Torque on a sign hung by a hinge

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The discussion centers on the misunderstanding of torque in a physics problem involving a sign suspended by a hinge. The user incorrectly applied the torque equation t = F*d*cosx, leading to an erroneous angle calculation of 68 degrees. The key issue is recognizing that the weight of the sign also exerts a torque, which must be considered alongside the hinge force. The correct approach involves understanding that the torque from the weight acts at a distance from the hinge, influencing the sign's equilibrium position. Ultimately, the user seeks clarification on the concepts of torque and the forces at play in this scenario.
Physical_Fire
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Homework Statement
A uniform square sign of weight 40 N is suspended vertically from its top edge by a horizontal hinge, as shown. The hinge is not frictionless. When the sign is displaced from the vertical by an external force and then released, it does not return to the vertical position. The maximum torque exerted by the hinge on the sign is 6.0 N m. The sign is displaced by 90 degrees so that it is horizontal and then gradually released. At which angle to the vertical does the sign hang after release? Write out your thought process and understanding as well. (5)
Relevant Equations
T= F*D
Since I understood that torque = net moment, and as it said angle from the vertical, I used the equation t = F*d*cosx. So, 6 = 40*0.4*cosx, x = 68 degrees. I got a zero out of 5 in this question because the thought process was wrong as well as the answer. I do not understand this problem beyond the equations involved. Why is the answer wrong and what is the real concept and understanding behind these problems?

Thanks
 

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Physical_Fire said:
Homework Statement: A uniform square sign of weight 40 N is suspended vertically from its top edge by a horizontal hinge, as shown. The hinge is not frictionless. When the sign is displaced from the vertical by an external force and then released, it does not return to the vertical position. The maximum torque exerted by the hinge on the sign is 6.0 N m. The sign is displaced by 90 degrees so that it is horizontal and then gradually released. At which angle to the vertical does the sign hang after release? Write out your thought process and understanding as well. (5)
Relevant Equations: T= F*D

Since I understood that torque = net moment, and as it said angle from the vertical, I used the equation t = F*d*cosx. So, 6 = 40*0.4*cosx, x = 68 degrees. I got a zero out of 5 in this question because the thought process was wrong as well as the answer. I do not understand this problem beyond the equations involved. Why is the answer wrong and what is the real concept and understanding behind these problems?

Thanks
What was your thought process? Specifically, what does the equation t = F*d*cosx represent? If you set this equal to the maximum torque exerted by the hinge, is the force exerted by the hinge on the sign applied at distance d from the hinge? I think not.

There is an additional force that exerts a torque on the sign. What is its origin?
 
kuruman said:
There is an additional force that exerts a torque on the sign. What is its origin?
Are you referring to weight?
 
Physical_Fire said:
Are you referring to weight?
Yes.
 
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