Solving Beam & Magnitude Reactions - 25kg Beam

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The discussion revolves around calculating the reaction forces at points A and B for a uniform beam supported by a roller and a pin, subjected to specific forces and dimensions. The user attempted to solve the problem using equations derived from static equilibrium but found their answers, Ra = 47.1 and Rb = 88.6, to be incorrect. They expressed uncertainty about their approach and noted that the figure referenced in the homework statement was not attached. Additionally, the user mentioned posting the same question in another forum for further assistance. Clarification on the correct methodology and verification of the calculations is sought.
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Homework Statement



As shown, a roller at point A and a pin at point B support a uniform beam that has a mass 25.0 kg . The beam is subjected to the forces f1 = 50.0N and f2= 79.0N . The dimensions are l1= 0.750m and l2= 2.30m . (Figure 2) What are the magnitudes and of the reaction forces and at points A and B, respectively? The beam's height and width are negligible.

Please see attachment for figure.


Homework Equations





The Attempt at a Solution



(79*cos(15)*3.05)+(50*0.75) = Rb *3.05
Rb=88.6

Ra * sin(53.13010)*3.05 = 50*2.3
Ra=47.1311

I am not sure if this is how you do this type of question..?
Also the answers are not correct as when I go to see if they are correct they come back as incorrect. The ansers i have tryed which are incorrect are, Fa = 47.1 Fb=88.6
 
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Uh ... what attached figure is that ?
 
wilson11 said:

Homework Statement



As shown, a roller at point A and a pin at point B support a uniform beam that has a mass 25.0 kg . The beam is subjected to the forces f1 = 50.0N and f2= 79.0N . The dimensions are l1= 0.750m and l2= 2.30m . (Figure 2) What are the magnitudes and of the reaction forces and at points A and B, respectively? The beam's height and width are negligible.

Please see attachment for figure.

Homework Equations



The Attempt at a Solution



(79*cos(15)*3.05)+(50*0.75) = Rb *3.05
Rb=88.6

Ra * sin(53.13010)*3.05 = 50*2.3
Ra=47.1311

I am not sure if this is how you do this type of question..?
Also the answers are not correct as when I go to see if they are correct they come back as incorrect. The answers i have tryed which are incorrect are, Fa = 47.1 Fb=88.6
There is no attached figure.
 
It says that I have already uploaded the image. I have also posted this question in a different forum. Can you please have a look at this one.

https://www.physicsforums.com/showthread.php?t=637033

Above is the same question with figure.

Sorry for the confustion.

Thanks
 
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