Question

Two cables are tied together at Cand are loaded with a force P-440 lb as shown. 50° 30° References eBook & Resources Section Break Difficulty Easy value: 2.00 points Determine the tension in cable AC. (Round the final answer to the nearest whole number.) The tension in cable AC is lb.

Answer 1

$\dpi{80} \fn_jvn Suppose\,\,F_{CA} \,\,and\,\,F_{CB}\,\,are\,\, the\,\,tension\,\,in\,\,cables\,\,AC\,\,and\,\,BC\,\,respectively.$

$\dpi{80} \fn_jvn Now\,\,by\,\,force\,\,balance,$

$\dpi{80} \fn_jvn \sum F_x\,\,=0\\\\ -F_{CA}\cos50^{\circ}\,\,+\,\,F_{CB}\cos30^{\circ}=0\\\\ F_{CB}=F_{CA}\frac{\cos50^{\circ}}{\cos30^{\circ}}$$\dpi{80} \fn_jvn .......(1)$

$\dpi{80} \fn_jvn \sum F_y\,\,=0\\\\ P-F_{CA}\sin50^{\circ}\,\,-\,\,F_{CB}\sin30^{\circ}=0\\\\ F_{CA}\sin50^{\circ}+F_{CA}\frac{\cos50^{\circ}} {\cos30^{\circ}}\sin30^{\circ}=440 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[putting \,\,F_{CB}\,\,from\,\,eq.(1)] \\\\ F_{CA} \left (\sin 50^{\circ}+\cos50^{\circ}\tan30^{\circ} \right )=440\\\\ F_{CA}=\frac{440}{\sin 50^{\circ}+\cos50^{\circ}\tan30^{\circ}}\\\\ F_{CA}=386.930\,\,lb$

$\dpi{80} \fn_jvn \,\,Pluging\,\,the\,\,value\,\,of\,\,F_{CA}\,\,in\,\,eq.(1)\\\\ F_{CB}= F_{CA}\frac{\cos50^{\circ}} {\cos30^{\circ}}\\\\ F_{CB}=386.930\frac{\cos50^{\circ}} {\cos30^{\circ}} \\\\ F_{CB} =287.190\,\,lb$

$\dpi{100} \fn_jvn \small \\ \therefore The\,\,tension\,\,in\,\,cable\,\,AC\,\,is\,\,387\,\,lb\\ \&\,\,\,the\,\,tension\,\,in\,\,cable\,\,BC\,\,is\,\,287\,\,lb\\$

$\dpi{80} \fn_jvn {\color{DarkBlue} [For\,\,any\,\,doubt\,\,please\,\,leave\,\,a\,\,comment]}$