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Old Oct 18th 2008, 08:54 AM   #1
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reaction at a fix support

Hi all;

An L-shaped bar is constrained by a fixed support at A while the other extreme C is free. Determine the reaction at the point A when the bar is loaded by the depicted forces.



Data:
$\displaystyle AB=3m, BC=2m, CD=1m$
$\displaystyle R_B=3N$
$\displaystyle R_{Cx}=6N, R_{Cy}=2N, R_{Cz}=3N$

Either I'm doing something wrong or there's more than enough data in the problem. I assume we just have to write the three equations of equilibrium:

$\displaystyle \sum{R_{ix}}=R_{Ax}-R_B-R_{Cx}=0$

$\displaystyle \sum{R_{iy}}=R_{Ay}-R_{Cy}=0$

$\displaystyle \sum{R_{iz}}=R_{Az}-R_{Cz}=0$

Then we just calculate their resultant in order to find $\displaystyle R_B$. If so, then why would we need all those distances?

Maybe my assumptions are plain wrong, I'm really confused... Help appreciated. Thanks.

Last edited by disclaimer; Oct 18th 2008 at 09:33 AM.
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Old Oct 18th 2008, 08:04 PM   #2
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Originally Posted by disclaimer View Post
Hi all;

An L-shaped bar is constrained by a fixed support at A while the other extreme C is free. Determine the reaction at the point A when the bar is loaded by the depicted forces.



Data:
$\displaystyle AB=3m, BC=2m, CD=1m$
$\displaystyle R_B=3N$
$\displaystyle R_{Cx}=6N, R_{Cy}=2N, R_{Cz}=3N$

Either I'm doing something wrong or there's more than enough data in the problem. I assume we just have to write the three equations of equilibrium:

$\displaystyle \sum{R_{ix}}=R_{Ax}-R_B-R_{Cx}=0$

$\displaystyle \sum{R_{iy}}=R_{Ay}-R_{Cy}=0$

$\displaystyle \sum{R_{iz}}=R_{Az}-R_{Cz}=0$

Then we just calculate their resultant in order to find $\displaystyle R_B$. If so, then why would we need all those distances?

Maybe my assumptions are plain wrong, I'm really confused... Help appreciated. Thanks.
Maybe you should consider the second condition for equilibrium in rotational motion.
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Old Oct 19th 2008, 03:56 AM   #3
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Originally Posted by werehk View Post
Maybe you should consider the second condition for equilibrium in rotational motion.
Thanks. My question is, are there any other reactions present in the system that I didn't consider? If we have only $\displaystyle R_{Ax},R_{Ay},R_{Az}$, then I don't see why we would have to consider also the second condition for equilibrium. But then again, the problem seems too simple with only three equations of equilibrium...
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Old Oct 21st 2008, 02:14 AM   #4
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Ah, silly me, I should also consider three equlibrium equations for torques...
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