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

Question

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?.

Answer 1

One approach: Split the bent bar into two pieces. Let one piece be the horisontal part together with a massless version of the vertical part, and let the other piece be a massless version of the horisontal part together with the original vertical part.

Each of the two pieces needs some green mass to stay in equilibrium at the position shown in the drawing. Add these two masses to find which mass the pieces need to stay in equilibrium when they are together.

Another approach: The center of mass of the entire bent bar is the (weighted) average of the means of the two pieces mentioned above. Then you can use that to calculate the torque caused by gravity from that center of mass.

Look at the drawing below:

If the horizontal part has mass $m_h$ and the vertical part has mass $m_v$, then the coordinates of the center of mass of the whole bent pipe is given by $$ \frac{m_h(a, b) + m_v(c, d)}{m_v + m_h} $$

Answer 2

Hint:

The rod is equilibrated wrt the gravity force when

Now add an additional horizontal $x$ force, plus a vertical force $x/2$ to equilibrate the additional momentum

in such a way as to get the resultant in A with the given angle .

Answer 3

We have

$$ 2m\left(10, 0 \right)+m\left(20, 5 \right)= 3m\left(x_g,y_g \right) $$

now calling

$$ \cases{ G =(x_g, y_g)\\ O = (0,0)\\ P = (20,10)\\ W = (0,-3mg)\\ F = f_0(\cos\phi,\sin\phi) } $$

we have the main equilibrium equation

$$ (G-O)\times W + (P-O)\times F = 0 $$

that can be solved for $f_0$

Answer 4

Every 2D system in equilibrium must satisfy three conditions: $\sum F_x = 0$, $\quad$ $\sum F_y=0$, $\quad$ $\sum M =0$ $\quad$ where $M$ is "momentum" product of force by distance, the torque. We use distances to the suport in the joinment to the wall.

I suppose your system is full-rigid, no deformations. Otherwise another considerations must be taken into account.

The bar can be considered as a unique element, with its center of mass located at $x= \frac {massHor·distHor+massVer·distVer}{massHor+massVer}$. But it's easier if we split the bar into two pieces: horizontal and vertical.

The mass of each "subbar" is proportional to its length, namely $4kg$ and $2kg$.

The actions of the bars over the joinment at wall are:
$\qquad$ $F_{xh} = 0$ $\qquad$ $F_{yh} = -4g$ $\qquad$ $M_{h}=F_{yh}·20/2 = -40g$
$\qquad$ $F_{xv} = 0$ $\qquad$ $F_{yv} = -2g$ $\qquad$ $M_{v}=F_{yv}·20 = -40g$

The action of the wire can be decomposed into X,Y directions:
$\qquad$ $F_{xT} = mg·cos(37º)$ $\qquad$ $F_{yT} = mg·sin(37º)$
Using these forces to calculate the momentum:
$\qquad$ $M_T= -F_{xT}·10 + F_{yT}·20 = -10mg·cos(37º)+20mg·sin(37º)= 10mg(-cos(37º)+2sin(37º))$ $\qquad$ Pay attention to the sign, positive as counter clockwise rotation.

The final action is that of the joinment to the wall. Intutively we can see that it has a X-component, balancing the system, otherwise the system would translate. So we have
$reaction_x = -F_{xh}-F_{xv}-F_{xT}= -mg·cos(37º)$

For the required equilibrium, because the joinment allows rotations (no momentum in reaction), to avoid them we use $\sum M =0$

$M_h + M_v + M_T = -40g - 40g + 10mg(-cos(37º)+2sin(37º)) = 0$

$m= \frac{80/10}{-cos(37º)+2sin(37º)} = 19.75 kg$

The last calculus is the Y-reaction in the support, using $\sum F_y=0$:

$reaction_y= -F_{yh}-F_{yv}-F_{yT}= 4g+2g-mg·sin(37º) = -5.89g = -58.9N$

This reaction points down (negative). This means that the mass in the hoist is used strictely to avoid the rotation of the bar, a the price of pulling up the support.

Answer 5

To determine the mass m. First of all you take moments about the hinge that is after you determine the cg of the l shaped beam.

Lets find the cg of the beam so we split the beam up into two parts the length of the horizontal section is 20m and the length of the vertical section is 10m.

So to determine the cg of the L shaped beam

Taking moments about the hinge we get

20 × 10 + 10 × 20 = (20 + 10)x

=200 + 200 = 30x

=400. = 30x

x. = 400/30

x. = 13.33m from the hinge

Therefore the cg of the beam is 13.33m from the hinge

Now to determine the mass for m we resolve the two components for mass m we get for F horizontal mg cos and Fvertical mg sin

Taking moments about the hinge we get

60 × 13.33 + mg cos 37 = mg sin 37 × 20

799.8 + mg × 0.799 × 10 = mg × 0.602 × 20

799.8 + 7.99 × mg = mg × 12.04

799.8. = mg × 12.04 - 7.99 × mg

799.8 = 4.05 × mg

mg 799.8/ 4.05

mg. 197.5.N. 19.75 kg

Now to determine the reaction on the hinge

Taking moments about the end of the beam at B we get

60(20-13.33) +Rv × 20 = 197.5 × 10 cos 37

60 × 6.67. + Rv × 20 = 19.75 × 10 × 0.799

400.2.+ Rv × 20 = 197.5. × 7.99

400.2 + Rv × 20. = 1578

Rv × 20. = 1578 - 400.2

Rv. = 1177.8/20

Rv. = 58.89N

Using pythagoras to determine the resultant reaction in the hinge

Horizontal reaction Rh. = - 157.8N

Therefore R = (Rh^2 + Rv^2)^1/2

               Rr =  (157.8^2 + 58.89^2)^1/2

               Rr = 168.4 N