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Centre of mass of special shapes question :D Please Help if you can. Watch

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    The uniform triangular lamina ABC has weight 30N and is right-angled at B; AB = 30 cm and BC = 15 cm. The lamina is suspended by vertical light strings PA and QB, and hangs in equilibrium in a vertical plane with AB horizontal and BC vertical. Find the tensions in the strings.

    The string QB is now cut, and the lamina settles in equilibrium supported only by the string PA. What angle does AB make with the vertical?
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    The centre of mass of a triangular lamina is 2/3 along median from vertex. Then take moments from A and then B to find tension in each string

    The centre of mass will hang vertically below PA. Therefore work out the distance from the centre of mass to the side AB and also to A. That'll give you a right-angled triangle with opposite and adjacent sides to the angle you want.
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    (Original post by mattypb)
    The centre of mass of a triangular lamina is 2/3 along median from vertex. Then take moments from A and then B to find tension in each string

    The centre of mass will hang vertically below PA. Therefore work out the distance from the centre of mass to the side AB and also to A. That'll give you a right-angled triangle with opposite and adjacent sides to the angle you want.
    So would I have to find centre of mass x and y. How do I do that.?
 
 
 
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