The speed of the point of intersection of two uniformly moving lines

[brotherbobby]

Homework Statement

Two intersecting straight lines move translationally in opposite directions with velocities $v_1$ and $v_2$ perpendicular to the corresponding lines. The angle between the lines is $\alpha$. Find the speed of the point of intersection of these lines.

Relevant Equations

Two vectors $\vec P$ and $\vec Q$ that make an angle $\theta$ between them have a resultant $\vec R$ whose magnitude is equal to $R=(P^2+Q^2+2PQ\cos\theta)^{\frac{1}{2}}$

This is the question as it appeared in the text.

Attempt : I start by drawing an image of the situation.

Two lines 1 and 2 have an angle $\alpha$ between them. The move perpendicular to themselves with velocities $v_1$ and $v_2$. I am required to find the velocity of their point of intersection P : $v_P=?$

The point P, lying on both lines 1 and 2, will have the velocities of 1 and 2 simultaneously.
It's velocity will be the resultant of the velocities.

Since the lines move in opposite directions, the angle between their velocities is $\pi-\alpha$. This can be seen by remembering that the angle between the perpendiculars of two lines is the same as the angle between them. This is shown in the diagram via the dotted arc $\alpha$. Being in opposite directions however, these perpendiculars are aligned at an angle supplementary to $\alpha$ or $\pi-\alpha$.

Hence, the velocity of the point of intersection $\boxed{v_P=\sqrt{v_1^2+v_2^2-2v_1v_2\cos\alpha}}\quad \Huge{\color{red}\times}$

The answer is wrong and doesn't agree with that of the text.

Request : Where am I mistaken with my reasoning?

 

[jbriggs444]

It's velocity will be the resultant of the velocities.

Have you ever squeezed a watermelon seed between your slowly moving fingers?

 

[brotherbobby]

Have you ever squeezed a watermelon seed between your slowly moving fingers?

No am afraid.

 

[jbriggs444]

No am afraid.

OK. :-) My midwestern upbringing led me to think that everyone had done that at some point. Watermelon seeds are fairly flat, sized like the tip of a finger and are slippery. If you squeeze them, they will squirt out from between your fingers and can be used in this manner to annoy one's siblings.

Let's skip the watermelon seeds. Think instead about an arrangement like a guillotine. An angled blade comes down toward a flat surface. Think about the point of intersection between blade and surface.

If the blade edge is nearly vertical, the point of intersection advances slowly.

If the blade edge is nearly horizontal, the point of intersection advances rapidly. More rapidly than the blade itself.

In the extreme, this sort of arrangement can be used to create super-luminal scissors.

 

[brotherbobby]

OK. :-) My midwestern upbringing led me to think that everyone had done that at some point.

Think instead about an arrangement like a guillotine. An angled blade comes down toward a flat surface. Think about the point of intersection between blade and surface.

If the blade edge is nearly vertical, the point of intersection advances slowly.

If the blade edge is nearly horizontal, the point of intersection advances rapidly. More rapidly than the blade itself.

Does it look like this?

Despite the image, I am struggling to follow your thinking.

The blade comes down with a certain speed, say $v_1$. The edge is at some angle to the horizontal. The most "advanced" point of the blade, the left, would connect with the surface first.

I don't see how that point of intersection with the surface should move with a speed different from the blade.

 

[jbriggs444]

View attachment 347897Does it look like this?

Despite the image, I am struggling to follow your thinking.

The blade comes down with a certain speed, say $v_1$. The edge is at some angle to the horizontal. The most "advanced" point of the blade, the left, would connect with the surface first.

I don't see how that point of intersection with the surface should move with a speed different from the blade.

Yes, this is what I have in mind. A good picture.

Let us say that the blade edge makes an angle of 26.5 degrees with the horizontal. The tangent of 26.5 degrees is about 0.5

Let us say that the blade is about 1 meter wide, post to post.

Let us say that the left edge is just touching a chosen horizontal line.

So the right edge of the blade is about 0.5 meters above that chosen line.

If the blade descends by 0.5 meters, where will the point of intersection be then?

In the time that it takes for the blade to descend by 0.5 meters, how far has the point of intersection moved?

What does this say about the velocity of the point of intersection compared to the velocity of the blade?

 

[brotherbobby]

Yes, this is what I have in mind. A good picture.

Let us say that the blade edge makes an angle of 26.5 degrees with the horizontal. The tangent of 26.5 degrees is about 0.5

Let us say that the blade is about 1 meter wide, post to post.

