How to find the torque and mass of an object hanging from a bended bar shaped as an L?

Jun 2017
399
6
Lima, Peru
The problem is as follows:

The figure from below shows a homogeneous bar with the shape of an $L$ which mass is $6\,kg$ and its length is $30\,cm$. Find the mass of the block which is hanging from a wire and pulley tied to the end of the bar such that it is in equilibrium in the position shown. You may use the gravity $g=10\frac{m}{s^2}$.



The alternatives given are as follows:

$\begin{array}{ll}
1.&4\,kg\\
2.&10\,kg\\
3.&15\,kg\\
4.&20\,kg\\
5.&24\,kg\\
\end{array}$

I'm not sure how to make the right interpretation of the torque in this problem. How should I make the vector decomposition?. The figure from below shows how I attempted to use those vectors.



However I don't know where should I put the center of mass in this weird object. Is it in the middle?. Is it at $15\,cm$ going from the wall where the joint is put?.

From the drawing I could spot that the torque for the system would be as follows:

I'm assuming that the force on $x-axis$ will not generate torque.

$-60(15)+10m(\sin 37^{\circ})(20)=0$

$120m=900$

$m=7.5\,kg$

Although I arrived to an answer it does not check with any of the alternatives. Can someone help me to find where exactly did I made the mistake?. Can someone help me with a solution using trigonometry approach and vector decomposition?. I would like that an answer could include a method also to calculate or find the center of mass in such a figure. Will this be relevant for the solution of this problem?.
 
Jun 2017
399
6
Lima, Peru
I made a mistake in the drawing, the order of the lengths is reversed. The bigger portion of the L is $20\,cm$ and the smaller one is $10\,cm$ as it is shown in the second sketch.
 
Jun 2019
493
262
USA
For the CoM, you can use the definition I gave you before, $\displaystyle \vec{r}_CM = \int \vec{r} dm$.
...Or you could make it really easy, and treat the L bar as two straight bars, each with its own weight and centre of mass. 4 kg in the middle of the long portion, plus 2 kg in the middle of the short section.
 
Jun 2019
493
262
USA
Now that I've stopped to look at your work so far, I have a couple more notes.

1) Moments/torques are only defined about a certain point. You need to make it clear where your origin is. You can define moments about the pin joint, or moments about the end of the bar, or even moments about the bend in the beam, and solve the problem any of those ways.

2) The neat (and useful) thing about vector decomposition of forces, is that you can treat it as a single force, or as two independent forces -- whichever best suits your needs. Let me show you what I mean.

I will define right in your diagram as $+x$ ($\hat{i}$ unit vector), up as $+y$ ($\hat{j}$), and towards us as $+z$ ($\hat{k}$).
We can then write the rope tension as $\vec{T} = T_x \hat{i} + T_y \hat{j}$.
$T_x$ produces a moment about the pin joint, $\vec{M}_{T_x} = \vec{r} \times \vec{F} = (10~cm~\hat{j})* \times (T_x ~\hat{i}) = -(10~cm)T_x ~\hat{k}$.
$T_y$ produces a moment about the same location, $\vec{M}_{T_y} = \vec{r} \times \vec{F} = (20~cm~\hat{i})* \times (T_y ~\hat{j}) = +(20~cm)T_y ~\hat{k}$.

If you add these moments together, you get the same answer as if you did $\vec{M}_T = \vec{r} \times \vec{F} = (20~cm~\hat{i}+10~cm~\hat{j}) \times \vec{T}$.


*Note, because of the cross product, we can ignore any part of $\vec{r}$ that is parallel to the force in question.


P.S. My previous post was supposed to say $\vec{r}_{CM}$, as in the location of the centre of mass.