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    hey could anyone help me with this mechanics question, i have attached the scan of it. its jus part a i need help with, i keep gettin the thrust to be zero somehow! thanks
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    please someone
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    I'll give it a go...
    First a bit of trig to find the size of angle alpha:
    Right angled triangle so we get tan(alpha)=o/a=1/2.4 gives alpha=22.6°(3 s.f.).
    Then moments take about the bottom of the display board (the bit touching the ground)
    clockwise moments = anticlockwise moments
    1.5m x component of weight of board = 2.4m x thrust
    1.5 x 26cos(alpha) = 2.4 x thrust
    thrust = (1.5 x 26cos(alpha))/2.4
    thrust = 15N
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    the book says 1.5 but i guess thats jus an error in decimal point doh, seein as u got the same as me on my third attempt! thanks
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    hmm, I'm not sure.
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    i got 15N as well
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    If it's from one of those aim higher books, perhaps could be a reason.
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    oh wow how did u know?! yeah it is from that book, are most of the answers wrong then?!
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    We use the M1 one, and it does have quite a lot of wrong answers in it.
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    yeah i have noticed quite a lot of wrong answers in the M3 one...it's not good, thanks for the help tho.
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    I got 15N!
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    15n
 
 
 
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Updated: June 13, 2005
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