Hi i was just wondering if its possible to get the T from the following:

ΣMA = - 4.66 × (2.5 – 0.12) − 10 × (5-1.5-0.12) + T×cos25o×0.25

+ T×sin25o×(5-0.12) = 0

If so how?

thanks

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- Nov 3rd 2007, 01:57 AMtaurusFind unknown
Hi i was just wondering if its possible to get the T from the following:

ΣMA = - 4.66 × (2.5 – 0.12) − 10 × (5-1.5-0.12) + T×cos25o×0.25

+ T×sin25o×(5-0.12) = 0

If so how?

thanks - Nov 3rd 2007, 03:32 AMtopsquark
You are trying to make it too hard.

Let B = - 4.66 × (2.5 – 0.12)

Let C = − 10 × (5-1.5-0.12)

Let D = cos25o×0.25

Let E = sin25o×(5-0.12)

Then your equation reads:

which I'm sure you can solve for T.

Don't get lost in the numbers. No matter how complicated they get, they're still just numbers.

-Dan - Nov 3rd 2007, 04:03 AMticbol
ΣMA = - 4.66 × (2.5 – 0.12) − 10 × (5-1.5-0.12) + T×cos25o×0.25

+ T×sin25o×(5-0.12) = 0

If that is

ΣMA = - 4.66*(2.5 – 0.12) − 10*(5 -1.5 -0.12) + Tcos(25degrees)*(0.25)

+ Tsin(25deg)*(5 -0.12) = 0,

then,

4.66(2.38) -10(3.38) +T(0.25)cos(25deg) +T(4.88)sin(25deg) = 0

11.0908 -33.8 +T(0.226576947) +T(2.062377117) = 0

-22.7092 +T(2.288954064) = 0

T = 22.7092 / 2.288954064

T = 9.921212643 ---------------------answer.