# Mechanics: Moments of force in a rigid beam

1. Sep 23, 2011

### JohanM

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$\vec{M} = \vec{r} \times \vec{F}$

3. The attempt at a solution
For part A, I figured that I can calculate the magnitude of FB by adding all of the moments about point A and separating |Fb|

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 FB points in the negative x-direction?

Thanks,
Johan

Last edited: Sep 23, 2011
2. Sep 23, 2011

### 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.