Does distance affect torque in hanging sign problem?

[paulimerci]

Homework Statement

A sign is suspended from two chains L and R. The chains are attached to a bar B which is supposed by a cable C and held at one end by a hinge H as shown. The mass of the sign is 2.6kg. The two chains have negligible mass, are 0.30m apart and are equidistant from the center of the sign. The bar has a uniform density, has a mass of 3.4kg and is 1.6m long. The cable is attached to the bar 1.2m from the hinge. The left chain L is positioned 1.0m from the hinge.
A) calculate the magnitude of the tension in just the left chain L.
B) determine the tension in the cable C.

The chains that support the sign are attached in such a way that they can slid towards or away from the wall, while still maintaining equal distance from each other and remain vertical.
C) if the sign is moved farther away from the wall, does the magnitude of the force on the hinge H increase, decrease, or stay the same? Justify your answer.
D) sketch a graph of the magnitude of the tension in cable C as the sign is moved further away from the wall. The scale of the y axis is not important, and does not need to be fill in. Only the shape and the location of the intercepts will be considered. The closest the left chain can get to the hinge is 0.30m away, and the maximum distance is 1.3m.

Relevant Equations

Torque = F x d

Part -B

$$\sum \tau_{cw} = \sum\tau_{ccw}$$

$$\tau_B=\ torque\ of\ the\ beam $$
$$\tau_S =\ torque\ of\ the\ sign\ board$$
$$\tau_C = \ torque\ of\ the \ cable$$
$$\tau_B+\tau_S = \tau_C$$
$$F_B\cdot d_1 + F_S\cdot d_2 = F_C \cdot d_3$$
Since the tension in the left and right chains are evenly distributed I took $F_S$ as 13N, 13N and their corresponding distances.
$$ 34\cdot 0.8+13+13\cdot 1.3 = T_C\cdot 1.2$$
$$ T_C = 47.58 ~\rm{N}$$

Now, Part -A of the question,

Where, $T_L$ = Tension in the left chain, $T_R$ = Tension in the right chain
Since the tensions of the left and right chains are evenly distributed,
I take $T_L$ = $T_R$ = T
$$ T_L +T_R = \Weight \of \the \sign \board $$
$$ 2T = 26 ~\rm{N}$$
$$ T =13 ~\rm{N}$$

Part -C
As the sign is moved farther away from the wall, the magnitude of the force on the hinge will increase because torque produced depends on the magnitude of the force and the perpendicular distance between the point about which torque is calculated and the point of application of force. And so as the distance increases magnitude of the force on the hinge will increase. So, mathematically torque is represented as:
$$ \tau = Fr\sin\theta$$

 

[paulimerci]

I don’t know whether I’ve done them all right or I need to do any corrections.

 

[kuruman]

If you wish us to check your answers, I can only speak for myself that I would prefer not to do that. That's because I would have to solve the problem from the start and put in numbers to compare with your numbers. It would be easier to check your work if you provided your answers in symbolic form and the given numerical constants, e.g. $d_1$, $d_2$ etc. Also, that would be the way to answer parts (c) and especially (d). Once you have the symbolic expressions, you can put in the numbers. I will be happy to compare my symbolic expressions with yours.

 

[BvU]

In addition, you have a long story and a lot of symbols which are not all explained. In part C, what is $\theta$ ? What is $r$ ?

 

[Lnewqban]

...
Part -C
As the sign is moved farther away from the wall, the magnitude of the force on the hinge will increase because torque produced depends on the magnitude of the force and the perpendicular distance between the point about which torque is calculated and the point of application of force. And so as the distance increases magnitude of the force on the hinge will increase.

I would reconsider that statement.
Consider the point of connection of the cable to the bar as a pivot and do a balance of moments about that point.
You can then find the direction and magnitude of the reaction on pivot H for each of both conditions.

Also, in your diagram, it seems that you incorrectly label F times g as a force.

 

[haruspex]

I would reconsider that statement.
Consider the point of connection of the cable to the bar as a pivot and do a balance of moments about that point.
You can then find the direction and magnitude of the reaction on pivot H for each of both conditions.

Not quite. Taking moments about there will exclude the horizontal component of the reaction at the hinge. Another equation will be needed.

 

[paulimerci]

If you wish us to check your answers, I can only speak for myself that I would prefer not to do that. That's because I would have to solve the problem from the start and put in numbers to compare with your numbers. It would be easier to check your work if you provided your answers in symbolic form and the given numerical constants, e.g. $d_1$, $d_2$ etc. Also, that would be the way to answer parts (c) and especially (d). Once you have the symbolic expressions, you can put in the numbers. I will be happy to compare my symbolic expressions with yours.

