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

Click For Summary

Homework Help Overview

The discussion revolves around calculating the torque produced by a force on a cylinder, specifically focusing on the angle between the position vector and the force vector.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the stated angle between the position vector (r) and the force vector (F), with some suggesting it should be 120 degrees instead of 30 degrees. There are discussions about the relationship between the sine and cosine of these angles in the context of torque calculation.

Discussion Status

The discussion is active, with participants providing insights on vector relationships and clarifying misunderstandings about the angles involved. There is no explicit consensus on the angle, but multiple interpretations are being explored.

Contextual Notes

There seems to be confusion regarding the correct angle to use in the torque calculation, as well as the implications of using sine and cosine functions in the context of the problem.

MatinSAR
Messages
673
Reaction score
204
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?
 
Physics news on Phys.org
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
 
  • Like
Likes   Reactions: MatinSAR and hutchphd
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}).##
 
Last edited:
  • Like
Likes   Reactions: MatinSAR
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}).##
 
  • Like
Likes   Reactions: MatinSAR and kuruman
Steve4Physics said:
You probably meant to write ##rF\sin(120^{\circ})=rF\cos(30^{\circ}).##
Yes, of course. Good catch.
 
  • Like
Likes   Reactions: MatinSAR
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.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
998
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
788
Replies
22
Views
2K
  • · Replies 14 ·
Replies
14
Views
1K
  • · Replies 20 ·
Replies
20
Views
3K