Hello jegues Originally Posted by

**jegues** So,

$\displaystyle

2T\sin27^o+T\cos\alpha = 0

$

and

$\displaystyle

2T\cos27^o+T\sin\alpha = 0

$

How do you go about solving this?

Is it simply,

$\displaystyle

T(2\sin27^o+\cos\alpha) = 0

$

T cannot be zero so we solve for the part in brackets?

Thanks again!

Yes, but (apologies!) the equation should read$\displaystyle 2T\sin27^o - T\cos \alpha=0$ (Of course!)

So you get:$\displaystyle \cos\alpha = 2\sin27^o$

$\displaystyle \Rightarrow \alpha = 24.8^o$

But your second equation isn't right. You've forgotten the weight force, which is $\displaystyle -200\times 9.8$ N. Add this into your equation, substitute the value you have found for $\displaystyle \alpha$, and solve for $\displaystyle T$.

Grandad