There is tension on the right-hand side, and on the left-hand side, there is the force of the man pulling on the rope. And yes, I think it has something to do with the fact that the man is standing on the platform with the pulley.
Yeah.. That is a much more rigorous approach to it!
That's one of the easier parts. Think about the displacement of m1 being in terms of the length of rope that has moved over into m2's subsystem with its pulley. If you add "X" meters of rope to that system, how much does m2 actually move...
Oh. I'd sort of intuited that, but I was pretty doubtful of myself. I was imagining the two lengths of the rope to the right and left of the lower pulley, but I couldn't decide what I thought about the tension. Thinking about it again, though... it makes perfect sense.
Thanks for all the help!
After thinking about it for a while today, I figured out part a..
The distance that m2 travels downward as m1 travels to the right is 1/2 the distance traveled by m1, i.e.
x_1 = 1/2 x_2
because of the pulley situated above m2 and its connection to a stationary object on the other side of...
Well.. If I'm reading the responses, so far, correctly... I was on the right track. If you know that a1 = 2*a2, then you can use that identity to get an equation out of the force equations for each of the masses. I still don't understand where that identity comes from, though.
But forces are vectors :(
I looked at the problem some more tonight, and I'm just lost. I setup an appointment with a tutor tomorrow (fortunately, our department has free personal tutoring). I'm hoping they can help shed some light on it.
Well, I just thought about how "Direction" in the case of acceleration is whether or not the velocity is decreasing or increasing... Not so much the "direction" of the acceleration. I guess you could say that if something is slowing down, it's accelerating in the opposite direction that it's...
Why? That's completely counterintuitive.
If you think of going UP as positive acceleration on the y-axis and going DOWN as negative acceleration (as I think most of the physics world does), then g is -9.8m/s2 and not 9.8m/s2. Hence,
Fy = 2*T - m2g
Homework Statement
In the figure, the pulleys and the cords are light, all surfaces are frictionless, and the cords do not stretch.
(a) How does the acceleration of block 1 compare with the acceleration of block 2? (Use a_2 for a2.)
(b) The mass of block 2 is 1.22 kg. Find its acceleration as...