Solve Breaking Tension Puzzle: Min Breaking Tension Revealed

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To solve the homework problem regarding the minimum breaking tension of a rope during a swing, it's essential to understand that the tension in the rope varies throughout the swing. The maximum tension occurs at the lowest point of the swing, where gravitational and centripetal forces act on the girl. To prevent the rope from breaking, the calculated maximum tension must be less than the minimum breaking tension of the rope. The relevant formulas involve calculating the forces at the bottom of the swing, factoring in the girl's mass and the length of the rope. Understanding these dynamics is crucial for determining the safe limits for the rope's tension.
radiobear
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I am brand new to this website so I hope somebody can help. I have a homework problem that goes like this: A girl of mass M is taking a picnic lunch to her grandmother. She ties a rope of length R to a tree branch over a creek and starts to swing from rest at a point that is a distance R/2 lower that the branch. What is the minimum breaking tension for the rope if it is not to break and drop the girl into the creek?
I do not even know where to begin in this problem. My textbook does not mention minimum breaking tension at all. If anybody knows a formula to use I would really appreciate it. Thanks. :confused:
 
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What would be the tension on the rope at the bottom of her swing?
 
radiobear:
With "minimum breaking tension" is meant that value of the tension which will cause the rope to break (ideal rope dynamics becomes invalid when that happens).
Clearly then, in a problem where the tension might be varying, it is crucial that the "minimum breaking tension" is larger than the maximal tension you may compute from your problem (in which you assume that the rope does NOT break).
Follow Tide's instruction.
 
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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