Question #241167

The tension in cable BC is 725 Newton. Determine the resultant of the three forces exerted at point B of beam AB.

Expert's answer

For the cable BC

We calculate the forces along the x and y direction as below

"(F_{BC})_x=-725N(\\frac{840}{1160})"

=-525N

"(F_{BC})y=725N(\\frac{800}{1160})"

=500N

For the 500N force,

"(F_{500})_x=-500N(\\frac{3}{5})"

=-300N

"(F_{500})_y=-500N(\\frac{4}{5})"

=-400N

For the 780N force,

"(F_{780})_x=780N(\\frac{12}{13})"

=720N

"(F_{780})_y=-780N(\\frac{5}{13})"

=-300N

Total force along x direction is;

"F_x=(F_{BC})_x+(F_{500})_x+(F_{780})_x"

=-525N-300N+720N

=-105N

Total force along y direction is;

"F_y=(F_{BC})_y+(F_{500})_y+(F_{780})_y"

=500N-400N-300N

=-200N

The resultant force is;

R="\\sqrt{\\smash[b]{(F_x)^{2}+(F_y)^{2}}}"

"=\\sqrt{\\smash[b]{(-105N)^{2}+(-200N)^{2}}}"

=225.89N

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