Ah thanks so much, I think I get it now :)
So if considering the top of the rope (where up is + and down is -)
##\vec F_{net} = \vec T + \vec F_g = m\vec a##
##\vec T - (4+5)|\vec g| = (4+5)(3.52)##
##\vec T = 120 N##
And if considering the middle of the rope...
Thanks for the quick reply and welcome :)
If I consider the rope and the lower box as a single mass, I can find that the net force acting on it is Fnet = ma = (4+5)(3.52) = 31.68 N [upwards]. The problem I'm having is that I'm not 100% sure what forces have to add up to that net force. The way...
Homework Statement
Two masses are connected by a heavy uniform rope (mass = 4 kg) and a upward force of 200N (Fa) is applied to the block on top; essentially the two masses and the rope are pulled up vertically. The mass on top (m1) weighs 6 kg, the mass on the bottom (m2) weighs 5 kg.
200...