Calculate the torque that is produced by this force on a cylinder

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SUMMARY

The discussion centers on the calculation of torque produced by a force on a cylinder, specifically addressing the angle between the position vector (r) and the force vector (F). Participants clarify that the correct angle is 120 degrees, not 30 degrees, when determining torque using the formula τ = rFsin(θ). The confusion arises from the misinterpretation of vector orientation, emphasizing the importance of placing vectors tail-to-tail to accurately assess angles.

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  • Understanding of torque calculations in physics
  • Familiarity with vector representation and operations
  • Knowledge of trigonometric functions, specifically sine and cosine
  • Basic principles of rotational dynamics
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  • Study the derivation of the torque formula τ = rFsin(θ)
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  • Investigate common misconceptions in physics regarding angles and forces
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Students of physics, educators teaching mechanics, and anyone involved in engineering applications requiring torque calculations.

MatinSAR
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Homework Statement
Please take a look at the picture :
Relevant Equations
torque = r F sin(r, F)
IMG_20230903_232642_013.jpg


Why it said that angle between r and F is 30?
I guess it should be 120 degrees... Am I wrong?
 
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As far as I can see, the length of the arm of the force respect to the center of rotation should be ##rcos30##.

63ff8d0b4fb6b3f197f8872f_moment1_perpendicular_distance.svg
 
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MatinSAR said:
Why it said that angle between r and F is 30?
It didn't say that. The angle between two vectors is obtained by putting them tail-to-tail. If you do that with r and F, you will get an angle of 120°. Then note that ##rF\sin(120^{\circ})=rF\cancel{\sin}\cos(30^{\circ}).##
 
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kuruman said:
It didn't say that. The angle between two vectors is obtained by putting them tail-to-tail. If you do that with r and F, you will get an angle of 120°. Then note that ##rF\sin(120^{\circ})=rF\sin(30^{\circ}).##
You probably meant to write ##rF\sin(120^{\circ})=rF\cos(30^{\circ}).##
 
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Steve4Physics said:
You probably meant to write ##rF\sin(120^{\circ})=rF\cos(30^{\circ}).##
Yes, of course. Good catch.
 
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kuruman said:
It didn't say that. The angle between two vectors is obtained by putting them tail-to-tail. If you do that with r and F, you will get an angle of 120°. Then note that ##rF\sin(120^{\circ})=rF\cancel{\sin}\cos(30^{\circ}).##
So the book is wrong since sin30 isn't equal to sin120 degrees... @Lnewqban
@kuruman
@Steve4Physics
Thanks for your help and time.
 

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