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The diagram shows a uniform rectangular sign ABCD weighing 50N which is being fixed to a vertical frame. Lengths AD and AB are 120 cm and 80 cm respectively. The corner A is already attached to the frame by a horizontal nail, so that the sign can rotate about A in a vertical plane. While the corner B is being attached an assistant temporarily holds the sign steady at C by a force at 40 degrees to the horizontal. Find the size of this force

Attachment not found


What I did:

Fsin40x80=50(40)Fsin40 x 80= 50(40)

F50(40)80sin40)F \dfrac{50(40)}{80sin40)}

F=39N

But F=14N




Please help
(edited 6 years ago)
Original post by joyoustele
The diagram shows a uniform rectangular sign ABCD weighing 50N which is being fixed to a vertical frame. Lengths AD and AB are 120 cm and 80 cm respectively. The corner A is already attached to the frame by a horizontal nail, so that the sign can rotate about A in a vertical plane. While the corner B is being attached an assistant temporarily holds the sign steady at C by a force at 40 degrees to the horizontal. Find the size of this force

Attachment not found



Please help


You don't know the magnitude or direction of the force at A , so take moments about A. If you find it easier, split the force at C into vertical and horizontal components before you take moments.
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joyoustele
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@RDKGames @Notnek
(edited 6 years ago)
Original post by joyoustele


What's confusing exactly? Tiny Hobbit has suggested to take moments about A.
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joyoustele
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Original post by RDKGames
What's confusing exactly? Tiny Hobbit has suggested to take moments about A.


I did this. I have shown my workings above. Where have I gone wrong?
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Notnek
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Original post by joyoustele

Please respond to a previous post before tagging someone else.
Original post by joyoustele
I did this. I have shown my workings above. Where have I gone wrong?


There is also 120Fcos⁡(40)120F \cos(40) that should be on the LHS. That force also makes the sign go anticlockwise.
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joyoustele
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Original post by RDKGames
There is also 120Fcos⁡(40)120F \cos(40) that should be on the LHS. That force also makes the sign go anticlockwise.


Oh, Thank you. XD. bad mistake from me.

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