Let us say that the left edge is just touching a chosen horizontal line.

So the right edge of the blade is about 0.5 meters above that chosen line.

If the blade descends by 0.5 meters, where will the point of intersection be then?

In the time that it takes for the blade to descend by 0.5 meters, how far has the point of intersection moved?

What does this say about the velocity of the point of intersection compared to the velocity of the blade?

My drawing on the right. Let PQ be the edge of the blade, descending to P'Q'. I hope I have got the image right - I am considering the entirely blade to move straight down.

P will move to P', which is 0.5 m below. Q to Q', which would make Q touch the surface. The velocity of the point of intersection P is the same as that Q and that of the blade.

 

[Steve4Physics]

Homework Statement: Two intersecting straight lines move translationally in opposite directions with velocities $v_1$ and $v_2$ perpendicular to the corresponding lines. The angle between the lines is $\alpha$. Find the speed of the point of intersection of these lines.

Here’s a an alternative approach as a backup...

Without loss of generality, we can choose a coordinate system such that at time t=0:
Line-1 lies on the x-axis (y=0) and is moving in the +y direction with speed $v_1$.
Line-2 is $y = mx$ and moves in the direction perpendicular to it with speed $v_2$. Note that $m = \tan \alpha$.
So at time t=0, the point of intersection is is O(0,0).

At time t=1:
Where is line-1? What is its equation?
Where is line-2? What is its equation?
(It helps to draw the axes and lines.)

Find the coordinates of the new point of intersection (P).

The point of intersection has moved a distance OP in a time-interval of 1.

 

[jbriggs444]

View attachment 347898
My drawing on the right. Let PQ be the edge of the blade, descending to P'Q'. I hope I have got the image right - I am considering the entirely blade to move straight down.

P will move to P', which is 0.5 m below. Q to Q', which would make Q touch the surface. The velocity of the point of intersection P is the same as that Q and that of the blade.

The "point of intersection" is only momentarily P.
The "point of intersection" is also momentarily Q'.

The point of intersection moves from P to Q' while the blade moves from P to P' and from Q to Q'.

 

[brotherbobby]

The "point of intersection" is only momentarily P.
The "point of intersection" is also momentarily Q'.

The point of intersection moves from P to Q' while the blade moves from P to P' and from Q to Q'.

Got you and forgive me. Let's have the diagram again.

The blade moves down with a speed $u$. The intersection of the blade with the surface moves right with a speed $v$. From the diagram $\dfrac{PP'}{PQ'}=\tan\theta$, which implies, $PQ'=PP'\cot\theta$. Thus, $\boxed{v=u\cot\theta}$.

Since $\cot\theta=2$, the point of intersection moves twice as fast as the blade, with this inclination.

 

[jbriggs444]

View attachment 347900
Got you and forgive me. Let's have the diagram again.

The blade moves down with a speed $u$. The intersection of the blade with the surface moves right with a speed $v$. From the diagram $\dfrac{PP'}{PQ'}=\tan\theta$, which implies, $PQ'=PP'\cot\theta$. Thus, $\boxed{v=u\cot\theta}$.

Since $\cot\theta=2$, the point of intersection moves twice as fast as the blade, with this inclination.

Yes.

You should be a bit careful though. The scenario in the original problem is not exactly the same.

In the original problem, the speed of the moving line is measured perpendicular to itself. Here in the guillotine case, we are measuring in the vertical direction -- which is not perpendicular to the blade edge.

 

[brotherbobby]

In post #8 above, @Steve4Physics has given a method of approach that seems doable.

Yes, the guillotine case is a bit different. One line, the surface, is stationary, relative to the ground frame. That in and of itself is not a problem, but the relative velocity there would have to be found out between the lines and matched to that here.
The other line moves not perpendicular to itself, which makes it a different problem.

 

[brotherbobby]

Here’s a an alternative approach as a backup...

Without loss of generality, we can choose a coordinate system such that at time t=0:
Line-1 lies on the x-axis (y=0) and is moving in the +y direction with speed $v_1$.
Line-2 is $y = mx$ and moves in the direction perpendicular to it with speed $v_2$. Note that $m = \tan \alpha$.
So at time t=0, the point of intersection is is O(0,0).

At time t=1:
Where is line-1? What is its equation?
Where is line-2? What is its equation?
(It helps to draw the axes and lines.)

