Measuring the length of a moving Beam in two different directions

[kimiko333]

Homework Statement

A wagon is carrying a long, wooden beam with constant speed in one direction. If we go along with the wagon, we find, that the wooden beam is 15 steps long (it takes 15 steps to go along the beam while the wagon is moving). If we go against the wagon, we find that the beam is 10 steps long. How many steps long is the beam?

Relevant Equations

v=s/t

v_g+v_e=15/t_1
v_g-v_e=10/t_2
v_g=s/(t_1-t_2)

But there are too many unknowns. What am I missing?

 

[berkeman]

Welcome to PF.

What does it mean to "measure" the beam by walking in two different directions? Are you walking along the beam itself (in which case the two measurements will be the same)? Or are you walking along the ground pacing it off as the beam passes you in two different directions?

 

[kimiko333]

We are walking along on the ground

 

[berkeman]

We are walking along on the ground

Ah, okay, thanks. So if you assume each step takes 1 second, then it takes 15 seconds for the beam to pass you when you are walking in the same direction, and 10 seconds when you are walking in the opposite direction. What does that mean for how long it would take you if the beam were stationary with respect to the ground instead?

 

[kimiko333]

Now that's a very good question! :D Could you give me something that can help me reach the IDEA how to solve this? Thank you!

 

[PeroK]

But there are too many unknowns. What am I missing?

What about letting $L$ be the length of the beam, $u$ be the speed of the wagon and $v$ be the speed of the walker. Then we have as a starting point:$$L = (v - u)t_1 \ \text{and} \ L = (v+u)t_2$$

 

[kimiko333]

What about letting $L$ be the length of the beam, $u$ be the speed of the wagon and $v$ be the speed of the walker. Then we have as a starting point:$$L = (v - u)t_1 \ \text{and} \ L = (v+u)t_2$$

I restarted solving this problem, from the exact same approach, and ended up here. But still, we have too many unknowns and I don't know how to proceed further.

 

[PeroK]

I restarted solving this problem, from the exact same approach, and ended up here. But still, we have too many unknowns and I don't know how to proceed further.

You want the next step as well?

Okay, what's the relationship between $t_1$ and $t_2$?

 

[kimiko333]

You want the next step as well?

Okay, what's the relationship between $t_1$ and $t_2$?

Well. t₁ is the time that it takes for the walker to pass by the beam while moving with the beam. So this is the maximum time. While in the other case, t₂ is the minimum time it takes to pass by the beam. Now the connection I think is that in both cases we go with the same speed and the beam is traveling with its same speed as well. Only the distance ON THE GROUND changes. Maybe we can write, that
$u=\frac {L} {(t1-t2)},$
?

 

[kuruman]

Say you start at one end and, as soon as you reach the other end, you immediately turn back.

The key to this problem is in finding the ratio of the two speeds, v/u (I am using the notation suggested by @PeroK). Your round trip relative to the ground is 25 steps. Where do you end up relative to your starting point? That is the distance that the beam has covered in the same amount of time. So the ratio of the speeds is $\dots$ Once you have that ratio, you can find the answer using either the "go" trip time or the "return" trip time.

 

[PeroK]

Well. t₁ is the time that it takes for the walker to pass by the beam while moving with the beam. So this is the maximum time. While in the other case, t₂ is the minimum time it takes to pass by the beam. Now the connection I think is that in both cases we go with the same speed and the beam is traveling with its same speed as well. Only the distance ON THE GROUND changes. Maybe we can write, that
$u=\frac {L} {(t1-t2)},$
?

I don't see that.

Given the implicit assumption that the walker walks at constant speed with identical steps, then are the number of steps a measure of time?

 

[kimiko333]

The key to this problem is in finding the ratio of the two speeds, v/u (I am using the notation suggested by @PeroK). Your round trip relative to the ground is 25 steps. Where do you end up relative to your starting point? That is the distance that the beam has covered in the same amount of time. So the ratio of the speeds is $\dots$ Once you have that ratio, you can find the answer using either the "go" trip time or the "return" trip time.

