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

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SUMMARY

The discussion focuses on calculating the torque exerted by a force F on a rectangular plate at an angle of 60 degrees above the horizontal, specifically about point A, the lower left corner. The correct expression for torque is derived by splitting the force into its x and y components, resulting in the formula: torque = x*F*sin(60) - y*F*cos(60). Participants clarify that the components of force act in opposite directions around point A, necessitating the inclusion of a minus sign in the final torque equation. This understanding resolves the confusion regarding the correct application of torque principles.

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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
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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
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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

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Last edited:
  • #13
Thank you! makes sense now
 

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