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Question: 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/mm2.
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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)

 modified 1 day ago by written 9 weeks ago by
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Stresses in wire 1 and 2 is 120 N/mm^2. And area as A1=200 mm^2 and A2=250 mm^2 Now considering the stresses as c and σ2 in wire 1 and 2. Let forces be F1 and F2 in wire 1 and 2 respectively.

Now, Stress=Force/Area. Therefore, σ1=F1/A1 120=F1/200, F1=24,000 N=24KN & 120=F2/250,F2=30,000 N=30KN

For wire by Sine Rule; F1/sin(45)=W/sin(70) 24/sin(45)=W/sin(70) W=31.89 KN......(1)

F2/sin(65)=W/sin(70) 30/sin(65)=W/sin(70) W=31.10 KN......(2)

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

 written 9 weeks ago by
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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, $W=31.10KN$