Okay!
Solution for Part -B,
A hinged bar of mass $M$ 3.4kg attached to a wall and is supported by a cable C. A sign is suspended from two chains and is attached to a bar. The mass of the sign $m$ is 2.6kg.

The bar is assumed to be uniform. Therefore, its weight force is applied at the half way point (0.8m). By taking the left-hand side of the rod (H) as our axis of rotation, the torque from force R is zero, leaving us with the following equilibrium statement for torques.

$d_B = 0.8m$ is the moment arm distance from the hinge to the COM of the bar, $d_S=1.15m$ is the moment arm distance from the hinge to the COM of the sign board, and $d_C = 1.2m$ is the moment arm distance from the hinge to the cable. The sum of all torques through contact point H must be zero, both weights produce clockwise torques! solving for $T_C$ in the below equation, we get

$$\sum\tau = Mg\cdot d_B + mg\cdot d_S - T_C\cdot d_C$$
$$ = 34\cdot 0.8 +26\cdot 1.15 - T_C\cdot 1.2$$
$$ T_C = 47.58~\rm{N}$$
I did only for part - B for now, just want to check whether I did it right at-least for this one.

 

[haruspex]

Okay!
Solution for Part -B,
A hinged bar of mass $M$ 3.4kg attached to a wall and is supported by a cable C. A sign is suspended from two chains and is attached to a bar. The mass of the sign $m$ is 2.6kg.

The bar is assumed to be uniform. Therefore, its weight force is applied at the half way point (0.8m). By taking the left-hand side of the rod (H) as our axis of rotation, the torque from force R is zero, leaving us with the following equilibrium statement for torques.

$d_B = 0.8m$ is the moment arm distance from the hinge to the COM of the bar, $d_S=1.15m$ is the moment arm distance from the hinge to the COM of the sign board, and $d_C = 1.2m$ is the moment arm distance from the hinge to the cable. The sum of all torques through contact point H must be zero, both weights produce clockwise torques! solving for $T_C$ in the below equation, we get

$$\sum\tau = Mg\cdot d_B + mg\cdot d_S - T_C\cdot d_C$$
$$ = 34\cdot 0.8 +26\cdot 1.15 - T_C\cdot 1.2$$
$$ T_C = 47.58~\rm{N}$$
I did only for part - B for now, just want to check whether I did it right at-least for this one.

Good so far.

 

[paulimerci]

Good so far.

Any hints on how to solve for Part A?

 

[kuruman]

Any hints on how to solve for Part A?

Draw a free body diagram (FBD) of the sign.

 

[paulimerci]

Okay!
Solution for Part -B,
A hinged bar of mass $M$ 3.4kg attached to a wall and is supported by a cable C. A sign is suspended from two chains and is attached to a bar. The mass of the sign $m$ is 2.6kg.

The bar is assumed to be uniform. Therefore, its weight force is applied at the half way point (0.8m). By taking the left-hand side of the rod (H) as our axis of rotation, the torque from force R is zero, leaving us with the following equilibrium statement for torques.

$d_B = 0.8m$ is the moment arm distance from the hinge to the COM of the bar, $d_S=1.15m$ is the moment arm distance from the hinge to the COM of the sign board, and $d_C = 1.2m$ is the moment arm distance from the hinge to the cable. The sum of all torques through contact point H must be zero, both weights produce clockwise torques! solving for $T_C$ in the below equation, we get

$$\sum\tau = Mg\cdot d_B + mg\cdot d_S - T_C\cdot d_C$$
$$ = 34\cdot 0.8 +26\cdot 1.15 - T_C\cdot 1.2$$
$$ T_C = 47.58~\rm{N}$$
I did only for part - B for now, just want to check whether I did it right at-least for this one.

Draw a free body diagram (FBD) of the sign.

The sign is in a state of equilibrium and thus all the forces acting upon it must be balanced. The leftward pull of cable L must balance the rightward pull of cable R and the sum of the upward pull of cable L and cable R must balance the weight of sign.
Where the upward pull of left cable is taken as $T_L$ and rightward pull of cable is taken as $T_R$.