How do we know that the round trip relative to the ground is 25 steps? (I'm not sure whether I understand what you mean by "round trip")

 

[kimiko333]

I don't see that.

Given the implicit assumption that the walker walks at constant speed with identical steps, then are the number of steps a measure of time?

At this point, I'm not so sure :D

 

[kuruman]

How do we know that the round trip relative to the ground is 25 steps? (I'm not sure whether I understand what you mean by "round trip")

I edited my post to explain that, but here it is anyway. You start at one end of the beam and move in the same direction as the cart. You take 15 steps to get to the front end. As soon as you are there, you turn back immediately and reach the end from which you started after taking another 10 steps. Your total trip is 25 steps.

 

[PeroK]

At this point, I'm not so sure :D

Well, it is a perfectly good way to measure time. Counting the walker's steps is as good a way to measure time as counting the swings of a pendulum or the oscillations of a quartz crystal or orbits of the Earth round the Sun! Do you see?

 

[kimiko333]

Well, it is a perfectly good way to measure time. Counting the walker's steps is as good a way to measure time as counting the swings of a pendulum or the oscillations of a quartz crystal or orbits of the Earth round the Sun! Do you see?

Oh, okay. That makes sense!

 

[kimiko333]

I edited my post to explain that, but here it is anyway. You start at one end of the beam and move in the same direction as the cart. You take 15 steps to get to the front end. As soon as you are there, you turn back immediately and reach the end from which you started after taking another 10 steps. Your total trip is 25 steps.

Great! Okay! Now I feel closer to the solution. Now I can imagine what's going on, and I was amazed by this solving strategy! Still. I have to figure out how to put that into equations.

 

[kuruman]

Great! Okay! Now I feel closer to the solution. Now I can imagine what's going on, and I was amazed by this solving strategy! Still. I have to figure out how to put that into equations.

Assume that you are moving with speed $v$, the beam with speed $u$ and that the beam's length is $L$. Can you find an expression for the time it takes from one end of the beam to the other? Look at post #6 by @PeroK.

 

[kimiko333]

Is it
$L=t_2*(u+v)$
and
$L=t_1*(v-u)$
?

 

[kimiko333]

Somehow I found out, that the ratio between v and u is 5
$\frac {v} {u} =5$

Is that correct? :D

 

[PeroK]

Somehow I found out, that the ratio between v and u is 5
$\frac {v} {u} =5$

Is that correct? :D

It is indeed!

 

[kimiko333]

It is indeed!

Greeaaat! Finally I found out something :D Now, if I understood correctly. Somehow, I should follow the instructions of the 10th post, right? (and the stress here is on "somehow" :D )

 

[PeroK]

Greeaaat! Finally I found out something :D Now, if I understood correctly. Somehow, I should follow the instructions of the 10th post, right? (and the stress here is on "somehow" :D )

You don't need @kuruman's clever ideas! Just substitute $u = \frac v 5$ into the appropriate equation.

 

[kimiko333]

You don't need @kuruman's clever ideas! Just substitute $u = \frac v 5$ into the appropriate equation.

Yeah, now I'm thinking about the appropriate equations. But I don't think those are the ones in 19th thread. Because if I substitute in, I get that v=0 :D.

 

[PeroK]

Yeah, now I'm thinking about the appropriate equations. But I don't think those are the ones in 19th thread. Because if I substitute in, I get that v=0 :D.

I don't see that at all. Why not sub $u = \frac v 5$ into:

$L=t_2*(u+v)$

 

[kimiko333]

I don't see that at all. Why not sub $u = \frac v 5$ into:

Oh, yeah. I forgot to mention that I supposed that t2 is equal to 15 and t1 is equal to 10

 

[PeroK]

Oh, yeah. I forgot to mention that I supposed that t2 is equal to 15 and t1 is equal to 10

That would be fine. It shouldn't lead to $v = 0$.

