Man pulling a boat along the water with a rope from above on a steep bank

[brotherbobby]

Homework Statement

A man stands on a steep shore of a lake and pulls a boat in the water by a rope which he takes up at a constant speed of $v$. What will the speed of the boat be at the moment when the angle between the rope and the surface of the water is $\alpha$?

Relevant Equations

1. The horizontal component of a vector $\vec A$ making an angle $\theta$ with the positive direction of the $x$ axis is $A_H=A\cos\theta$
2. Instantaneous velocity $v = \dfrac{dx}{dt}$
3. In a right-angled triangle, $\cos\theta=\dfrac{\text{Adjacent}}{\text{Hypotenuse}}$

Attempt : Let me copy and paste the question from the text alongside.

Method 1 : I draw an image of the problem situation to the right.

Boat B moves to the right pulled by a rope from the point O where the man is located. The angle the boat makes with the horizontal $\alpha$ changes with time.
From the principle of horizontal and vertical components of a vector, the speed of the boat $\quad\boxed{u=v\cos\alpha}\quad (1) \quad{\huge{\color{red}\times}}$

This method was my first reaction to the problem. Turns out, the answer is incorrect.

Method 2 : I consider infinitesimal increments of the boat and the rope. The diagram is to the right where I represent the boat B as a point ($\color{brown}{\bullet}$) to save space.

The boat B moves from B to B', a distance of $dx$ in a time $dt$. B'P is perpendicular to BO.
In $\triangle BB'P, BP=dx\cos\alpha$. This is the distance moved by the rope, in a time $dt$. We note that PO = B'O.
Hence, the velocity of the rope $v=\dfrac{BP}{dt}=\dfrac{dx\cos\alpha}{dt}=u\cos\alpha$, where $u=\dfrac{dx}{dt}$ is the velocity of the boat.
Hence, the velocity of the boat, $u=\dfrac{v}{\cos\alpha}\Rightarrow\boxed{u=v\sec\alpha}\quad{\huge{\color{DarkGreen}\checkmark}}\quad (2)$

This answer agrees with the text.

Doubt(s) : I have two of those, for the moment.

(1) What is wrong with method 1, given it seems intuitive enough?

(2) If indeed $u=v\sec\alpha$, then, as $\alpha$ is increasing with time, so is $u$, as $\sec\alpha$ is an increasing function in the first quadrant. This can be shown as $\alpha\uparrow\;\rightarrow\; u\uparrow$.

Isn't that counter-intuitive? I would imagine that the steeper the rope got (increasing $\alpha$), the slower the boat should move horizontally along the water surface, not faster.

 

[kuruman]

Method 1 does not follow that condition of the problem that the man "takes up the rope at a constant speed of $v$". You have a right triangle with the vertical side constant whilst the horizontal side $a$ and hypotenuse $h$ are varying.

Method 1 does not address the question that is being asked.

Method 2 assumes that $\dfrac{dh}{dt}$ is constant and is consistent with what the problem is asking.

More compactly, from the Pythagorean theorem, $~h^2=a^2+b^2$
Take the time derivative on both sides noting that $b$ is the constant vertical side,
$\cancel{2}h\dot h=\cancel{2}a\dot a+0$
$h~v=a~u\implies u=\dfrac{h~v}{a}=\dfrac{v}{\cos\!\alpha}.$

 

[Lnewqban]

... I would imagine that the steeper the rope got (increasing $\alpha$), the slower the boat should move horizontally along the water surface, not faster.

What would slow the horizontal movement of the boat down?

 

[brotherbobby]

Method 1 assumes that da/dt is constant and does not address the question that is being asked.

I did not follow that, though I did follow the rest of your argument. I attach the situation alongside.

Here $\dot h(t)=v\,(\text{rope speed})$ and $\dot a(t)=u\,(\text{boat speed})$.

In method 1, I claim $\boxed{u=v\cos\alpha}$ for a given value of $\alpha$. The speed of the rope $v=\dfrac{dh}{dt}$ is held constant. For a different $\alpha$, it's a different $u$ for the same $v$. Since $u=\dfrac{da}{dt}$, I don't see how this method assumes $\dfrac{da}{dt}$ is held constant.

 

[brotherbobby]

What slows the boat down?

What would slow the horizontal movement of the boat down?

Imagine you're pulling at a boat from above, that is on the surface of water below. Isn't it intuitive that the more horizontal your rope is, the faster the boat would travel along the water?
Or, conversely, the steeper the rope is, the slower it would travel horizontally?