Therefore,
$$ T_L + T_R = mg $$
Since leftward pull must balance the rightward pull, we can write $T_L = T_R = T$. Therefore, substituting $T_L$and $T_R$ as $T$ in the above equation we get,
$$ T + T = mg$$
$$ 2T = 2.6 \cdot 10$$
$$ T = 13~\rm{N}$$

 

[paulimerci]

The sign is in a state of equilibrium and thus all the forces acting upon it must be balanced. The leftward pull of cable L must balance the rightward pull of cable R and the sum of the upward pull of cable L and cable R must balance the weight of sign.
Where the upward pull of left cable is taken as $T_L$ and rightward pull of cable is taken as $T_R$.

Therefore,
$$ T_L + T_R = mg $$
Since leftward pull must balance the rightward pull, we can write $T_L = T_R = T$. Therefore, substituting $T_L$and $T_R$ as $T$ in the above equation we get,
$$ T + T = mg$$
$$ 2T = 2.6 \cdot 10$$
$$ T = 13~\rm{N}$$

We can conclude each cable pulls upwards with a force of $13~\rm{N}$. I.e., $T_L= 13~\rm{N}$ & $T_R = 13~\rm{N}$.

 

[kuruman]

That is correct. You have solved part (a). Proceed to part (b) with a FBD of the bar. Please use symbols T_L and T_R for the tensions even though you now have numbers for them.

 

[haruspex]

The leftward pull of cable L must balance the rightward pull of cable R

Please explain exactly what you mean by that. It is not at all clear that you have arrived at your result validly. What physical law are you invoking?

 

[paulimerci]

Please explain exactly what you mean by that. It is not at all clear that you have arrived at your result validly. What physical law are you invoking?

Since the sign is in a state of equilibrium, the tension in the left cable is same as the tension in the right cable.

 

[paulimerci]

That is correct. You have solved part (a). Proceed to part (b) with a FBD of the bar. Please use symbols T_L and T_R for the tensions even though you now have numbers for them.

The system is in static equilibrium and so the net forces are equal to zero. The reaction force $R$ is the response of the wall to the bar and acts at an angle $\phi$. Thus, we state that
$$\sum F_x = R_x =0$$
$$\sum F_y = R_y + T_c +T_L+T_R - Mg$$
$$R_y = Mg -T_c-T_L-T_R$$
I'm getting reaction force along the vertical component as negative.

 

[kuruman]

I'm getting reaction force along the vertical component as negative.

If a chain exerts a force on the sign directed up, what is the direction of the force that the chain exerts on the bar?

 

[paulimerci]

If a chain exerts a force on the sign directed up, what is the direction of the force that the chain exerts on the bar?

Oh yes, I never thought that. The direction of the force that the chain exerts on the bar is down.
$$\sum F_y = R_y + T_c- Mg - T_L -T_R$$
$$ R_y = Mg + T_L + T_R-T_c$$
$$ R_y = 36 + 13 +13 -47.58$$
$$R_y = 14.42~\rm{N}$$
And so the $R$ force from the wall is entirely vertical.

 

[kuruman]

Good. Onto the next task. You can answer (c) and (d) if you change the distance of the left chain from 1.0 m to $x$ and find an equation for the reaction force in terms of $x$.

 

[haruspex]

Since the sign is in a state of equilibrium, the tension in the left cable is same as the tension in the right cable.

That is still missing a crucial part of the argument. How does being in equilibrium imply the tensions are equal? There is a specific piece of information in the question statement that you need to use.

 

[paulimerci]

That is still missing a crucial part of the argument. How does being in equilibrium imply the tensions are equal? There is a specific piece of information in the question statement that you need to use.

The two chains have negligible mass, and are equidistant from the center of the sign.

 

[paulimerci]

Good. Onto the next task. You can answer (c) and (d) if you change the distance of the left chain from 1.0 m to $x$ and find an equation for the reaction force in terms of $x$.

Thanks, how can I find reaction force for $x$ when the horizontal component is zero?

 

[haruspex]

The two chains have negligible mass, and are equidistant from the center of the sign.

Right! And how does the equality of the distances imply the tensions are the same?

 

[kuruman]

Thanks, how can I find reaction force for $x$ when the horizontal component is zero?

The same way you found it before except that the left chain is at some distance $x$ from the hinge and the right change is at distance ($x+0.3$ m) from the hinge. The idea is to get an expression for $R_y$ as you did before and see whether it increases or decreases as $x$ increases for part (c) and then plot $R_y$ vs. $x$ for part (d).

However, before you do all that, it is important to answer the question that @haruspex has been trying to get you to answer, "How does the equality of the distances from the edge of the chain imply that th tensions are the same?" To help you, here is another way to look at this is by answering the following question.

