Oh no, of course you can't tell me the answer; that would defeat the purpose. I hope you didn't take the statement in my last reply as resentment!
Okay, I'll try again.
Since the mass is at rest, the tension all along rope B is equal. So the tension to the left of the pulley and to the...
I'm feeling pretty dense right now... I don't get exactly what you're trying to imply (I do you know you want me to come to my own conclusion though, rather than spell it out for me).
I'm taking it as the peg attached to the end of the rope is equivalent to the rope attached to the ground in...
Not exactly a homework question, but I need to know this before I can answer the question
There's a pulley suspended from a rope A. Hanging from the pulley from one side is a mass m. "Hanging" off the other side of a pulley is simply the same rope (which is attached to...
Show that if mass = 0, then E=pc and u=c.
E^2 = (pc)^2 + (mc^2)^2
B = u/c = pc/E
The Attempt at a Solution
I understand that if m=0, then E^2=(pc)^2 => E=pc.
But isn't p = Ymu? then:
E^2 = (Ymuc)^2 + (mc^2)^2
Plugging in m=0 sets E=0...