Find the coordinates of the new point of intersection (P).

The point of intersection has moved a distance OP in a time-interval of 1.

I'd proceed according to your suggestions presently. But I had a doubt to clarify.

In my original post (#1), I had assumed that the point of intersection $\color{red}P$ (shown as $\large{\color{red}\bullet}$) moves with the combined velocity of the two lines. This sounded plausible to me since $\color{red} P$ lies on both the lines and hence its velocity should be the vector sum of the velocity of the lines, implying $\vec{v}_P=\vec{v}_1+\vec{v}_2$. But it leads to the wrong answer.

Why is this not true?

 

[jbriggs444]

I'd proceed according to your suggestions presently. But I had a doubt to clarify.
View attachment 347903
In my original post (#1), I had assumed that the point of intersection $\color{red}P$ (shown as $\large{\color{red}\bullet}$) moves with the combined velocity of the two lines. This sounded plausible to me since $\color{red} P$ lies on both the lines and hence its velocity should be the vector sum of the velocity of the lines, implying $\vec{v}_P=\vec{v}_1+\vec{v}_2$. But it leads to the wrong answer.

Why is this not true?

One easy answer is because there is no reason that it should be true. It is an easy calculation. But not all easy calculations yield the desired answer.

Another reason that makes sense to me is that $\vec{v_1}$ and $\vec{v_2}$ are not an orthonormal basis for the vector space that they span.

Here is an explanation that might work...

The colored red point P must have a component of motion perpendicular to line 1 that matches $\vec{v_1}$. Otherwise, it will not stay on line 1.

The colored red point P must have a component of motion perpendicular to line 2 that matches $\vec{v_2}$. Otherwise, it will not stay on line 2.

If the lines were perpendicular to each other, the resultant, $\vec{v_1 + v_2}$ would indeed fit both criteria.

But if they are not perpendicular then $\vec{v_1}$ has a non-zero component perpendicular to line 2. If you add the two vectors together, the component perpendicular to line 2 will not be equal to $\vec{v_2}$. But as above, that perpendicular component must be equal to $\vec{v_2}$.

Similarly, the component of $\vec{v_1 + v_2}$ perpendicular to line 1 will not be right because of the non-zero contribution from $\vec{v_2}$. So the resultant is the wrong answer in that respect as well.

This is sort of why orthogonality is such a useful thing. You can mess about with one thing without worrying about the others.

 

[Steve4Physics]

... hence its velocity should be the vector sum of the velocity of the lines, implying $\vec{v}_P=\vec{v}_1+\vec{v}_2$. But it leads to the wrong answer.

Why is this not true?

Just to add to what @jbriggs444 said (Post #14), it might help if you imagine a particular case, where $v_1$ and $v_2$ are small and $\alpha$ is almost zero.

If you can imagine this, you will 'see' that the intersection point moves fast even though the speeds of the lines are slow. $\vec{v}_P=\vec{v}_1+\vec{v}_2$ makes no sense.

Minor edit.

 

[haruspex]

since $\color{red} P$ lies on both the lines and hence its velocity should be the vector sum of the velocity of the lines,

I do not see the logic of that. If the lines had the same velocity then their intersection would have that velocity too.
My approach was to parameterise the lines, like $(\hat p_iv_it+\hat q_i\lambda_i )\forall \lambda_i$, $i=1,2$, where the two unit vectors in each case are orthogonal and $\hat p_1\cdot\hat p_2=\cos(\alpha)$, find where the two lines intersect at time t and differentiate.

 

[brotherbobby]

If the lines were perpendicular to each other, the resultant, v1+v2→ would indeed fit both criteria.

Sorry for coming in late.

I had to draw in order to see how your statement above is true. Below are my workings for the orthogonal case.

Lines 1 and 2 have the same velocities and differing velocities but remain orthogonal. Their intersection point P moves with their combined velocities : $v_P=\vec v_1+\vec v_2$ in each case.

However, when lines 1 and 2 are non-orthogonal, their intersection P moves with a velocity that is not a vector sum of the velocities of the lines : $\vec v_P\ne \vec v_1+\vec v_2$. I drew similar diagrams and some calculations to convince myself that such is the case, the details of which are fairly involved and I won't bother you with them.

I still haven't solved the problem - far from it. I don't know what is the right way, only what is the wrong way.

I see an approach in post #8 above which looks promising. Get back to you soon.