The sign is 0.4 m wide and has weight 26 N. The left chain is 0.1 m from the left edge and the right chain is right on the edge at 0.4 m from the left chain. Are the tensions $T_L$ and $T_R$ equal? Why? If they are not equal, find the teo tensions.

Related to all this is the equation you posted in #7 $$\begin{align}\sum\tau & = Mg\cdot d_B + mg\cdot d_S - T_C\cdot d_C \nonumber \\ & = 34\cdot 0.8 +26\cdot 1.15 - T_C\cdot 1.2 \nonumber \\T_C & = 47.58~\rm{N} \nonumber \end{align}$$I can see that in the second line 26 is the weight of the sign, but where did the $1.15$ come from? One chain is at 1.00 m and the other at 1.30 m from the hinge.

 

[paulimerci]

The same way you found it before except that the left chain is at some distance $x$ from the hinge and the right change is at distance ($x+0.3$ m) from the hinge. The idea is to get an expression for $R_y$ as you did before and see whether it increases or decreases as $x$ increases for part (c) and then plot $R_y$ vs. $x$ for part (d).

However, before you do all that, it is important to answer the question that @haruspex has been trying to get you to answer, "How does the equality of the distances from the edge of the chain imply that th tensions are the same?" To help you, here is another way to look at this is by answering the following question.

The sign is 0.4 m wide and has weight 26 N. The left chain is 0.1 m from the left edge and the right chain is right on the edge at 0.4 m from the left chain. Are the tensions $T_L$ and $T_R$ equal? Why? If they are not equal, find the teo tensions.

Related to all this is the equation you posted in #7 $$\begin{align}\sum\tau & = Mg\cdot d_B + mg\cdot d_S - T_C\cdot d_C \nonumber \\ & = 34\cdot 0.8 +26\cdot 1.15 - T_C\cdot 1.2 \nonumber \\T_C & = 47.58~\rm{N} \nonumber \end{align}$$I can see that in the second line 26 is the weight of the sign, but where did the $1.15$ come from? One chain is at 1.00 m and the other at 1.30 m from the hinge.

That's right but for the $d_s$ I took $1.15m$ which is the moment arm distance from the hinge to the COM of the sign and not separating the chains.

 

[paulimerci]

Right! And how does the equality of the distances imply the tensions are the same?

I don't know, may be the tensions are at equal distances from the center of sign.

 

[haruspex]

I don't know, may be the tensions are at equal distances from the center of sign.

There are two valid answers to my question.
Symmetry argument: since the system of sign and support chains looks exactly the same viewed from the other side, there is no reason for either tension to be greater than the other.
Balance of Moments: let the tensions be T_L, T_R and their horizontal displacements from the sign's mass centre be x_L, x_R. Since the system is static, T_Lx_L+T_Rx_R=0.
We are told x_L=- x_R, so T_L=T_R.

 

[paulimerci]

There are two valid answers to my question.
Symmetry argument: since the system of sign and support chains looks exactly the same viewed from the other side, there is no reason for either tension to be greater than the other.
Balance of Moments: let the tensions be T_L, T_R and their horizontal displacements from the sign's mass centre be x_L, x_R. Since the system is static, T_Lx_L+T_Rx_R=0.
We are told x_L=- x_R, so T_L=T_R.

Text book doesn’t teach these concepts. Thanks for clarifying them. What does these arguments teach us? I'm unable to grasp them.

 

[paulimerci]

That's right but for the $d_s$ I took $1.15m$ which is the moment arm distance from the hinge to the COM of the sign and not separating the chains.

I don’t understand how to find for $R_y$ in terms of x. Can you give some hints?

 

[haruspex]

Text book doesn’t teach these concepts. Thanks for clarifying them. What does these arguments teach us? I'm unable to grasp them.

Does the textbook not teach the moments equivalent of ΣF=ma, namely, Στ=Iα (the sum of the moments about an axis equals the moment of inertia about that axis multiplied by the angular acceleration)?
In the special case of rotational stasis, the angular acceleration is zero, so it reduces to Στ=0, the net moment about the axis is zero.

 

[paulimerci]

Does the textbook not teach the moments equivalent of ΣF=ma, namely, Στ=Iα (the sum of the moments about an axis equals the moment of inertia about that axis multiplied by the angular acceleration)?
In the special case of rotational stasis, the angular acceleration is zero, so it reduces to Στ=0, the net moment about the axis is zero.

Yes, they do teach on balance of moments. Why there is a need to understand about symmetry? Can you explain it in detail. Just curious to learn!

