# Measuring the length of a moving Beam in two different directions

• kimiko333
In summary, the relationship between t1 and t2 is that t1 is the time it takes for the walker to pass by the beam while moving with the beam, and t2 is the minimum time it takes to pass by the beam. However, because the distance the walker travels on the ground changes with each trip, the two times are not always the same.
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?

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?

We are walking along on the ground

kimiko333 said:
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?

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!

kimiko333 said:
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:$$L = (v - u)t_1 \ \text{and} \ L = (v+u)t_2$$

PeroK said:
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.

kimiko333 said:
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##?

PeroK said:
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)},##
?

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.

kimiko333 said:
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?

kuruman said:
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")

PeroK said:
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

kimiko333 said:
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.

kimiko333 said:
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?

PeroK said:
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!

kuruman said:
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 and PeroK
kimiko333 said:
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.

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

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

Is that correct? :D

PeroK
kimiko333 said:
Somehow I found out, that the ratio between v and u is 5
##\frac {v} {u} =5##

Is that correct? :D
It is indeed!

PeroK said:
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 )

kimiko333 said:
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.

PeroK said:
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.

kimiko333 said:
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:
kimiko333 said:
##L=t_2*(u+v)##

PeroK said:
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

kimiko333 said:
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?

PeroK said:
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?

kimiko333 said:
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!

PeroK said:
That's not two solutions. That's the solutuion and a mistake!
:O Where?

kimiko333 said:
:O Where?
How do you get ##L = 3##?

kimiko333 said:
:O Where?
The length of the beam cannot be less than both 10 and 15 steps.

PeroK said:
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
kimiko333 said:
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!

PeroK said:
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

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