A doubt respect to the sign of the torsor (torque)

In summary: It seems that the book's solution shows F2=10N in the sumatory of moments in C, while in the sumatory of moments in F, F2 is calculated to be 416.7 N. This discrepancy is causing confusion. Additionally, there is a question about the sign of the torsor along the axis CD, as the solution takes it to be positive while the action-reaction law suggests it should be negative.
  • #1
Queren Suriano
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0

Homework Statement



upload_2015-5-1_22-50-1.png


Actually I have two doubs, the first one is Why in the solution of this problem, in the sumatory of moments in C, Force 2=10?? If in the sumatory of moments in F, Force 2 is 416.7 N. Actually I calculate with 416.7 N, and T in C = 100 N. The other doubt is with the sign of the torsor along the axis CD. Because the action-reaction law, the torsor in the axis will enter in the axis, thus it will NEGATIVE, but the solution takes positive!. Why??
upload_2015-5-1_22-55-32.png


Homework Equations



upload_2015-5-1_22-56-11.png
upload_2015-5-1_22-56-19.png

The Attempt at a Solution


It is in the The problem statement, at the picture
 

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  • #3
phinds said:
I can't answer your questions but I would point out to you that as far as I can tell, you don't have any doubts at all, what you have is questions.

https://www.physicsforums.com/showthread.php?t=607274

Ok, thank you for correcting my grammar. It's true, I have 2 questions
 
  • #4
I understood what you meant by "doubt"... :smile:

You are saying that the book's solution shows F2 = 10N? That seems way off...

And as for the direction of the net torque on the center gear, I think you have labeled the direction incorrectly. The force F2 on the bottom of the gear is larger, so the net torque will be towards the back of the apparatus (clockwise as seen from the front)...
 
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  • #5
berkeman said:
I understood what you meant by "doubt"... :smile:

You are saying that the book's solution shows F2 = 10N? That seems way off...

And as for the direction of the net torque on the center gear, I think you have labeled the direction incorrectly. The force F2 on the bottom of the gear is larger, so the net torque will be towards the back of the apparatus (clockwise as seen from the front)...
Thank you
 

Related to A doubt respect to the sign of the torsor (torque)

1. What is a torsor?

A torsor is a mathematical concept that represents a physical quantity with magnitude and direction, but without a fixed point of application. It can be thought of as an "unattached" vector or force.

2. What is the sign of a torsor?

The sign of a torsor refers to the direction in which it acts. It can be either positive or negative, depending on the direction of the torsor relative to a chosen reference direction.

3. How is the sign of a torsor determined?

The sign of a torsor is determined by the right-hand rule. This means that if you curl your fingers in the direction of the torsor, your thumb will point in the positive direction. If your thumb points in the opposite direction, the torsor has a negative sign.

4. What is the significance of the sign of a torsor in physics?

In physics, the sign of a torsor is important because it indicates the direction of the force or torque being applied. This information is necessary for accurately describing and predicting the motion or behavior of an object.

5. Can the sign of a torsor change?

Yes, the sign of a torsor can change depending on the reference direction chosen. For example, if you switch the direction of your reference axis, the sign of the torsor will change accordingly. However, the magnitude of the torsor will remain the same.

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