Hang on. It's $t_1 = 15, t_2 = 10$, surely?

 

[kimiko333]

That would be fine. It shouldn't lead to $v = 0$.

Hang on. It's $t_1 = 15, t_2 = 10$, surely?

This is how I'm thinking:
I. $L=t_1*(\frac {v} {5}+v)$
II. $L=t_2*(v- \frac {v} {5})$
$t_1=10$
$t_2=15$

I. $L=3$
II. $L=12$

Now I'm not so sure what this says to me in a physical sense. Why are there two solutions? I happen to know, that 12 is the correct answer (I had the answer sheet) but why are there two solutions?

 

[PeroK]

This is how I'm thinking:
I. $L=t_1*(\frac {v} {5}+v)$
II. $L=t_2*(v- \frac {v} {5})$
$t_1=10$
$t_2=15$

I. $L=3$
II. $L=12$

Now I'm not so sure what this says to me in a physical sense. Why are there two solutions? I happen to know, that 12 is the correct answer (I had the answer sheet) but why are there two solutions?

That's not two solutions. That's one solution and one mistake!

 

[kimiko333]

That's not two solutions. That's the solutuion and a mistake!

:O Where?

 

[PeroK]

:O Where?

How do you get $L = 3$?

 

[kuruman]

:O Where?

The length of the beam cannot be less than both 10 and 15 steps.

 

[kimiko333]

How do you get $L = 3$?

If I substitute 10 into the first equation, I get:
$L=t_1*(\frac {v} {5} +v)$
$L=2v+v=3v$

 

[PeroK]

If I substitute 10 into the first equation, I get:
$L=t_1*(\frac {v} {5} +v)$
$L=2v+v=3v$

I'm not impressed by that!

 

[kimiko333]

I'm not impressed by that!

but why is this so? I'm supposed to get 12. And I get 12 with the second equation

 

[PeroK]

but why is this so? I'm supposed to get 12. And I get 12 with the second equation

It's just totally wrong! You multiply something bigger that $v$ by $10$ and get $3v$? Come on!

 

[kimiko333]

It's just totally wrong! You multiply something bigger that $v$ by $10$ and get $3v$? Come on!

Oh Jesus. I forgot to multiply v by 10. Aaaahhhh. Told you... I'm tired :D

 

[kimiko333]

Now it's 12, okay. But if I substitute in, I get 12v, not just simply 12. How do I know, that the answer is in steps?

 

[PeroK]

Now it's 12, okay. But if I substitute in, I get 12v, not just simply 12. How do I know, that the answer is in steps?

That's the units you used when you set $t_1 = 10$. That's time in units of "time for a step". In those units $v = 1$ (meaning $1$ step per time for a step).

When you come to study Special Relativity, you'll find this is an amusing analogy to setting the speed of light $c = 1$!

Alternatively, you could have left time in unspecified units, whereby:

$L = \frac{6}{5}vt_1$

And $vt_1 = 10$ steps.

 

[kimiko333]

That's the units you used when you set $t_1 = 10$. That's time in units of "time for a step". In those units $v = 1$ (meaning $1$ step per time for a step).

When you come to study Special Relativity, you'll find this is an amusing analogy to setting the speed of light $c = 1$!

Alternatively, you could have left time in unspecified units, whereby:

$L = \frac{6}{5}vt_1$

And $vt_1 = 10$ steps.

Ooooh, okay. Totally makes sense! Great! Thank you! :D Now I just have to figure out, why is it that I can easily solve problems in electrodynamics, thermodynamics, etc, but don't know how to deal with these... :D (I'm a teacher-student, studying to be an English and Physics teacher)

 

[PeroK]

Ooooh, okay. Totally makes sense! Great! Thank you! :D Now I just have to figure out, why is it that I can easily solve problems in electrodynamics, thermodynamics, etc, but don't know how to deal with these... :D (I'm a teacher-student, studying to be an English and Physics teacher)

You started off with the assumption that the quantities could be determined. But, actually, the only things that could be determined were the relationships between quantities. You end up with 12 steps, but that's true whatever the speed of the walker and beam. All we've found is that one is 5 times the other - but, they could be anything.