 

[kuruman]

In method 1, I claim $\boxed{u=v\cos\alpha}$ for a given value of $\alpha$.

Can you justify this claim? It is not supported by any math or reasoning that you have shown.

If what you say is correct, then
$u=v\cos\alpha$ becomes
$\dot a=\dot h\dfrac{a}{h}$ which means $~h\dot a=a\dot h.$
This implies that
$\dfrac{a}{h}=\text{const.}~$ (Take the time derivative if you don't believe me.)

But $\dfrac{a}{h}=\cos\alpha$ and cannot be constant. Your claim leads to an absurd result.

 

[haruspex]

From the principle of horizontal and vertical components of a vector

That principle applies to finding a component of motion of a given object. The part of the string attached to the block is moving horizontally at speed u, while the part almost at O is moving at speed v at angle α to the horizontal. (An interesting challenge is to find how string points in between are moving.)
So near O, you can find the the horizontal component of motion, $v\cos(\alpha)$, and near the boat, find the component in the direction of the string, $u\cos(\alpha)$.
Since the parts of the string remain at constant separation, they all have the same component in the direction of the string, $v=u\cos(\alpha)$. They do not have the same horizontal component.

I would imagine that the steeper the rope got (increasing α), the slower the boat should move horizontally along the water surface, not faster.

Intuition has a habit of confusing cause and effect. The steeper the rope the harder you will have to pull to achieve the same boat motion against the water resistance. But if you do manage to keep taking up rope at the same speed then the boat will accelerate.

 

[brotherbobby]

Can you justify this claim? It is not supported by any math or reasoning that you have shown.

If what you say is correct, then
$u=v\cos\alpha$ becomes
$\dot a=\dot h\dfrac{a}{h}$ which means $~h\dot a=a\dot h.$

I follow your argument thus far.

This implies that
$\dfrac{a}{h}=\text{const.}~$ (Take the time derivative if you don't believe me.)

This is where I don't follow you. If $h\dot a=a\dot h\Rightarrow \dfrac{a}{h}=\dfrac{\dot a}{\dot h}\quad (1)$.
Why are you saying $\dfrac{a}{h}=\text{constant}$?
If I say $u=v\cos\alpha$, $u\;(=\dot a)$ is not a constant. It is changing with changing $\alpha$.
Hence the right side of (1) $\dfrac{\dot a}{\dot h}$ is not a constant.

 

[kuruman]

What is $$\frac{d}{dt}\left[\frac{a(t)}{h(t)}\right]~?$$ Hint: Quotient rule.

 

[brotherbobby]

What is $$\frac{d}{dt}\left[\frac{a(t)}{h(t)}\right]~?$$ Hint: Quotient rule.

Yes. $\dfrac{d}{dt}\left\{\dfrac{a(t)}{h(t)}\right\}=\dfrac{h\dot a-a\dot h}{h^2}=\dfrac{uh-av}{h^2}=0$, since this folows from $u=v\cos\alpha$.
Thus $\dfrac{a}{h}=\text{constant}$, according to my claim of $u=v\cos\alpha$.
But $\dfrac{a}{h}=\cos\alpha$ where $\alpha$ is changing.
Hence the contradiction. Many thanks.

 

[brotherbobby]

That principle applies to finding a component of motion of a given object. The part of the string attached to the block is moving horizontally at speed u, while the part almost at O is moving at speed v at angle α to the horizontal. (An interesting challenge is to find how string points in between are moving.)
So near O, you can find the the horizontal component of motion, $v\cos(\alpha)$, and near the boat, find the component in the direction of the string, $u\cos(\alpha)$.
Since the parts of the string remain at constant separation, they all have the same component in the direction of the string, $v=u\cos(\alpha)$. They do not have the same horizontal component.

Intuition has a habit of confusing cause and effect. The steeper the rope the harder you will have to pull to achieve the same boat motion against the water resistance. But if you do manage to keep taking up rope at the same speed then the boat will accelerate.

Both your points are deep and require some serious thinking. Many thanks.

 

[haruspex]

Further to my post #7, consider a point on the rope at distance z from O.
Resolving its motion into components along and normal to the rope, it is moving at speed $v$ along the rope and at speed $z\dot\alpha$ normal to it.
Since the block (at z=h) moves horizontally, we have
$v\sin(\alpha)=h\dot\alpha\cos(\alpha)=H\cot(\alpha)\dot\alpha$, where H is the height of O
$u=v\cos(\alpha)+h\dot\alpha\sin(\alpha)=H\dot\alpha$.
Whence $u\cos(\alpha)=v$.