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

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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:
 
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Thank you!
but I'm still kinda lost...
 
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I'll do a new drawing as I don't think my first was very helpful. Back soon.
 
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.
 
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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.
 
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.
 
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)
 
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