Doh, I'm sorry, that was a horrible use of varibles on my part.
It should have been written as m= mass of block 2
the final answer reads mass of black 2 = 4.0 kg
as for the Fn=Fg cos 53.1 , another miswriting on my part,
as sumF in the y direction = Fn - Fg cos 53.1 = 0, therefore Fn=...
Thanks!
So after going down the line, and fixing that trig error, i got to this
Fn=ma*cos53.1
Fn=12kg*a
Fgsin53.1+(uk*12kg*a)=2*ma
(ma*.8)+(4.8kg*a)=2*ma
[cancel out the acceleration varible]
.8m+4.8kg=2m
4.8kg=1.2m
and
4.0kg=m
Does this help at all?
I'm pretty sure I'm going in the right direction, I just don't know what to do next.
https://webmail.unm.edu/Session/36232-acFPEU5bQ2jidPBk2CKY-kmbdsse/MessagePart/INBOX/970-01-B/0927011758.jpg
Homework Statement
Two blocks are connected by a string (through a frictionless massless pully) block 1 is on an inclined plane, while block two is hanging from the other end of the string.
m1 = 20kg, theta =53.1, friction between the block and the inclinde is \muk =.4
what is m2 such that...