Mechanics: Moments of force in a rigid beam

[JohanM]

Homework Statement

Consider the rigid beam below, which is loaded by five forces shown. The forces at A and B act to keep the total force on the bar equal to zero and the sum of the moments about point A equal to zero.
a) Find the forces at A and B that satisfy the conditions stated above.
b) Find the sum of the moments about point B.

Homework Equations

\vec{M} = \vec{r} \times \vec{F}

The Attempt at a Solution

For part A, I figured that I can calculate the magnitude of F_B by adding all of the moments about point A and separating |F_b|

Force C: creates 0 moment about point A since its direction is towards A (theta=180)

Force D:

\vec{r} = \left\langle 1000,-200 \right\rangle
\vec{F} = \langle 25 cos(20^\circ), 25 sin(20^\circ) \rangle

\vec{M}=\left| \begin{array}{ccc}\hat{i}&\hat{j}\\1000&-200 \\ 25 cos(20^\circ)&25 sin(20^\circ) \end{array} \right| \approx 3852.04 \hat{k}

Force B:

\vec{r} = \left\langle 1000,-650 \right\rangle
\vec{F} = \langle |\vec{F_{B}}|, 0 \rangle

\vec{M}=\left| \begin{array}{ccc}\hat{i}&\hat{j}\\1000&-650 \\ |\vec{F_{B}}|&0\end{array} \right| = 650 |\vec{F_{B}}| \hat{k}


\sum \vec{M}=O\hat{k}+3852.04...\hat{k}+650|\vec{F_{B}}|\hat{k}=0

|\vec{F_{B}}| \approx -5.92

Before I go any further I just want to make sure that I am approaching this correctly... Does it make sense that F_B points in the negative x-direction?

Thanks,
Johan

 

[Spinnor]

I think so, if the drawing is roughly to scale the 25kN force tends to rotate the beam counter clockwise therefore for the beam not to move the force at B must tend to rotate the beam clockwise and can only do so if the force at B acts to the left.