The geometry is the first part of the Problem. To ease our work, let us name the points C, D, E, and G as shown in the figure. The unit vectors from C to D and from E to G are essential to the location of points A and B. The diagram shown contains two free bodies plus the pertinent geometry. The unit vectors from C to D and from E to G are given by
Using the coordinates of points C, D, E, and G from the picture, the unit vectors are
where CA D 3 m. From these equations, we find that the location of point A is given by A (2.13, 2.34, 0) m. Once we know the location of point A, we can proceed to find the location of point B. We have two ways to determine the location of B. First, B is 3 m from point A along the line AB (which we do not know). Also, B lies on the line EG. The equations for the location of point B based on line AB are:
The equations based on line EG are:
We have six new equations in the three coordinates of B and the distance EB. Some of the information in the equations is redundant. However, we can solve for EB (and the coordinates of B). We get that the length EB is 2.56 m and that point B is located at (0, 1.53, 1.96) m. We next write equilibrium equations for bodies A and B. From the free body diagram for A, we get
From the free body diagram for B, we get
We now have two fewer equation than unknowns. Fortunately, there are two conditions we have not yet invoked. The bars at A and B are smooth. This means that the normal force on each bar can have no component along that bar. This can be expressed by using the dot product of the normal force and the unit vector along the bar. The two conditions are
Solving the eight equations in the eight unknowns, we obtain
F = 36.6 N.
Other values obtained in the solution are EB = 2.56 m,
NAx = 145 N, NAy = 116 N, NAz = -112 N,
NBx = -122 N, NBy = 150 N, and NBz = 112 N.
