T=MLxawhere I is the tension at point C. SoF1−T=MLxF1−F2M=F1−F2xLor T=F1−F1−F2xLor T=F11−xL+F2xL.