Question

A cable suspends across a gap of 200 m and carries a load of 15 kN/m vertically. The central sag is 6 m. The maximum tension developed in cable is _____kN.

• A. 12589

Answer 1

The correct option is A 12589

$\sum F_y=0$

$V_A=V_B=\frac{15\times 200}{2}=1500kN$

$\sum M_c=0$

$V_A\times 100=6H_A+15\times 100\times 50$

$1500\times 100=6H_A+75000$

$H_A=12500kN$

$\text{Maximum tension}=\sqrt{V_A^2+H_A^2}$

$=\sqrt{(1500{)}^2+(12500{)}^2}=12589.68kN$