Solving a Beam Question: 3000 kg, 6000 kg*m, Calculate Force & Tension

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The discussion revolves around solving a beam problem involving a 3000 kg beam attached to a wall with a cable and subjected to a moment of 6000 kg*m. Participants seek clarification on the graphical method for calculating forces and tension, emphasizing the need to accurately represent lengths and angles. There is uncertainty regarding the units of measurement, specifically whether to use kilograms force or convert to Newtons. The reaction forces at point A are acknowledged to exist in both the X and Y axes. The main focus is on how to incorporate the applied moment into the force balance, with a note that the moment affects the sum of moments equation rather than being directly equated to force times distance.
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ok, i don't know how to start this problem :blushing:

the beam in the drawing weighs 3000 kg. It is attached to the wall by means of a simple union and without friction in point A. It is attached to the wall by means of a cable CD. The moment applied in point B is 6000 kg*m. Calculate graphically the force in point A and tension in cable CD:


lots of questions here:

0) when they say graphically... what exactly do the mean? you know, that you have to draw the lines with their exact length and their angle and with a ruler measure the resultant force...

1) UNITS: the kilograms here are kilograms force?, or should i convert them to Newtons?

2) REACTION IN POINT A: I assume there will be a reaction in the X and Y axis.

3) the main question here is: the momentum applied, how should i consider it when balancing the forces, does this thing have to do with flexural moments and that stuff? or i just translate the momentum applied to F*d?

thanks in advance for any help you can give me
 

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Remember that unlike the force, that can only move along its line of action without doing a force-couple (force momentum) system, the momentum is a free vector and you can put it anywhere.

The momentum will only affect the sum of moments equation.

\sum \tau also represented as \sum M

Notice calculating sum of momentum on point B, won't make it 0 (It's a Free Vector).
 
thanks a lot, i really appreciate it

and i really hope you're Somewhere in the Sunny Caribbean

:cool:
 
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