You good with math?
Originally posted by SurfnSoCal@August 19, 2005, 12:01 PM
Answer: There is the implicit assumption here that the top of the ladder remains resting against the wall. However, that is not always true. Once the ladder has reached a sufficiently small angle to the horizontal, your pulling of the bottom away from the wall will actually cause the top to pull away from the wall too. When this happens, there is no longer the relationship , because x, y, and L no longer form the sides of a closed right triangle.
good job bro, you got it.
Answer: There is the implicit assumption here that the top of the ladder remains resting against the wall. However, that is not always true. Once the ladder has reached a sufficiently small angle to the horizontal, your pulling of the bottom away from the wall will actually cause the top to pull away from the wall too. When this happens, there is no longer the relationship , because x, y, and L no longer form the sides of a closed right triangle.
good job bro, you got it.
rrobello, the ladder does not get extended, the triangle's hypotenuse will stay the same length. The angles will change as the ladder is drawn away from the wall. You can see this because y= (L²-x²)^1/2 is the equation for a circle. The sum of the 2 changing angles will always be 90°. Essentially, the velocity of the ladder falling against the wall will approach zero, not infinity.
Shelby GT350 Member
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Originally posted by Enfynet@August 19, 2005, 11:20 AM
It's all relative though. The ladder WILL be horizontal when the top starts to move away from the wall unless your given velocity is enough to overcome the acceleration due to gravity. The assumption made in that answer is that the ladder is being pulled rapidly away from the wall. What if it was moving at ~1mm/sec? Against the frictionless wall, the ladder would lay against the ground before it moved away from the wall.
It's all relative though. The ladder WILL be horizontal when the top starts to move away from the wall unless your given velocity is enough to overcome the acceleration due to gravity. The assumption made in that answer is that the ladder is being pulled rapidly away from the wall. What if it was moving at ~1mm/sec? Against the frictionless wall, the ladder would lay against the ground before it moved away from the wall.
Eh...not entirely true. its not to "overcome the acceleration due to gravity" but rather once the force pushing the top of the ladder against the wall becomes less than the force with which you are sliding the bottom of the ladder away from the wall, your force will prevail and you will pull the ladder away from the wall. Once this happens, the top will be in free-fall and no longer governed by the same set of equations.
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Originally posted by Enfynet@August 19, 2005, 12:20 PM
rrobello, the ladder does not get extended, the triangle's hypotenuse will stay the same length. The angles will change as the ladder is drawn away from the wall. You can see this because y= (L²-x²)^1/2 is the equation for a circle. The sum of the 2 changing angles will always be 90°. Essentially, the velocity of the ladder falling against the wall will approach zero, not infinity.
rrobello, the ladder does not get extended, the triangle's hypotenuse will stay the same length. The angles will change as the ladder is drawn away from the wall. You can see this because y= (L²-x²)^1/2 is the equation for a circle. The sum of the 2 changing angles will always be 90°. Essentially, the velocity of the ladder falling against the wall will approach zero, not infinity.
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Originally posted by SurfnSoCal@August 19, 2005, 12:42 PM
Eh...not entirely true. its not to "overcome the acceleration due to gravity" but rather once the force pushing the top of the ladder against the wall becomes less than the force with which you are sliding the bottom of the ladder away from the wall, your force will prevail and you will pull the ladder away from the wall. Once this happens, the top will be in free-fall and no longer governed by the same set of equations.
Eh...not entirely true. its not to "overcome the acceleration due to gravity" but rather once the force pushing the top of the ladder against the wall becomes less than the force with which you are sliding the bottom of the ladder away from the wall, your force will prevail and you will pull the ladder away from the wall. Once this happens, the top will be in free-fall and no longer governed by the same set of equations.
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Originally posted by Enfynet@August 18, 2005, 2:16 PM
I've got a couple math-riddles too.
Here's one that is appropriate with my username...
++Hotel Infinity:
One of the qualifications for a hotel clerk at Hotel Infinity is a working knowledge of infinity. Paul applied, was interviewed, and started work the following evening. He wondered why the hotel required that all its clerks know about infinity, infinite sets, and transinfinite numbers. He figured since it was an infinitely roomed hotel it would be no problem finding rooms for its guests. After his first night on the job, he was glad he had that knowledge.
When he relieved the day clerk, she informed him that there were an infinite number of rooms presently occupied. As she left, a new guest walked in with a reservation. He needed to decide which room to give the guest. He thought for a moment, then moved each occupant to a room with the next highest number, and was therefore able to vacate room #1. He felt good about his solution, but just then an infinite bus load of new guests arrived. How would he give then their rooms?
