## College Physics (7th Edition)

For part of below most wire, $T_{2}=\frac{mg}{2cos30^{\circ}}=\frac{10\times9.8}{2cos30^{\circ}}=57N$ For upper most part, $T_{1}=\frac{mg}{sin45^{\circ}}=69N$ $T_{AB}=69cos45^{\circ}-57sin30^{\circ}=20N$ And finally $T=mg=98N$