A force F is exerted on the top right corner of a rectangular plate at

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Homework Help Overview

The problem involves calculating the torque exerted by a force on a rectangular plate, specifically focusing on the force's components and their respective distances from a pivot point. The subject area is mechanics, particularly the concepts of torque and force decomposition.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the decomposition of the force into its components and how to apply these to calculate torque. Questions arise regarding the placement of angles and the correct interpretation of the torque equation. Some participants express confusion about the signs and directions of the components.

Discussion Status

The discussion is ongoing, with participants providing hints and drawing diagrams to clarify their understanding. There is a recognition of differing interpretations regarding the torque calculation, particularly concerning the signs of the components. Some participants indicate that they are beginning to understand the concepts, while others remain uncertain.

Contextual Notes

Participants note potential confusion stemming from the angle's placement and the relationship between the components of the force and their distances from the pivot. There is also mention of a drawing that may not have effectively communicated the necessary information.

cmkc109
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Homework Statement



A force F is exerted on the top right corner of a rectangular plate at an angle 60o
above the horizontal, as shown below. The magnitude of the torque of the force F
about point A, the lower left corner is given by

See picture please : http://tinypic.com/view.php?pic=2njv9xt&s=6


Homework Equations



Both y and x are being pulled upwards, so that is counterclockwise..
but answer is Fx sin 60 + Fy cos 60
I don't understand torque at all.. can someone please help.


The Attempt at a Solution



No idea..
 
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Hint:
 

Attachments

  • Torque.png
    Torque.png
    2.7 KB · Views: 1,398
Thank you!
but I'm still kinda lost...
 
Last edited:
one question, where do i put the angle in?
 
I'll do a new drawing as I don't think my first was very helpful. Back soon.
 
ok thanks, check ur msg too please , thanks
 
Try this drawing. The idea is to split the force F into two component forces. Then multiply each component force by it's distance from the pivot to give two torques (Torque = force * distance). Finally you simply add the two torques.
 

Attachments

  • Torque.png
    Torque.png
    26.7 KB · Views: 1,309
this is the drawing i had, but from this picture, it shows that it is Fy sin 60 and Fx cos 60 , but the answer is opposite.
 
Seems ok to me.

The two components are

F*Cos(60)
F*Sin(60)

Multiply by their distance from the pivot..

y*F*Cos(60)
x*F*Sin(60)

then add.

x*F*Sin(60) + y*F*Cos(60)

Thats' the answer you gave in your first post.
 
  • #10
it still seems to be Fx cos 60 + Fy sin 60 to me :/
 
  • #11
cmkc109 said:
it still seems to be Fx cos 60 + Fy sin 60 to me :/
You need to multiply the x by the component of F that's perpendicular to it, so that's the Y component, F sin(60o).
However, none of the offered answers is correct. The X and Y components of F act in opposite sense around A, so there should be a minus sign somewhere. E.g. consider the case where y = x tan(60o). There should be no moment.
 
  • #12
The X and Y components of F act in opposite sense around A, so there should be a minus sign somewhere.

Oops I missed that.

So answer should be either...

x*F*Sin(60) - y*F*Cos(60)

or

y*F*Cos(60) - x*F*Sin(60)
 

Attachments

  • Torque.jpg
    Torque.jpg
    12.5 KB · Views: 926
Last edited:
  • #13
Thank you! makes sense now
 

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