Recent content by goodOrBad

Yeah and when I see it now I see it would have been easier to just use the angles close to the x-axis

To find forces in member 1 and 2 that is all, yes I just choose S as a symbol

Yeah I wrote it wrong it aint G1 and G2 but G and Q, I was working on another problem so it sticked with me. The balance is about all of it remaining in their respective positions they won't move. Since I made S2=G and G is 250N, then I used it in the Capstan equation and just divided it with...

really? Because I thought it goes like this

Okay but without the unfiorm descent would this be it

or should the angle of T1 be 50 due to the slope being 30deg and the rope 20deg so 50deg and that means angle around the pulley is 50+90=140deg

it looks more like this

It is given as 20 but my drawing is not faithful

This is my attempt of solving it

Thank you now I have one more question, what if almost everything was the same but... I was given weight of B and needed to find A, would I be able to use T2=T1e^(µ2β)

Great then this should be correct

then beta should be 60 right?