Recent content by jalalapeno

Thank you for the tips and very clear explanation. I think the way you have done it by not substituting in numbers until the end is the best way to do as to not get confused or make any mistakes. Really appreciate your help :D

I have, and one of the solutions satisfies the equation. The angle I have calculated isn’t 13.8 degrees. I am trying to work out if this makes sense in the real world.

That’s my bad Using both results from the quadratic I do get an answer that works, and that is 13 degrees but I don’t know if that makes sense in a physical sense.

I have compared my results to others in the class and nobody really knows how to do it...

It does, but I’ve decided to ignore that for now as this question has given me enough stress :D

sinΘ = 73.7 satisfies sinΘ - 4/5 = -(√3/3cosΘ) Thank you both for your help, really appreciate it! :partytime::partytime::partytime:

Might be a bit hard to follow apologies

Of course Resolving Vertically for mass B gives: T2 = 5g Resolving Vertically for mass A gives: T1sin(30) + T2sinΘ = 4g Subbing T2 = 5g back: 1/2T1 + 5gsinΘ = 4g sin30 = 1/2 Resolving Horizontally for mass A gives: 5gcosΘ = T1cos30 Rearranging for T1 : T1 = 5gcosΘ/cos(30) Subbing T1...

:oldfrown::oldfrown::oldfrown: I just keep getting different answers. My latest answers are Θ=13.8 and Θ=73.7 From those it should be 73.7 because the angle should be larger than 30 (Θ1 ) Can anyone also do it to double check? or just tell me I'm wrong haha

I got those equations by considering the system as a whole, resolving vertically and horizontally. So you're saying T2=5g?

What I've attempted to do is resolve vertically and horizontally: T1sin(30)+T2sinΘ+T2=9g T2cosΘ=T1cos30 After that... I'm completely stuck. Any help would be appreciated, I've spent so long on this I bet I'm just missing one simple thing. Thanks in advance