 

[haruspex]

I see an approach in post #8 above which looks promising.

You can take that a bit further by working in the frame of one line so that there is only one velocity. Of course, you have to add back in the ignored velocity at the end.

 

[brotherbobby]

I have done the problem. To complete the thread, I'd like to post the solution. To refresh, let's have the problem statement from the text again.

Problem statement :

Solution : I draw the image of the problem alongside.

Lines 1 and 2, marked in blue and red, move with speeds $v_1$ and $v_2$ perpendicular to themselves, in opposite directions. The lines make an angle $\alpha$ between them. Line 1 goes to 1' and 2 goes to 2' after a time of 1 unit. The lines intersect at O at time $t=0$ and at P at time $t=1$. We have to find the velocity of this intersecting point. Since the time increment is 1, it amounts to finding the distance $\text{OP=?}$.

$\text{At time}\; t=0 :$ Line 1 has equation $y_1=0$ and line 2 $y_2=x\tan\alpha$.

$\text{At time}\; t=1 :$ The equation of line 1' is $\underline{y'_1=-v_1}$. But that of line 2' has to be found. We have $\text{OQ}=v_2$. In $\triangle \text{OQR}, \text{OR}=v_2\sec\alpha$. Hence the equation of line 2' : $\underline{y'_2=x\tan\alpha+v_2\sec\alpha}$.
These dotted lines intersect at P. The $y$ co-ordinate of P is $y_P=-v_1$, clearly. To find the $x$ co-ordinate, we put this value of $y$ in the equation of line 2' :
$-v_1=x_P\tan\alpha+v_2\sec\alpha\Rightarrow x_P=\dfrac{-v_1-v_2\sec\alpha}{\tan\alpha}$.
Hence the coordinates of the point P are : $P\rightarrow \left( \dfrac{-v_1-v_2\sec\alpha}{\tan\alpha}, -v_1 \right)$.
The distance OP can be found using the distance formula.
$\small{OP=\sqrt{v_1^2+\left( \dfrac{v_1+v_2\sec\alpha}{\tan\alpha}\right)^2}=\sqrt{v_1^2+\dfrac{v_1^2+2v_1v_2\sec\alpha+v_2^2\sec^2\alpha}{\tan^2\alpha}}=\dfrac{\sqrt{v_1^2\sec^2\alpha+2v_1v_2\sec\alpha+v_2^2\sec^2\alpha}}{\tan\alpha}}$, which simplifies to the velocity of the intersection point $\boxed{v_P=\dfrac{1}{\sin\alpha}\sqrt{v_1^2+v_2^2-2v_1v_2\cos\alpha}}\quad{\color{green}{\huge\checkmark}}$

The answer matches with that of the text.

 

[TSny]

Here's an approach using vectors.

Let $\vec v_p$ be the velocity of $P$ in the given frame of reference where line $1$ has velocity $\vec v_1$ and line $2$ has velocity $\vec v_2$.

$\vec v_p$ can be written as $\vec v_p = \vec v_1 + \vec v_{p/1}$, where $\vec v_{p/1}$ is the velocity of $p$ relative to line $1$.
$\vec v_1$ and $\vec v_{p/1}$ are perpendicular to each other.

Similarly, we can write $\vec v_p = \vec v_2 + \vec v_{p/2}$.

Thus, $$\vec v_1 + \vec v_{p/1} = \vec v_2 + \vec v_{p/2}.$$
Take components of this equation along the direction of $\vec v_2$ to get $$-v_1 \cos \alpha + v_{p/1} \sin \alpha = v_2.$$ Solve for $v_{p/1}$: $$v_{p/1} = \frac{v_2+v_1 \cos \alpha}{\sin \alpha}.$$
Then, $$v_p = \sqrt{v_1^2+v_{p/1}^2} = \left[v_1^2+\frac{(v_2+v_1\cos\alpha)^2}{\sin^2 \alpha} \right]^{1/2} = \frac{\sqrt{v_1^2+v_2^2+2v_1v_2 \cos \alpha}}{\sin \alpha}$$

 

[TSny]

$\small{OP=\sqrt{v_1^2+\left( \dfrac{v_1+v_2\sec\alpha}{\tan\alpha}\right)^2}=\sqrt{v_1^2+\dfrac{v_1^2+2v_1v_2\sec\alpha+v_2^2\sec^2\alpha}{\tan^2\alpha}}=\dfrac{\sqrt{v_1^2\sec^2\alpha+2v_1v_2\sec\alpha+v_2^2\sec^2\alpha}}{\tan\alpha}}$, which simplifies to the velocity of the intersection point $\boxed{v_P=\dfrac{1}{\sin\alpha}\sqrt{v_1^2+v_2^2-2v_1v_2\cos\alpha}}\quad{\color{green}{\huge\checkmark}}$

Looks like a typo in the final expression where the minus sign should be a plus sign.

