Question

Consider the cable arrangement as shown in the figure. Cable can support a maximum tension of 75 kN. The largest load P that can be applied is _______kN.

• A. 38.58 kN
• B. 72 kN
• C. 64.311 kN
• D. 75 kN

Answer 1

The correct option is B 72 kN



$V_A+V_B=P+50$

$B{M}_D=0$

$-V_B\times 3+H\times 5=0$

$V_B=\frac{5H}{3}$

$B{M}_D=0$

$+V_A\times 7-H\times 3-P\times 4=0$

$V_A=\frac{3H+4P}{7}$

$R_B=\sqrt{V_B^2+H^2}$

${\left(\frac{5H}{3}\right)}^2+H^2={75}^2$

$[\frac{25}{9}+1]H^2={75}^2$

$H^2=\frac{9}{34}\times {75}^2$

$H=38.58kN$

$V_B=\frac{5\times 38.58}{3}$

$=64.311kN$

$\frac{3\times 38.58+4P}{7}+64.311=P+50$

$16.534+\frac{4P}{7}+64.311=P+50$

$30.845=(P-\frac{4P}{7})$

$P=71.97\underset{–}{\sim }72kN$