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 $u\cos(\alpha)=v$.

 

[Lnewqban]

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?

If the problem is asking for the magnitude of the speed of the boat at a point beyond the initial one, giving no value of drag force, it seems safe to assume zero resistive force from the water.
If that is the case, shouldn't the value of that later velocity be at least equal to the initial one?

The pulling force is free to change directions, but the boat is only free to move horizontally.
As the pulling force vector tends to the vertical direction, the needed force magnitude to keep a constant pulling velocity grows greater and greater.

When the pulling force gets closer to be vertical, the pulling force tends to an infinite value, until the pulling velocity suddenly becomes zero (unless you can lift the boat out of the water).

 

[kuruman]

I don't think that you simulation shows what is being described by the problem. A man is at constant height $b$ above the river pulling on the rope. A right triangle is formed by right side $b$, right side $a$ which is the instantaneous distance from the point directly below the man to the boat, and the hypotenuse $h$ which is the instantaneous length of the rope. As the man pulls gathers the rope, the hypotenuse gets shorter at a constant rate. I can see a right triangle in your simulation, however its blue hypotenuse is constant and the vertical side (not drawn) varies from zero to the length of the hypotenuse. Nice simulation, though.

 

[Lnewqban]

I don't think that you simulation shows what is being described by the problem. A man is at constant height $b$ above the river pulling on the rope. A right triangle is formed by right side $b$, right side $a$ which is the instantaneous distance from the point directly below the man to the boat, and the hypotenuse $h$ which is the instantaneous length of the rope. As the man pulls gathers the rope, the hypotenuse gets shorter at a constant rate. I can see a right triangle in your simulation, however its blue hypotenuse is constant and the vertical side (not drawn) varies from zero to the length of the hypotenuse. Nice simulation, though.

That is absolutely correct.
As the hypotenuse $h$ is reduced at a constant rate, and right side $b$ remains constant, the right side $a$ progressively increases in the same period of time.

The shown simulation is just an approximation to show the acceleration of the boat.
It actually represents a typical Scott Russell linkage.

 

[kuruman]

An interesting challenge is to find how string points in between are moving.

Interesting but straightforward given the Pythagorean theorem and some elementary trigonometry. Refer to the right triangle shown in post #4.

The hypotenuse is given by $~h(t)=h_0-vt~$ where $h_0$ is its length at $t=0.$

Let B be a point on the hypotenuse at distance $b$ from the boat. It is at distance $~d=h_0-b~$ from the hands of the man and reaches them at time $T=\dfrac{h_0-b}{v}.$

The horizontal right side is given by $$a(t)=\sqrt{h(t)^2-b^2}=\sqrt{(h_0-vt)^2-b^2}~~~~~~~(0 \leq t \leq T).$$Let $\theta$ be the angle of the rope relative to the vertical. Then $$\sin\!\theta=\frac{a(t)}{h(t)}=\frac{\sqrt{(h_0-vt)^2-b^2}}{h_0-vt}~;~~\cos\!\theta=\frac{b}{h(t)}=\frac{b}{h_0-vt}.$$ Let the origin of coordinates be the hands of the man. A point on the rope at distance $l$ from the man's hands has coordinates (assuming positive axes up and to the right) $$y=-l \cos\!\theta=-l\frac{b}{h_0-vt}~;~~x=-l\sin\!\theta=-l\frac{\sqrt{(h_0-vt)^2-b^2}}{h_0-vt}.\tag{1}$$

The plot on the right shows the trajectory of point B at $l=h_0-b$, i.e. the point that reaches the man's hands at $t=T$.

The extension of the dashed line and the horizontal line at $y=-b$ intersect at the initial position of the boat. The man's hands are at the origin, upper right corner.

The plot parameters are $h_0=10~\text{m}$, $b=5~\text{m}$ and $v=0.1~\text{m/s}.$ The time interval of the "trip" is $T=13.4~\text{s}.$

The plot below is a "zoom out" of the first plot. It shows successive snapshots of the section of the rope from the boat to point B.

Edited for equation typo, to tag equation (1) and to change the aspect ratio of the second plot to a realistic 1:1.

 

[Lnewqban]

The plot below is a "zoom out" of the first plot. It shows successive snapshots of the section of the rope from the boat to point B.
View attachment 348463

Why does the diagram show that hand's elevation y is changing by 1 meter at the upper end?

 

[kuruman]

Why does the diagram show that hand's elevation y is changing by 1 meter at the upper end?

