Analysis of a Figure to Calculate Force and Moment

[Guillem_dlc]

Homework Statement

The bent bar $ABDE$ is supported by spherical plain bearings at $A$ and $E$ and by the cable $BF$. If a load of $600\, \textrm{N}$ is applied at $C$, as shown in the figure, determine the tension in the cable.

Solution: $853\, \textrm{N}$

Relevant Equations

Static equations, trigonometry

Figure:

My attempt at a Solution:

$$\overrightarrow{TFD}=TFD\dfrac{(-0,16\widehat{i}+0,11\widehat{j}-0,08\widehat{k})}{0,21}$$
View from above:

We calculate $D$:
$$\sigma =90-\arctan \left( \dfrac{0,07}{0,240}\right)=73,74\, \textrm{º}$$
$$d=0,16\cdot \sin (\sigma)=0,1536\, \textrm{m}$$
$TF_x$ does not make moment and $TF_z$ does not make time $\rightarrow \alpha =16,26\, \textrm{º}$
$$\sum M_{EA}=TF_x\cdot \sin \alpha \cdot d+TF_z\cdot \cos \alpha \cdot d+TF_yd -600\cdot d'$$
$$\rightarrow TFD\dfrac{0,16}{0,21}\sin \alpha d+TFD\dfrac{0,08}{0,21}\cos \alpha d+TFD\dfrac{0,11}{0,21}d=600d'\rightarrow$$
$$\rightarrow TFD=\dfrac{0,21\cdot 600d'}{d}\cdot \left( \dfrac{1}{0,16\sin \alpha +0,08\cos \alpha +0,11}\right)=405,2\, \textrm{N}$$
Could you have a look at this one?

 

[erobz]

What do you get for the system if you just apply $\sum_{x,y,z} F = 0$ $\sum_{A,E} M = 0$ using the coordinate system already in place?

 

[erobz]

I find $T \approx 853 \rm{N}$. You will only actually need 3 equations in the end to answer this question by using a coordinate system offset through the point $E$ to generate the remaining two equations (you already have the third for the tension $T$ in terms of its components).

 

[haruspex]

$TF_x$ does not make moment and $TF_z$ does not make time $\rightarrow \alpha =16,26\, \textrm{º}$
$$\sum M_{EA}=TF_x\cdot \sin \alpha \cdot d+TF_z\cdot \cos \alpha \cdot d+TF_yd -600\cdot d'$$

You state correctly that the x component of tension has no moment about AE, but you include such a term in the equation. Similarly the z component (but "does not make time"?).

 

[Guillem_dlc]

Homework Statement:: The bent bar $ABDE$ is supported by spherical plain bearings at $A$ and $E$ and by the cable $BF$. If a load of $600\, \textrm{N}$ is applied at $C$, as shown in the figure, determine the tension in the cable.

Solution: $853\, \textrm{N}$
Relevant Equations:: Static equations, trigonometry

Figure:
View attachment 316451

My attempt at a Solution:
View attachment 316452
$$\overrightarrow{TFD}=TFD\dfrac{(-0,16\widehat{i}+0,11\widehat{j}-0,08\widehat{k})}{0,21}$$
View from above:
View attachment 316453
We calculate $D$:
$$\sigma =90-\arctan \left( \dfrac{0,07}{0,240}\right)=73,74\, \textrm{º}$$
$$d=0,16\cdot \sin (\sigma)=0,1536\, \textrm{m}$$
$TF_x$ does not make moment and $TF_z$ does not make time $\rightarrow \alpha =16,26\, \textrm{º}$

I've ended up doing it this way now and I got it:
$$\sum M_{EA}=TF_yd-600\cdot d'=0\rightarrow TFD\dfrac{0,11}{0,21}d=600d'\rightarrow$$
$$\rightarrow \boxed{TFD=853,13\, \textrm{N}}$$