Author Topic: Statics of a Particle Q? (Read 833 times) Tweet Share

luken93 · « **on:** September 15, 2010, 07:45:55 pm »

See attached for the question and thanks in advance

Also, are there any sort of formulas when doing this chapter?
The book doesnt really say much...

jasoN- · « **Reply #1 on:** September 16, 2010, 05:56:46 pm »

Refer to the diagram

$x=\cos^{-1}(\frac{10^2+12^2-15^2}{2(10)(12)})$
$x=85.459$
$a=\cos^{-1}(\frac{12^2+15^2-12^2}{2(12)(15)})$
$a=41.650$
$b=180-(90-a)$
$b=131.650$
Therefore
$c=360-x-b$
$c=142.891$
Transform the graph into a triangle of forces (diagram) (I don't think year 11 uses Lami's theorem?)
Using the sin rule:
$\frac{20g}{180-x}=\frac{T1}{180-b}=\frac{T2}{180-c}$
Substitute and solve

You probably only need to know the sin/cosine formulas

EDIT: don't use g if the answer's in kgwt (just use 20), multiplying by g gives the answer in Newtons

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Author Topic: Statics of a Particle Q? (Read 833 times) Tweet Share

luken93

Statics of a Particle Q?

jasoN-

Re: Statics of a Particle Q?

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