(adsbygoogle = window.adsbygoogle || []).push({}); 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

[itex]\vec{M} = \vec{r} \times \vec{F}[/itex]

3. 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:

[itex]\vec{r} = \left\langle 1000,-200 \right\rangle[/itex]

[itex]\vec{F} = \langle 25 cos(20^\circ), 25 sin(20^\circ) \rangle [/itex]

[itex]\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}[/itex]

Force B:

[itex]\vec{r} = \left\langle 1000,-650 \right\rangle[/itex]

[itex]\vec{F} = \langle |\vec{F_{B}}|, 0 \rangle [/itex]

[itex]\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}[/itex]

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

[itex]|\vec{F_{B}}| \approx -5.92[/itex]

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 thenegativex-direction?

Thanks,

Johan

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# Mechanics: Moments of force in a rigid beam

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