Two chests connected by a rope in tension

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Homework Statement


Two chests are connected by a 15 meters long rope attached to a ceiling hook hanging 4 meters high. The chest (1) at 5 meters from the hook has a velocity of 1/2 m/s away from the other chest (2). The chests remain flat on the ground and the rope is under tension.
What is the speed of the other chest?

Homework Equations

The Attempt at a Solution


Let c1,2 be the chests, let h be the hook, let centre be the perpendicular projection of the hook on the ground.
We know the distance c1-centre (right triangle) at any time (1/2 m/s). We also know the distance c1-h (right triangle), h-c2 (length of the rope), hence we know the distance c2-centre (right triangle) at any time. The time derivative of this function at t=0 solves the problem.
Is there a shortcut, another approach? Can you think of any interesting physical or mathematical considerations? I don't think my solution is very satisfactory.
 
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Due to the Darwinian environment of this forum and with all the heavy competition going on, I'd like to bring my post back to the top of the food chain.
Perhaps my question is a little too vague?
 
There are two right angled triangles and for each you know the hypotenuse and base length. In time δt the base of one of the triangles changes by a certain factor, and therefore the whole triangle is scaled by that factor. Can you see how to work out the scaling factor for the other triangle in that time?
 
Scaled triangles? The h(eight) of the two triangles is a constant, but it's an interesting take on a different problem.
 
JanEnClaesen said:
The h(eight) of the two triangles is a constant
Oops, that's a good point.
 
You could consider components of velocities in the directions of the attached ropes.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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