Mechanical Energy and Newton's Laws

AI Thread Summary
A block of mass m is dropped from point A, and when it reaches point B, the tension in the rope is T = 2mg, causing the rope to break. The solution manual uses mechanical energy principles and centripetal force to determine that the height h where the rope broke is 4 cm. A participant initially miscalculated the height as 3 cm by incorrectly assuming the net force was horizontal. Upon further analysis, they recognized that the tension and weight vectors must be accurately represented, confirming that the resultant vector cannot be horizontal. The correct interpretation leads to the conclusion that the height where the rope broke is indeed 4 cm.
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Homework Statement



A block of mass ## m ##, attached to a rope, is dropped at the point A.
When the block reaches the point B, the tension is ## T = 2 \cdot m \cdot g ## and the rope broke at that same point.
If the length of the rope is ## L = 6cm ##, evaluate the height ## h ## where the rope broke.

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Homework Equations


The solution's manual uses mechanical energy and the centripetal force ( ## T - m \cdot g \cdot sin\theta = \frac{m \cdot v^{2}}{2} ##). Since ## h = sin \theta \cdot L##, we find the height ## h = 4cm ##.

The Attempt at a Solution


I understand and agree with that solution, but I can't figure out why my solution is wrong.
Since ## W = m \cdot g ## is perpendicular to the horizontal line, we have a vectorial triangle composed by ## \vec{T}, \vec{W}, \vec{F}## (the green, red and purple vectors in the figure below, respectively), where F is the resultant force of T and W, i.e. ## \vec{F} = \vec{T} + \vec{P}## .
In this triangle, we have ## sin \theta = \frac{W}{T} = \frac{W}{2\cdot W} = \frac{1}{2}## and ##sin \theta = \frac{h}{L}##. Therefore, ## h = \frac{L}{2} = 3cm ##. I couldn't find my mistake.
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Welcome to PF!

You are assuming the net force, F, on the block is horizontal. If that were true in general, would the block ever have any vertical acceleration?
 
Thank you very much - for the welcome and for the solution. :)

After your comment, I noticed that in my figure the tension and the weight have similar lenghts. I redone it with more precise lenghts (T = 2W) and, indeed, it's impossible to the resultant vector to be horizontal in terms of vectors. (R is the real resultant vector, and F is the wrong one at my first attempt.)

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OK, good.
 
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