Determine the maximum weight 'w' that can be supported by two wires as shown in figure if the stress in each wire is not exceed to 120N/mm^2 (5marks)

Stresses in wire 1 and 2 is $120 N/mm^2$. And area as $A1=200 mm^2$ and $A_2=250mm^2.$ Now considering the stresses as c and $\sigma_2$ in wire 1 and 2. Let forces be $F_1$ and $F_2$ in wire 1 and wire 2 respectively. Now, $stress=\frac{force}{area}$. Therefore, $\sigma_1=\frac{F_1}{A_1}$

$120=\frac{F_1}{200}$

$F_1=24000 N$

$N=24KN$ and $120=\frac{F_2}{250}$

$F_2=30000$

$N=30KN$

For wire by Sine rule, $\frac{F_1}{sin(45)}=\frac{W}{sin(70)}$

$\frac{24}{sin(45)}=\frac{W}{sin(70)}$

$W=31.89 KN$

$\frac{F2}{sin(65)}=\frac{W}{sin(70)}$

$\frac{30}{sin(65)}=\frac{W}{sin(70)}$

$W=31.10KN$

For safe load, we consider minimum of (1) and (2). Therefore, $W=31.10KN$