I think you misunderstood the second plot. The hand is fixed at the origin. The point that is displaced by about a meter at the upper end in the second plot is point B. It is initially at distance (along the rope) $d=h_0-b$ and finally at distance $d=0$ when point B reaches the hand. The upper closed circles in the second plot are representative points from the first plot.

 

[Lnewqban]

I think you misunderstood the second plot.

I did.
Thank you, kuruman.

 

[brotherbobby]

Sorry for coming in late.

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, vcos⁡(α), and near the boat, find the component in the direction of the string, ucos⁡(α).
Since the parts of the string remain at constant separation, they all have the same component in the direction of the string, v=ucos⁡(α). They do not have the same horizontal component.

I am struggling to understand the reasoning of the above point. For illustration, I draw an image of the problem situation taking three points along the rope - O and P at the ends and Q somewhere in between.

The boat B is pulled along the water surface by a rope at an angle $\alpha$ to the horizontal. The point where the rope is being pulled at is O at the top, distant $b$ vertically from the ground. As the boat B moves horizontally, it's distance from the "base" varies as some $a(t)$ and so does its lateral height $h(t)$ from O. Let's take a point P on the rope tied to the boat and another point Q at a distance of $z$ vertically from P.

All three points have the same velocity $v$ along the rope. We know that the horizontal velocity of P is $u_P=v\sec\alpha$. What if the boat was at Q? Its horizontal velocity wouldn't change, all else remaining the same, specially $v,\alpha$. Hence the horizontal velocity of Q : $u_Q=u_P=v\sec\alpha$. Q can be any point along the rope. But this cannot be true because the horizontal velocity of O is $u_O=v\cos\alpha$, leading to a discontinuity with that of Q. My understanding contradicts @haruspex's point above that all points along the rope have different horizontal velocities.


I'd like @haruspex or someone else to show me where am going wrong.

 

[kuruman]

All three points have the same velocity $v$ along the rope. We know that the horizontal velocity of P is $u_P=v\sec\alpha$. What if the boat was at Q? Its horizontal velocity wouldn't change, all else remaining the same, specially $v,\alpha$.

Why wouldn't the horizontal velocity change? Read post #18. Equation (1) gives you the $x$ and $y$ coordinates of a point on the rope at distance $l$ from the hands of the man. You can take derivatives to find the velocity and acceleration.

 

[haruspex]

What if the boat was at Q?

It is unclear what you have in mind.
Are you saying merely that b is less, and the block is still moving horizontally, or that there is a second block tied to the string at Q? If the latter, this second boat must have some vertical velocity since the part of the string it is attached to will eventually reach the pulley.

 

[brotherbobby]

Are you saying merely that b is less

Yes. If $b$ was less and the boat was at Q, what would be its horizontal speed? I am claiming that it would be the same as that at P : $u_Q=u_P=v\sec\alpha$. Obviously, I am wrong due to discontinuity of this if I carried it all the way to O ($u_O=u\cos\alpha$). But if $u_Q\ne u_P$, what is $u_Q=?$

 

[haruspex]

Yes. If $b$ was less and the boat was at Q, what would be its horizontal speed? I am claiming that it would be the same as that at P : $u_Q=u_P=v\sec\alpha$.

Yes.

if I carried it all the way to O ($u_O=u\cos\alpha$).

Why?

 

[brotherbobby]

Yes.

Why?

The velocity at O is $v$. Its horizontal component should be $v\cos\alpha$.
You said as much in post #7 above.

It is on line no. 4 from the cutting above.

 

[jbriggs444]

When the pulling force gets closer to be vertical, the pulling force tends to an infinite value, until the pulling velocity suddenly becomes zero (unless you can lift the boat out of the water).

The correct statement would be that the pulling velocity becomes undefined. If we are plotting horizontal position as a function of time, the function ceases to be defined at the time when the limiting position is reached. Else we violate a constraint.

The velocity at O is $v$. Its horizontal component should be $v\cos\alpha$.
You said as much in post #7 above.

View attachment 348558
It is on line no. 4 from the cutting above.

The velocity at O is $v$. Its horizontal component should be $v\cos\alpha$.
You said as much in post #7 above.

You have that backward. The velocity at O is $v$. If the horizontal velocity is $v_h$ then the component of the horizontal velocity in the direction of O ($v_h \cos \alpha$) must be equal to $v$. Solving for $v_h$, we get $v_h = \frac{v}{\cos \alpha}$

If instead of pulling a string, you imagine pushing a rod, you're going to have to push faster and faster to get a constant rate of rod extrusion as you get closer and closer to vertical.

Pushing slower and slower will not get the job done.

