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I know that this exact question has been asked here a number of times, but none of the answers sit right with me. The question says that the ends of the strings are pulled with a velocity of "u" units. We are to find the velocity of the block in terms of u. The answer is u/cos(θ) units. However, my approach included adding two vectors of magnitude "u" seperated by an angle of 2θ. Evaluating that gives an answer of 2ucos(θ), which is wrong. Where am I going wrong?

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  • $\begingroup$ It's the forces exerted by the two ropes on the block that add together. But here, you're considering how fast they're moving. There's no reason why those should add. The point is that each rope has to pull half as hard as a single rope would in order to maintain the same speed. $\endgroup$
    – march
    Apr 10 at 15:59

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The model you are assuming here is not correct for the following reason: Imagine you have a horse carriage that is pulled by one horse with a speed of $v$. Then (if the horse is strong enough) the carriage will travel at a speed of $v$. Now, if you put 10 horses in front of the carriage, the carriage will not travel at the speed of $10v$, but still at a speed of $v$. Otherwise it would overtake the horses and this is clearly not physical.

Remember that these are velocities and not forces!

This might be confusing since sometimes velocities are broken up in their coordinate components, one horizontal and one vertical component. However, this is a different setup here. The velocities are not components of a larger velocity, but the velocity of some points of the ropes. You can also think of the points of the ropes as the horses in front of the carriage!

Also, a good check is always taking the limiting case. Here this would correspond to $\theta\rightarrow0$. The physical situation is then that the mass is very far below from the setup which is outlined here. Then, if you pull by some distance $x$, you will lift the package by the same distance $x$. This is also exactly what you find with the formula since $\cos(0)=1$!

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  • $\begingroup$ How do I know when to add velocities vectorially? For example, the carriage situation you mentioned can be stated differently as: "A carriage is given a velocity of "u" in one direction and a velocity of "9u" in the same direction". I would be forgiven in adding the velocities as vectors upon reading that. So, when can I use vector addition to add velocities? $\endgroup$
    – user331392
    Apr 10 at 17:47