 

[paulimerci]

The same way you found it before except that the left chain is at some distance $x$ from the hinge and the right change is at distance ($x+0.3$ m) from the hinge. The idea is to get an expression for $R_y$ as you did before and see whether it increases or decreases as $x$ increases for part (c) and then plot $R_y$ vs. $x$ for part (d).

However, before you do all that, it is important to answer the question that @haruspex has been trying to get you to answer, "How does the equality of the distances from the edge of the chain imply that th tensions are the same?" To help you, here is another way to look at this is by answering the following question.

The sign is 0.4 m wide and has weight 26 N. The left chain is 0.1 m from the left edge and the right chain is right on the edge at 0.4 m from the left chain. Are the tensions $T_L$ and $T_R$ equal? Why? If they are not equal, find the teo tensions.

Related to all this is the equation you posted in #7 $$\begin{align}\sum\tau & = Mg\cdot d_B + mg\cdot d_S - T_C\cdot d_C \nonumber \\ & = 34\cdot 0.8 +26\cdot 1.15 - T_C\cdot 1.2 \nonumber \\T_C & = 47.58~\rm{N} \nonumber \end{align}$$I can see that in the second line 26 is the weight of the sign, but where did the $1.15$ come from? One chain is at 1.00 m and the other at 1.30 m from the hinge.

Do I need to find the net torque with this change in distances of left and right chain?

 

[kuruman]

Do I need to find the net torque with this change in distances of left and right chain?

Yes, and my hope is that you will get the same answer and see what is going on and why. If not, we can help you see it, but you have to do the work first.

 

[paulimerci]

Yes, and my hope is that you will get the same answer and see what is going on and why. If not, we can help you see it, but you have to do the work first.

Okay, should I take the left chain as $x$ and for right chain $x+3.0$?

 

[kuruman]

Yes. Find an expression for $R_y$ containing $x$. Then substitute $x=1.00$ m and see if you get the answer that you already got in part (b).

Correction to your work: The right chain should be $x+0.3~$m.

 

[paulimerci]

Yes. Find an expression for $R_y$ containing $x$. Then substitute $x=1.00$ m and see if you get the answer that you already got in part (b).

Correction to your work: The right chain should be $x+0.3~$m.

Okay, if there are changes in distances does the values of $T_c$, $T_L$, and $T_R$ remains the same? If there is no change, then in the torque equation can I substitute for the values i calculated earlier for $T_c$, $T_L$, $T_R$?

 

[paulimerci]

Okay, if there are changes in distances does the values of $T_c$, $T_L$, and $T_R$ remains the same? If there is no change, then in the torque equation can I substitute for the values i calculated earlier for $T_c$, $T_L$, $T_R$?

$$\sum\tau= T_c \cdot d_c-Mg \cdot d_B-T_L \cdot x-T_R \cdot (x+0.30)$$
$$1.2 \cdot T_c = 27.2+ 13 \cdot x + 13 \cdot(x+0.3)$$
$$T_c = \frac {31.1 +26 \cdot x} {1.2}$$
Substitute $T_c$ in the equation of $R_y$ we get,
$$ R_y = Mg +T_L+T_R-T_c$$
$$ R_y = 34+13+13- \frac{31.1+26\cdot x} {1.2}$$
$$ R_y = 60 - \frac{31.1+26\cdot x} {1.2}$$
have I done it right?

 

[kuruman]

Very good. Now you can answer parts (c) and (d) based on the equation you derived for $R_y.$

 

[paulimerci]

Very good. Now you can answer parts (c) and (d) based on the equation you derived for $R_y.$

Thanks for your guidance.
Part C,
Therefore, from the equation of $R_y$ we can justify that as the distance of the sign is moved farther away from the wall, the magnitude of the force on the hinge decreases.

 

[kuruman]

Thanks for your guidance.
Part C,
Therefore, from the equation of $R_y$ we can justify that as the distance of the sign is moved farther away from the wall, the magnitude of the force on the hinge decreases.

Correct. Now for the plot. Use a spreadsheet if you know how.

 

[paulimerci]

Correct. Now for the plot. Use a spreadsheet if you know how.

 

[kuruman]

I think you're done. Congratulations! I hope you learned something from this experience.

 

[paulimerci]

View attachment 317979

Looks like I plotted for $T_c$ vs $x$. S

I think you're done. Congratulations! I hope you learned something from this experience.

Definitely! It was a humongous problem and I didn't believe I can do it and I thank you so much for guiding me throughout! Thanks to @haruspex for his guidance too!