Ok I am not sure if I get this, is it a real question and is it a math one or a riddle? my best guess would be he would give them their rooms one by one or something like that, I dunno.
I've got a couple math-riddles too.
Here's one that is appropriate with my username...
++Hotel Infinity:
One of the qualifications for a hotel clerk at Hotel Infinity is a working knowledge of infinity. Paul applied, was interviewed, and started work the following evening. He wondered why the hotel required that all its clerks know about infinity, infinite sets, and transinfinite numbers. He figured since it was an infinitely roomed hotel it would be no problem finding rooms for its guests. After his first night on the job, he was glad he had that knowledge.
When he relieved the day clerk, she informed him that there were an infinite number of rooms presently occupied. As she left, a new guest walked in with a reservation. He needed to decide which room to give the guest. He thought for a moment, then moved each occupant to a room with the next highest number, and was therefore able to vacate room #1. He felt good about his solution, but just then an infinite bus load of new guests arrived. How would he give then their rooms?
Ok I am not sure if I get this, is it a real question and is it a math one or a riddle? my best guess would be he would give them their rooms one by one or something like that, I dunno.
Well, when the 1st guest arrived, he had everyone move to the next highest number. So when the infinite number of guests arrived, he had all the current guests double their room number, freeing up all the odd numbered rooms. 
It's a math-riddle...
---------
A farmer has to get his goat, wolf, and cabbage across a river. His boat will only hold him and one other. Assuming the wolf will eat the goat, and the goat will eat the cabbage, how does he get all of them across?
It's a math-riddle...
---------
A farmer has to get his goat, wolf, and cabbage across a river. His boat will only hold him and one other. Assuming the wolf will eat the goat, and the goat will eat the cabbage, how does he get all of them across?
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Originally posted by Enfynet@August 19, 2005, 4:12 PM
Well, when the 1st guest arrived, he had everyone move to the next highest number. So when the infinite number of guests arrived, he had all the current guests double their room number, freeing up all the odd numbered rooms.
It's a math-riddle...
---------
A farmer has to get his goat, wolf, and cabbage across a river. His boat will only hold him and one other. Assuming the wolf will eat the goat, and the goat will eat the cabbage, how does he get all of them across?
Well, when the 1st guest arrived, he had everyone move to the next highest number. So when the infinite number of guests arrived, he had all the current guests double their room number, freeing up all the odd numbered rooms.
It's a math-riddle...
---------
A farmer has to get his goat, wolf, and cabbage across a river. His boat will only hold him and one other. Assuming the wolf will eat the goat, and the goat will eat the cabbage, how does he get all of them across?
hmmm that was interesting couldnt he have just done pretty much anything though since there are an infinite number of rooms????
the next one the farmer first takes the goat, then comes back and gets the cabbage, takes that across and brings the goat back with him to get the wolf, takes the wolf across and leaves it with the cabbage and then goes back for the goat and crosses yet again thus getting all of them across.
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here is a real easy one (I dont really care for those find whats wrong with the proof since the whole premise is always wrong to begin with the questions never hold water and are just wrong in and of themselves to begin with, but always fun) so here is an easy one like I said to get us motivated again.....
A number of children are standing in a circle. They are evenly spaced and the 5th child is directly opposite the 20th child. How many children are there altogether?
A number of children are standing in a circle. They are evenly spaced and the 5th child is directly opposite the 20th child. How many children are there altogether?
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told you it was easy, here is another, still pretty easy but a little bit more complex, not much though....
A group of four people have to cross a bridge. It is dark, and they have to light the path with a flashlight. No more than two people can cross the bridge simultaneously, and the group has only one flashlight. It takes different time for the people in the group to cross the bridge:
Annie crosses the bridge in 1 minute,
Bob crosses the bridge in 2 minutes,
Chris crosses the bridge in 5 minutes,
Dorothy crosses the bridge in 10 minutes.
How can the group cross the bridge in 17 minutes?
A group of four people have to cross a bridge. It is dark, and they have to light the path with a flashlight. No more than two people can cross the bridge simultaneously, and the group has only one flashlight. It takes different time for the people in the group to cross the bridge:
Annie crosses the bridge in 1 minute,
Bob crosses the bridge in 2 minutes,
Chris crosses the bridge in 5 minutes,
Dorothy crosses the bridge in 10 minutes.
