Answer to Question #241167 in Mechanical Engineering for Parkey

Question

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

Answer 1

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