Chapter 3 - Equilibrium of a Particle - Section 3.3 - Coplanar Force Systems - Problems - Page 100: 15

$8.56Kg$

We know that $F=K\Delta x=30(5-3)=60N$ Now $\Sigma F_x=0$ $\implies Tcos(45)-\frac{4}{5}(60)=0$...eq(1) and $\Sigma F_y=0$ $\implies -W+Tsin(45)+\frac{3}{5}(60)=0$....eq(2) Solving eq(1) and eq(2), we obtain: $T=67.9N$ and $W=84.0N$ Now $m=\frac{W}{g}=\frac{84.0N}{9.81}=8.56Kg$