Solve Value of R: Moment & Equilibrium

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SUMMARY

The discussion centers on solving for the value of R in a moment and equilibrium problem involving a beam. The user initially attempted to resolve R into a perpendicular force but found confusion regarding its application. It was clarified that R, when applied perpendicularly, results in a torque value that is 8.00/8.67 times smaller than the represented R. The reactive force compensating for the applied force on point B is denoted as Ax, and all torque calculations reference the fulcrum A.

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maitake91
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Homework Statement
Solve the value of R based on the diagram attached.
Relevant Equations
Sum of moment in equilibrium = 0
Summary:: I need some help with moment when using equations of equilibrium

Edit: Sorry I forgot to say what the question was! The question was to solve the value of R based on the diagram below alone.

I was solving the question below and I tried to resolve R into a force that is perpendicular to the beam, but I couldn't. So then I looked at the workout for the answer, and apparently you don't have to do that, since they just used R to multiply 8. I don't understand why this is the case. By intuition, I imagine if I tried to push a beam with the same force but did it diagonally instead of at a position perpendicularly towards the beam, it makes more sense to me that it would spin slower. Please can someone explain to me why I am wrong? I would really appreciate the help.

スクリーンショット 2020-08-14 20.44.05.png
 
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Welcome, maitake91 :cool:

Your intuition is correct: If you "tried to push a beam with the same force but did it diagonally instead of at a position perpendicularly towards the beam", the value of the resulting moment or torque would be smaller.
For the extreme case of line of application of the force intercepting the fulcrum, the torque value would be zero.

The value of R, if perpendicularly applied to the beam, would be 8.00/8.67 times smaller than the represented R.
All torques are in reference to the fulcrum A.
The reactive force that compensates for the angular application of the 500 N applied on B is Ax.
The direction of the reactive force on B can only be perpendicular to the surface that supports the rocking fixture (note its rounded base).
 
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Lnewqban said:
Welcome, maitake91 :cool:

Your intuition is correct: If you "tried to push a beam with the same force but did it diagonally instead of at a position perpendicularly towards the beam", the value of the resulting moment or torque would be smaller.
For the extreme case of line of application of the force intercepting the fulcrum, the torque value would be zero.

The value of R, if perpendicularly applied to the beam, would be 8.00/8.67 times smaller than the represented R.
All torques are in reference to the fulcrum A.
The reactive force that compensates for the angular application of the 500 N applied on B is Ax.
The direction of the reactive force on B can only be perpendicular to the surface that supports the rocking fixture (note its rounded base).
Thank you so much for your reply, it was very helpful and I understand now!
 
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You are welcome :smile:
 

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