MHB How is Torque Calculated in Equilibrium Problems?

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The discussion focuses on how torque is calculated in equilibrium problems, specifically using the equation 600(200) - R_{AH}(600) = 0. The key point is that the sum of torques must equal zero for equilibrium, with the torque from the 600 N force at a distance of 200 mm from point A being positive. The horizontal force component at point C, denoted as H_C, generates a negative torque due to its counterclockwise effect. There is confusion regarding the notation used for the horizontal force at A, R_{AH}, which some believe may have been incorrectly labeled. Ultimately, the calculation of torque was clarified to be taken with respect to point C rather than A.
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I've seen someone's solution in which they did the following:


How did they arrive at that conclusion by taking the moment about $A$?
 

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Rido12 said:
https://www.physicsforums.com/attachments/3352
I've seen someone's solution in which they did the following:
$$600(200)-R_{AH} (600) = 0$$

How did they arrive at that conclusion by taking the moment about $A$?

Hey Rido! (Wave)

The third equilibrium condition is that the sum of the torques must be zero.

The torque of the vertical 600 N force with respect to A is

Apparently clockwise has been chosen to be positive, so this torque is positive.

In A and C there will be vertical force components and horizontal force components.
The vertical force components have a lever-arm of zero with respect to A, so the corresponding torques are zero.
The horizontal force component at A also has a lever-arm of zero.
So that leaves only the horizontal force component at C.

Let's use the symbol $H_C$ for the horizontal force component at C, which is pointing to the right.
Then the corresponding torque is:

This is negative, since it would make the structure turn counter clockwise.

So the total sum of torques is (1) and (2) together:


Btw, I don't know why the symbol $R_{AH}$ was chosen for $H_C$.
 
Thanks ILS! (Wave)

http://skule.ca/courses/exams/custom/20099/CIV102_2009__1329360871.pdf

The reason why I posted it was because the above link confused me. (See #2). I guess they used the wrong letter? (Presumably my 80yo+ prof)
 
Rido12 said:
Thanks ILS! (Wave)

http://skule.ca/courses/exams/custom/20099/CIV102_2009__1329360871.pdf

The reason why I posted it was because the above link confused me. (See #2). I guess they used the wrong letter? (Presumably my 80yo+ prof)

From your notes I can see that $R_{AH}$ is the force at $A$ that is $H$orizontal.
I'm guessing that he picked the letter $R$ for $R$eactive force.

Either way, it means the torque was calculated with respect to $C$ and not with respect to $A$. :eek:
Moreover, the direction of $R_{AH}$ has been picked to point to the left.
 
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