Velocity of boat connected by a pulley fixed at some height

AI Thread Summary
The discussion revolves around calculating the velocity of a boat connected to a pulley, where the string is pulled at a constant speed. The key equation derived is that the velocity of the boat, v_b, is equal to u sec(θ), where u is the speed at which the string is pulled and θ is the angle the string makes with the water. Confusion arises regarding whether the boat's motion is due to the horizontal component of the string's velocity or the overall velocity along the string. It is clarified that both radial and tangential motions of the string contribute to the boat's movement, but only the radial component affects the velocity over the pulley. The discussion emphasizes the importance of understanding the geometry and motion involved in the system to arrive at the correct conclusion.
cr7einstein
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Homework Statement


A pulley fixed on a wall of height h connects a toy boat with a man on the wall. The string is pulled by the man at a constant speed u m/s. Find the velocity of boat when the string makes an angle $$\theta$$ with the water.

Homework Equations


The question will be more comprehensible with the help of a diagram, but unfortunately, I don't know how to upload one here. I have tried my best to explain the problem in words. The diagram will be kind of like a right angled triangle with pulley and boat joining the vertices of hypotenuse, and wall forming the perpendicular.

The Attempt at a Solution


If I write the equations relating displacements, assuming length of string connected to boat (on the other side of pulley) to be y, and distance of boat from the base of the wall x, i get $$u=-dy/dt, v_{b}=-dx/dt$$(-ve because decreasing), and using x^2+h^2=y^2, I get $$v_{b}=usec\theta$$, which is right(according to the book). If I now use the fact that the velocities of points along a string must be equal, at the point connecting boat and string, the component of boat's velocity is $$v_{b}cos\theta$$, which must equal u (on the other end of string), and hence I get the same result-$$v_{b}=usec\theta$$.
Now, my problem is, the boat must be moving because of the horizontal component of the velocity of string pulling it(i.e.u). In other words, $$v_{b}=ucos\theta$$(i.e., the boat's motion is due to the horizontal component of string's velocity, which is inclined to it at an angle of theta at the point of contact. But the answer is $$usec\theta$$. What is wrong in this argument? Is the velocity with which the boat is being pulled not due to the horizontal component of string's speed?? The only thing which could cause the motion of boat must be the velocity component of string which is imparted to it, right? Obviously, something is wrong in my argument, for this gives the wrong answer, but what exactly?
Thanks in advance!
 
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cr7einstein said:
If I now use the fact that the velocities of points along a string must be equal, at the point connecting boat and string, the component of boat's velocity is
Can you justify this statement? It is trivially true for a string whose motion is directed along its length, but in this case the boat end of the string is constrained to travel in a direction which does not lie along the length of the string...
 
cr7einstein said:
Is the velocity with which the boat is being pulled not due to the horizontal component of string's speed??
Causality is certainly not relevant. You could equally envision it as the boat pushing a rod up over a roller. Then you would argue that the rod's movement was due to the the component of the boat's movement along the rod, and get the right answer.
 
But then which is the correct arguments-the boat's velocity is the same as the horizontal component of string's velocity, or that the velocity component of boat along the string is the same as the speed of string. The latter gives the same as the more rigorous calculus based approach, so I suppose it is correct. But then, what is wrong with the first argument?
 
cr7einstein said:
But then which is the correct arguments-the boat's velocity is the same as the horizontal component of string's velocity, or that the velocity component of boat along the string is the same as the speed of string. The latter gives the same as the more rigorous calculus based approach, so I suppose it is correct. But then, what is wrong with the first argument?
The string is not only moving towards the pulley. The end attached to the boat is also rotating around the pulley. This contributes to the boat's horizontal motion.
 
But then we can equally apply the argument in the reverse case(as you pointed out before); that we can say that the bushes the string. The two approaches should be equivalent, right? So, if we argue that the rotational motion of the pulley also affects the velocity of boat, the same should apply in reverse case too. But as you said earlier, this approach gives the right answer without any rotational constraints to worry about. (vcosx=u => v=usecx). Then, how can we say that the rotation of pulley will affect the motion of boat?
 
cr7einstein said:
But then we can equally apply the argument in the reverse case(as you pointed out before); that we can say that the bushes the string. The two approaches should be equivalent, right? So, if we argue that the rotational motion of the pulley also affects the velocity of boat, the same should apply in reverse case too. But as you said earlier, this approach gives the right answer without any rotational constraints to worry about. (vcosx=u => v=usecx). Then, how can we say that the rotation of pulley will affect the motion of boat?
I was not referring to the rotation of the pulley. The end of the string attached to the boat has both a radial motion, towards the pulley, and a tangential motion, a rotation around the point of contact of string with pulley. Both contribute to the movement of the boat, but only the first contributes to the velocity of string over pulley.
With the boat pushing string view, only the boat's component of motion towards the pulley contributes to the velocity of string over pulley.
 
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