How can the group cross the bridge in 17 minutes?
I'm assuming they cannot cross the bridge without the flashlight...
Hmm... That sounds like
Ax + By + Cz + Dw = 17
1x + 2y + 5z + 10w = 17
x,y,z,w must be non-zero positive integers.
w has to be 1 or it would take too long.
x + 2y + 5z + 10 = 17
x + 2y + 5z = 7
:scratch: Unless Annie only takes people partway across before turning around and going back, it doesn't look possible.
---
It could work if they're on opposite sides of the bridge... but that doesn't match the premis.
Hmm... That sounds like
Ax + By + Cz + Dw = 17
1x + 2y + 5z + 10w = 17
x,y,z,w must be non-zero positive integers.
w has to be 1 or it would take too long.
x + 2y + 5z + 10 = 17
x + 2y + 5z = 7
:scratch: Unless Annie only takes people partway across before turning around and going back, it doesn't look possible.
---
It could work if they're on opposite sides of the bridge... but that doesn't match the premis.
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Originally posted by Enfynet@August 19, 2005, 7:44 PM
I'm assuming they cannot cross the bridge without the flashlight...
Hmm... That sounds like
Ax + By + Cz + Dw = 17
1x + 2y + 5z + 10w = 17
x,y,z,w must be non-zero positive integers.
w has to be 1 or it would take too long.
x + 2y + 5z + 10 = 17
x + 2y + 5z = 7
:scratch: Unless Annie only takes people partway across before turning around and going back, it doesn't look possible.
---
It could work if they're on opposite sides of the bridge... but that doesn't match the premis.
I'm assuming they cannot cross the bridge without the flashlight...
Hmm... That sounds like
Ax + By + Cz + Dw = 17
1x + 2y + 5z + 10w = 17
x,y,z,w must be non-zero positive integers.
w has to be 1 or it would take too long.
x + 2y + 5z + 10 = 17
x + 2y + 5z = 7
:scratch: Unless Annie only takes people partway across before turning around and going back, it doesn't look possible.
---
It could work if they're on opposite sides of the bridge... but that doesn't match the premis.
How is that making it too complicated? All my equations are is Times per person & number of times across = total time. If only 2 people can cross at a time, someone has to bring the flashlight back to the next group of people. If they didn't need the flashlight it would take them 10 minutes to cross. D & C leave together, when C reaches the other side B would leave, then when B crosses A would leave and then they'd wait for D. Otherwise A would carry the flashlight and cross with each person... unless the flashlight can magically transport from one side of the bridge to the other. Am I missing something in the premis that would make it possible for the flashlight to get from one person to another without going back across the bridge?
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Because it's not so much a math puzzle as a logic puzzle.
There's a flaw in your reasoning as well...
You wrote:
x + 2y + 5z + 10 = 17
x + 2y + 5z = 7
x + 2y + 5z does not equal 7. Keep in mind when D crosses someone is travelling with her. If you plan it right, that someone does not have to cross the bridge more than once.
Which person would you pick for that?
Once you have that, figure out how many times you can get the other two back and forth within seven minutes.
There's a flaw in your reasoning as well...
You wrote:
x + 2y + 5z + 10 = 17
x + 2y + 5z = 7
x + 2y + 5z does not equal 7. Keep in mind when D crosses someone is travelling with her. If you plan it right, that someone does not have to cross the bridge more than once.
Which person would you pick for that?
Once you have that, figure out how many times you can get the other two back and forth within seven minutes.
:scratch: If A and D cross, that's 10+1 minutes, because A has to take the flashlight back to B and C. If B and D cross it's 10+2... Unless they don't need the flashlight to cross the whole bridge. It works the other way as well, because the flashlight needs to cross the bridge with each pair... The only way to get 17 is to have the flashlight transport itself back across the bridge, otherwise getting D across takes at least 11 minutes. If you have C and D cross together it will only take 10 minutes, but it would take at least 5 more to bring the flashlight back... Unless I'm reading the problem wrong, it can't be done in less than 19...
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I figured it out!
A&B cross together -2min
B goes back with flaslight-4min
C&D cross -14 min
A goes back with light- 15 min
A&B cross together- 17 min
Man that was a real good one, the fun part now is playing it on family members who feel smart lol.
A&B cross together -2min
B goes back with flaslight-4min
C&D cross -14 min
A goes back with light- 15 min
A&B cross together- 17 min
Man that was a real good one, the fun part now is playing it on family members who feel smart lol.
