Recent content by Tlurk

Well, yes, if the initial position of the middle mass makes the angle $$\beta = \arccos{\frac{m_2}{2m_1}}$$ but not if the initial position makes a different angle with the same masses. But as I said I'm getting to far ahead of myself now, and need to understand this fully at equilibrium first...

I do believe that we're supposed to test a few different values to see how the system moves, eventually graphing the position, velocity and acceleration of each mass as that's what we've been doing to simpler systems so far. But as it's been quite a while since the last time I've had any physics...

The problem is not in English, but translated it reads: The figure shows two pulleys and three masses. You choose the value for the mass m, the starting position for the masses and the distance between the pulleys. Disregard the friction in the pulleys and the air resistance. Is it possible to...

Thank you for answering! So I redid a) using symbols, I'm not sure why I didn't do that in the first place so thanks for reminding me: This matches my previous answer, but I can now change the masses without doing it over again. For part b) I added a length of the rope R, made the Y-axis go...

I used m1=m2=15,0kg System: FBD: Note: I believe I have solved a) correctly and am more confused about b). a) I started with drawing the FBD. Knowing that the net sum has to be zero for the system to be at rest I used the left mass to find the tension on the rope: S1 = G1 = 15kg*9,81m/s2 =...