vest4R

I can see exactly where you're coming from. I believe that this issue so systemic throughout my facilty that most (including myself until now) do not know we're writing it incorrectly. I can see your way as correct.

I guess we've essentially made our own language while being oblivious to it.

Thanks for your help guys. I appreciate the insight into mathmatical correctness.

This is a typical example of how engineers butcher math lol.

here's a good example from my lecturer. I just happen to have it open now.

To relate to the percentage example above,

"*100%"

I guess it could be written as "*10^-3kN"

interesting. I see how you're getting confused.

This is how my teaching team have been writing their expressions over the years the

*10^-3 is meant to signify the coversion from 3000N to 3kN. It's not suppose to be a part of the mathmatical problem.

I hope that makes sence

Sorry melody. I meant 1000N ://. I've amended the original post.

Thanks for your help pavlov. I've just found my error.

Upon calculating the term 132/dn * (dn-40). I used E as 200MPa for steel instead of 200GPa.

I appreiate your time.

unfortunately dn cannot exceed 460mm in this problem

ahhh yes. you're right

yeah I don't understand how it's 6.85 either. I was hoping someone could shed some light.

The third image shows how the 6.85dn was calculated

sorry, I wrote in the *10^-3 at the last second.

I've used 6852300 as Newtons

that's it!

Thank you, couldn't recall it at all!

Actually, I just realized how silly I've been for the past couple of hours trying to work this out.....

\(3Tcos\theta - 1.5Tsin\theta = 0 \\ 3Tcos\theta =1.5Tsin\theta \\ 3T = 1.5T tan\theta \\ \frac{3}{1.5} = tan\theta \)

I appreciate your help for waking me up!!

Thanks Melody,

Can I ask where: \(\frac{3cos\theta}{1.5cos\theta} = \frac{1.5sin\theta}{1.5cos\theta}\) came from?

is this just dividing the left hand side by only half of its self?

great! thank you. yes I got the same with: