How is Torque Calculated in Equilibrium Problems?

[Dethrone]

View attachment 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 $A$?

 

[I like Serena]

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
$$+600\text{ N} \cdot \frac{400\text{ mm}}{2} = +600\text{ N} \cdot 200\text{ mm}\tag 1$$
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:
$$-H_C \cdot (300 \text{ mm} + 300 \text{ mm}) = -H_C \cdot 600\text{ mm} \tag 2$$
This is negative, since it would make the structure turn counter clockwise.

So the total sum of torques is (1) and (2) together:
$${}_+^\curvearrowright\sum T_A = +600\text{ N} \cdot 200\text{ mm} - H_C \cdot 600\text{ mm}$$

Btw, I don't know why the symbol $R_{AH}$ was chosen for $H_C$.

 

[Dethrone]

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)

 

[I like Serena]

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 $A$.
Moreover, the direction of $R_{AH}$ has been picked to point to the left.

 

[CellCoree]

To understand how the solution was arrived at, we must first understand the concept of torque. Torque is a measure of the rotational force applied to an object, and it is calculated by multiplying the force applied by the distance from the point of rotation.

In this problem, the force applied is 600 and the distance from the point of rotation (which is point A) is 200. Therefore, the torque at point A would be 600(200) = 120,000.

Now, as per the problem, there is another force acting at point A, which is represented by $R_{AH}$. This force is causing a torque in the opposite direction, and we can calculate its value by multiplying it by the distance from point A, which is also 600.

Therefore, the torque caused by $R_{AH}$ would be $R_{AH} (600)$. Now, since the problem states that the system is in equilibrium, the total torque must be equal to zero. This can be written as:

120,000 - $R_{AH} (600) = 0$

Solving for $R_{AH}$, we get:

$R_{AH} = \frac{120,000}{600} = 200$

Hence, the solution provided by taking the moment about point A is correct. It is a simple application of the concept of torque and equilibrium.