Torque problem - Hanging sign

In summary, the hinged bar of mass 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 2.6kg. The bar is assumed to be uniform. Therefore, its weight force is applied at the half way point (0.8m). The equilibrium statement for torques is: ##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
  • #36
kuruman said:
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##?
 
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  • #37
paulimerci said:
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?
 
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  • #38
Very good. Now you can answer parts (c) and (d) based on the equation you derived for ##R_y.##
 
  • #39
kuruman said:
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.
 
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  • #40
paulimerci said:
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.
 
  • #41
kuruman said:
Correct. Now for the plot. Use a spreadsheet if you know how.
Screen Shot 2022-11-30 at 2.35.09 PM.png
 
  • #42
I think you're done. Congratulations! I hope you learned something from this experience.
 
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  • #43
paulimerci said:
Looks like I plotted for ##T_c## vs ##x##. S
kuruman said:
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!
 
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