This is the sort of problem that undermines the "plug-and-chug" approach - too many unknowns, you said. Well, everything is still an unknown, because we don't know how long a step is or how fast anything is moving.

Finally, there is a difference between a problem being elementary and being difficult. This is an elementary problem that really made you think. You might have an advanced EM problem that involves advanced concepts but only requires you to plug the given numbers into the correct formula - so it's advanced but easy.

 

[kimiko333]

You started off with the assumption that the quantities could be determined. But, actually, the only things that could be determined were the relationships between quantities. You end up with 12 steps, but that's true whatever the speed of the walker and beam. All we've found is that one is 5 times the other - but, they could be anything.

This is the sort of problem that undermines the "plug-and-chug" approach - too many unknowns, you said. Well, everything is still an unknown, because we don't know how long a step is or how fast anything is moving.

Finally, there is a difference between a problem being elementary and being difficult. This is an elementary problem that really made you think. You might have an advanced EM problem that involves advanced concepts but only requires you to plug the given numbers into the correct formula - so it's advanced but easy.

Great answers! Thank you for the explanations! :) I'm grateful!

 

[PeroK]

PS if the walker takes $s_1$ steps walking in the same direction as the beam is moving; and, $s_2$ steps walking in the opposite direction; then, the length of the beam in steps is $$\frac{2s_1s_2}{s_1 + s_2}$$
We can check that for $s_1 = 15$ and $s_2 = 10$, we do indeed get the answer of $12$ steps.

In general, I'm rarely content to solve a single numerical problem. I usually want to find a general formula like the one above.

 

[Steve4Physics]

I know it’s a bit late, but I think there are 2 possible solutions depending on which is faster, the walker or the wagon. The original question seems to allow both possibilities.

Take the unit of distance as 1-step and the unit of time as the time-for-1-step.
Speed of walker = 1.
Speed of wagon = v.

There are 2 cases:
Case 1; wagon is faster than walker: v>1
Case 2: wagon is slower than walker: v<1
___________________________

Case 1; wagon is faster than walker: v>1

When walker and beam move in the same direction:
Front of beam passes observer at $t=0$
Back of beam passes observer at $t= t_1$
$t_1 = \frac {L}{v-1}$ (since speed of walker relative to beam is v-1.)

When walker and beam move in opposite directions:
$t_2 = \frac {L}{v+1}$ (since speed of walker relative to beam is v+1.)

Eliminating v gives:
$L = \frac {2t_1t_2}{t_1 – t_2}$

Since the number of steps is the same as the number of time units:
$L = \frac{2s_1s_2}{s_1 – s_2}$

For $s_1 = 15$ and $s_2 = 10$
$L = \frac {2·15·10}{15 – 10} = 60$
____________________________________

Case 2: wagon is slower than walker: v<1

Observer passes back of beam at $t = 0$
Observer passes front of beam at $t = t_1$
$t_1 = \frac {L}{1-v}$ (since speed of walker relative to beam is 1-v.)

When walker and beam move in the opposite directions:
$t_2 = \frac {L}{v+1}$ (since speed of walker relative to beam is v+1.)

Eliminating v gives:
$L = \frac {2t_1t_2}{t_1 + t_2}$

Since the number of steps is the same as the number of time units:
$L = \frac{2s_1s_2}{s_1 + s_2}$

For $s_1 = 15$ and $s_2 = 10$
$L = \frac {2·15·10}{15 + 10} = 12$

 

[kuruman]

Good thinking outside the proverbial box. The statement of the problem invokes the stereotypical picture of a fast walker catching up with a slow-moving (ox)cart. It would have been different if the beam were on a flatbed truck on the highway. Either formulation would have two solutions.