 

[Lnewqban]

The correct statement would be that the pulling velocity becomes undefined. If we are plotting horizontal position as a function of time, the function ceases to be defined at the time when the limiting position is reached. Else we violate a constraint.

Sorry, I can't understand that.
What constrain gets violated?

How I see it:
The pulling velocity has a constant value as long as the length of the remaining cord is greater than the height differential.
It suddenly becomes zero at the instant both distances become equal.

It could regain its original value in the opposite direction (the hands let go the rope in this case) if the boat overshoots that point.

 

[haruspex]

The velocity at O is $v$. Its horizontal component should be $v\cos\alpha$.

But in post #23 you wrote

$u_O=u\cos\alpha$

So presumably you meant $u_O=v\cos\alpha$.

Let the height of O be H and consider a point on the rope at distance r from O.
Using my analysis in post #12, and that $\dot\alpha=\frac vH\sin(\alpha)\tan(\alpha)$, that point has horizontal velocity $v(\cos(\alpha)+(r/H)\sin^2(\alpha)\tan(\alpha))$ and vertical velocity $v(\sin(\alpha)-(r/H)\sin^2(\alpha))$.

At r=0 that reduces to $v\cos(\alpha)$ and $v\sin(\alpha)$ while at P, $r=H\csc(\alpha)$, it gives $v(\cos(\alpha)+\sin(\alpha)\tan(\alpha))=v\sec(\alpha)$ and vertically $v(\sin(\alpha)-\sin(\alpha))=0$.

What you are overlooking is that by making point Q move horizontally you have changed the way all points of the rope move. Taking the limit as a horizontally moving Q approaches O is not going to give the same result as when keeping P moving horizontally.

"Mankind is composed of two kinds of people - those who think Mankind is composed of two kinds of people, those who don’t think Mankind is composed of two kinds of people, and those who cannot count."

 

[Lnewqban]

I am struggling to understand the reasoning of the above point.
...
I'd like @haruspex or someone else to show me where am going wrong.

 

[haruspex]

You have that backward. The velocity at O is $v$. If the horizontal velocity is $v_h$ then the component of the horizontal velocity in the direction of O ($v_h \cos \alpha$) must be equal to $v$. Solving for $v_h$, we get $v_h = \frac{v}{\cos \alpha}$

No, @brotherbobby is referring to the motion of a point on the rope very near O, compared with P. Its motion is almost entirely towards O, with only a small component normal to the rope. So the horizontal component of its motion is roughly $v\cos(\alpha)$.
See posts #12 and #28.

 

[kuruman]

"Mankind is composed of two kinds of people - those who think Mankind is composed of two kinds of people, those who don’t think Mankind is composed of two kinds of people, and those who cannot count."

I've heard that Mankind is composed of 10 kinds of people, those who understand binary numbers and those who don't.

 

[kuruman]

I'd like @haruspex or someone else to show me where am going wrong.

Here is another way to look at the trajectory in space of a point $Q$ on the rope at distance $PQ$ from the boat along the rope. The key features to understand is that the rope always points towards the hands of the man at point O. This means that the trajectory of $Q$ in space is independent of whether the rope is pulled at a constant rate or whether the boat moves at constant speed. Point $Q$ will move along a uniquely defined path as angle $\alpha$ increases to $\frac{\pi}{2}.$

The figure on the right shows segment $PQ$ (in red) at two successive positions of the boat. In time interval $dt$ the boat advances to the right by amount $~dx=u(t)dt~$ and segment $PQ$ simultaneously rotates counterclockwise about point $P$ by an arc of length $~ds=(PQ)d\theta$. Note that $d\theta=-d\alpha$.

You can see from the drawing that
$d\mathbf x=dx~\mathbf{\hat x}$ and that
$d\mathbf s=-ds\sin\!\alpha~\mathbf{\hat x}+ds\cos\!\alpha ~\mathbf{\hat y}$ (Angle $\alpha$ is not shown to avoid clutter.) With $~ds=(PQ)~d\alpha$, the displacement of point $Q$ in time $dt$ is $$d\mathbf r=\left[dx-(PQ)\cos\alpha~d\alpha\right]~\mathbf{\hat x}+(PQ)~\sin\alpha~d\alpha~\mathbf{\hat y}.$$From this you can see that all points along the rope move simultaneously

• to the right by a the same amount $dx$
• to the left by an amount that is proportional to how far away ($PQ$) they are from the boat
• up by an amount that is also proportional to how far away they are from the boat.

(Edited to fix equation typos.)