Cable friction and tension problem

AI Thread Summary
The discussion centers on the equal tensions (T) in a massless pulley system as illustrated in example 2 from the provided link. The equal tensions arise from the nature of cables, which maintain consistent tension regardless of application points. Although torque is present, it cancels out due to the equal tensions and the assumption of no friction in the system. Initially, there may be a brief difference in tension, but it equalizes as tension is transferred along the cable. This understanding is crucial for analyzing the dynamics of the pulley system effectively.
Zeno's Paradox
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Read please example 2 of http://cnx.org/content/m14060/latest/ .

P1
fap6.gif


P2
fap7.gif


On picture 2 there are those two tension labeled by T which are equal. Why equal?

And shouldn't they exert just torque? I know we are considering the pulley to be massless, so no torques. So is it correct to consider those two tensions applied at the center of mass of the pulley?

Thanks in advance for your clarification. :smile:
 
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The cable has to have the same T, because that is the nature of strings and cable. It is irrelevant where you applie them. The magnitude and direction matter. There is torque on the pulley, but they cancel out.
 
That is because thay are the same cable and there is supposedly nor friction in the system. You might, for an instant initially get tension which is different from the other but you can imagine that the tension is 'transferred' along the cable until they are equal.
 
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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