Does distance affect torque in hanging sign problem?

[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!