How Do You Calculate Reaction Forces in Static Equilibrium?

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SUMMARY

The calculation of reaction forces in static equilibrium involves applying the equations of equilibrium: sum of forces in the x-direction (sum of Fx = 0), sum of forces in the y-direction (sum of Fy = 0), and sum of moments about a point (sum of moment about point O = 0). In this discussion, the forces R1 and R2 were determined to be equal at 21kN, while R3 was calculated to be 70kN. The direction of moments was established based on the convention of anticlockwise being positive. The calculations were confirmed through consistent results across different approaches.

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Kasthuri
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Homework Statement



Question: The plate shown, picture is attached, is subjected to force F and 3 reaction forces R1 R2 R3. Find R1

Homework Equations



equations of equilibrium:
sum of Fx = 0
sum of Fy = 0
sum of moment about point O = 0


The Attempt at a Solution



sum of Fx = 0: (going right is positive)

-R1 - R2 + 3/5 R3 = 0

sum of Fy = 0: (going up is positive)

-56kN + 4/5 R3 = 0
so, R3 = 70kN

sum of moment about point O = 0 (going anticlockwise is positive)

R1 x (0.11m) - R2 x (0.11m) = 0
so, R1 = R2

therefore, R1 = R2 = 21kN

I am unsure of whether I have done this question correctly, in particular the moment about O.
Also, how do you figure out which direction the moment is in (anticlockwise/clockwise). As in, I took R1 x 0.11m as positive due to anticlockwise and
R2 x 0.11m as negative due to clockwise moment.
is this correct?

Thanks for any help you can offer!
 

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R3=70Kn appears correct.

As for the moment...

R1*110 + (-R2*110) = 0

So

R1 = R2

Both same sign

Horizontally, if left is positive...

-R3*3/5 + R1 + R2 = 0

-R3*3/5 + R1 + R1 = 0

-R3*3/5 + 2R1 = 0

R1 = R3*3/(5*2)

= 21kN

which is a +ve number so to the left.

R1 = R2

so R2 also to the left.
 
Thank you!
 
To check whether you are correct, just take moments about any other point. The sum should be zero.
 

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