 

[brotherbobby]

Looks like a typo in the final expression where the minus sign should be a plus sign.

Thank you. Yes, I am afraid that is a typo. Since it's been several hours since my post, I cannot change it. However, if an admin can enable an edit of the post (or the site moderator), I'd be grateful.

The least I can do is to mention the problem statement and write the correct answer to it.

Problem statement :




Correct answer : Speed of the point of intersection P : $\boxed{v_P=\dfrac{1}{\sin\alpha}\sqrt{v_1^2+v_2^2+2v_1v_2\cos\alpha}}$

 

[brotherbobby]

Here's an approach using vectors.

Let $\vec v_p$ be the velocity of $P$ in the given frame of reference where line $1$ has velocity $\vec v_1$ and line $2$ has velocity $\vec v_2$.

$\vec v_p$ can be written as $\vec v_p = \vec v_1 + \vec v_{p/1}$, where $\vec v_{p/1}$ is the velocity of $p$ relative to line $1$.
$\vec v_1$ and $\vec v_{p/1}$ are perpendicular to each other.
View attachment 348215

Similarly, we can write $\vec v_p = \vec v_2 + \vec v_{p/2}$.

Thus, $$\vec v_1 + \vec v_{p/1} = \vec v_2 + \vec v_{p/2}.$$
Take components of this equation along the direction of $\vec v_2$ to get $$-v_1 \cos \alpha + v_{p/1} \sin \alpha = v_2.$$ Solve for $v_{p/1}$: $$v_{p/1} = \frac{v_2+v_1 \cos \alpha}{\sin \alpha}.$$
Then, $$v_p = \sqrt{v_1^2+v_{p/1}^2} = \left[v_1^2+\frac{(v_2+v_1\cos\alpha)^2}{\sin^2 \alpha} \right]^{1/2} = \frac{\sqrt{v_1^2+v_2^2+2v_1v_2 \cos \alpha}}{\sin \alpha}$$

Brilliantly done! Easier and more compact than my method, not to mention a sense of maturity about it.
I suppose most problems of this kind should be tried using the superior approach of vectors than trigonometry and co-ordinate geometry.

 

[TSny]

I suppose most problems of this kind should be tried using the superior approach of vectors than trigonometry and co-ordinate geometry.

I first worked the problem similarly to your solution. Only after playing around for a while did I come upon the method I posted. Yes, working with vectors can sometimes help to avoid some algebra.

 

[Steve4Physics]

If of interest, here is a short, quite simple, solution. But it uses a fairly obscure property of certain cyclic quadrilaterals – ones in which a diagonal is a diameter of the circumcircle:

In this case: $\sin(\angle BAD) = \sin (\angle BCD) = \frac {BD}{AC} ~~~~(1)$

(Hint for anyone wanting to prove the above – use Ptolemy’s theorem.)
_______________________________

On the diagram below, initially the point of intersection (P) is at A.

After 1 unit of time the new point of intersection is at C. The blue line has advanced a distance $DC= v_1$ and the red line has advanced a distance $BC = v_2$. P has moved a distance $AC = v_P$ where $v_P$ is the speed of P.

Let $\alpha = \angle BAD$, then $\angle BCD =180^o - \alpha$.

Apply the cosine formula to triangle BCD:
$BD^2 = DC^2 + BC^2 - 2 \cdot DC \cdot BC \cos(\angle BCD)$

Substituting $DC = v_1, BC= v_2$ and $\angle BCD = 180^o – \alpha$ and using the fact that $\cos(\theta) = -\cos(180^0 - \theta)$:

$BD^2 = v_1^2 + v_2^2 + 2v_1v_2 \cos\alpha$

From equation (1): $\sin(\alpha) =\frac {BD}{AC} = \frac {BD}{v_P}$

$v_P = \frac {BD}{\sin{\alpha}} = \frac 1{\sin \alpha} \sqrt{v_1^2 + v_2^2 + 2v_1 v_2 \cos \alpha}$