Tension with a with a massless rope

AI Thread Summary
The discussion revolves around calculating tensions in a system involving a massless pulley and ropes with boxes of different masses. Initially, with friction, the tensions are T1 = mg, T2 = 2mg, and T0 = 3mg. When friction is negligible and the system moves, the calculated tensions change to T1 = T2 = 4/3mg, while T0 remains at 3mg. A participant questions the correctness of T0, suggesting that if T1 + T2 equals T0, then T0 should be 8/3mg instead. The conversation highlights confusion regarding the equilibrium of the system and the relationships between the tensions.
kbka
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Hi...

I have a little assignment where I'm not quite sure, what would be the correct answer...

A massless pulley, is mounted to the ceiling with a massless rope. Over the pulley runs another massles rope, with a box with mass m in one end and a box with mass 2*m in the other.

Q1) In this first question, the friction holds the whole system in equilibrium.
What is the tension T0 between pulley and ceiling, T1 between pulley and box with mass m, T2 between pulley and box with mass 2*m

A1: T1=m*g, T2=2*m*g and T0=(m*2m)*g=3*m*g

Q2) Now, friction is negligible, and the system starts to move... What is the tension T0, T1 and T2
A2: I have calculated T1=T2=4/3*mg and T0=3*m*g (still)

Is the last answer for T0 correct? My assumption is that this part of the system (pulley+rope) is still in equilibrium, and therefore Fres=m*g-T=0
 
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kbka said:
Hi...

I have a little assignment where I'm not quite sure, what would be the correct answer...

A massless pulley, is mounted to the ceiling with a massless rope. Over the pulley runs another massles rope, with a box with mass m in one end and a box with mass 2*m in the other.

Q1) In this first question, the friction holds the whole system in equilibrium.
What is the tension T0 between pulley and ceiling, T1 between pulley and box with mass m, T2 between pulley and box with mass 2*m

A1: T1=m*g, T2=2*m*g and T0=(m*2m)*g=3*m*g

Q2) Now, friction is negligible, and the system starts to move... What is the tension T0, T1 and T2
A2: I have calculated T1=T2=4/3*mg and T0=3*m*g (still)

Is the last answer for T0 correct? My assumption is that this part of the system (pulley+rope) is still in equilibrium, and therefore Fres=m*g-T=0

If T1 + T2 = T0, then isn't T0 = 8/3mg ?
 
LowlyPion said:
If T1 + T2 = T0, then isn't T0 = 8/3mg ?

Thank you for the reply, but i highly doubt that's the case... I have drawn the situation to make it more clear...

http://peecee.dk/upload/view/136892
 
kbka said:
Thank you for the reply, but i highly doubt that's the case... I have drawn the situation to make it more clear...

http://peecee.dk/upload/view/136892

Is it your suggestion then that T1 + T2 ≠ T0 ?

Looks to me like T1 = m*(g + 1/3g) = T2 = 2m*(g - 1/3g)

2*T1 = 2*T2 = 8/3*m*g = T0
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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