Max Speed & Acceleration for Climbing Rope

AI Thread Summary
The discussion centers on the limitations of climbing a rope given specific weight and tension constraints. A man weighing m1g cannot climb the rope if the maximum tension T is only 2m1g/3, as this is insufficient to support his weight. The calculations indicate that the maximum acceleration while climbing would be negative, suggesting he cannot ascend without breaking the rope. Instead, he can slide down at a constant acceleration of (1/3)w. Overall, the consensus is that the man cannot climb the rope safely under these conditions.
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If a man, weighing m1g is tryingting to climb up or down a rope, and the max tension T that the rope can withstand is 2m1g/3, then find the maximum speed and acceleration with which the man can climb without breaking the rope.

My work:
2m1g-m1g=m1a
=>a=-1g/3

doesn't this mean that the rope will break?
 
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Apparrantly the man can not climb this rope. If he weighs w, and the rope can only hold (2/3)w, he can not even hang on to the rope, nevermind climbing it.

He can, however, slide down the rope at a constant acceleration of (1/3)w

BTW, "find the maximum speed" is nonsense for this problem unless the length of the rope is given.
 
He can climb down the rope just by letting go- no tension at all!

As Chi Meson said, if the rope can only support 2/3 of the man's weight, he cannot climb up at all- he can't even hang on.
 
thank you very much!
 
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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