mechanics moments | Revision Server | Page 1

verbal cargo Aug 31, 2024, 3:51 PM

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Help

atomic vale Aug 31, 2024, 3:55 PM

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verbal cargo Help

take moments around C

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0.6mg - 8g(0.15) = 0

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it's in equilibrium so anticlockwise = clockwise

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so total = 0

verbal cargo Aug 31, 2024, 4:01 PM

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Yeah I got it thanks

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How would u do part b I have a way but it feels really extra

atomic vale Aug 31, 2024, 4:23 PM

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verbal cargo How would u do part b I have a way but it feels really extra

the midpoint is 0.75

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and if AD is x

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then between AD and the mp

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is 0.75-x

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you now have a 5g particle

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now you can take moments around D and cancel the g's to find x, which will be AD

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also

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DB will be 1.5-x

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so in total

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5gx - 8g(0.75-x)-2g(1.5-x)=0

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@verbal cargo

verbal cargo Aug 31, 2024, 7:56 PM

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Yeah that’s what I did

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Thought there would be an easier way

atomic vale Aug 31, 2024, 8:52 PM

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verbal cargo Thought there would be an easier way

I mean as soon as you need to introduce a variable

#mechanics moments