How clever are you with mechanical stuff etc
#61
More importantly - Did you know that there is a little red x below your name and if you think it is appropriate to call people 'thickies' someone will press it and you will disappear. (OK - I am kidding, but there are a few sensitive (and thick) souls on here who might actually miss your smiley)
Anyway - Can you explain the 250kg masses with pulleys in terms of potential and kinetic energy and moments of inertia, cos I am 100% certain that it would sit tight until it go a push
It's actually nothing to do with energy or inertia. The easiest way to see it is to cut the problem in two straight down the middle. Imagine you cut the rope in the middle of the horizontal section and are left holding each end. On the right you have one pulley and a 250kg weight, on the left you have 2 pulleys and a 250 kg weight.
Obviously your right hand would be experiencing the full 250kg whereas your left hand would be experiencing only 125kg. If you then tied the ropes together again you would have 125kg difference acting to the right and the right hand weight would fall to the ground, raising the left hand weight by half the distance.
#62
It doesn't all necessarily follow the path of least resistance though. You know when the wife tells you you are going shopping. You can either agree (POLR*) or look less than overly excited (PTID**). Now if you changed the non stated background information to say that you had left her at the altar, you would have an option for stay at home drinking beer and watching ****, which would in this case be the new POTR*
*POLR - Path of least resistance
** PTID - Path terminating in death
*POLR - Path of least resistance
** PTID - Path terminating in death
#64
The thickies are welcome to click it all they want... they might find they get more than they bargained for.
It's actually nothing to do with energy or inertia. The easiest way to see it is to cut the problem in two straight down the middle. Imagine you cut the rope in the middle of the horizontal section and are left holding each end. On the right you have one pulley and a 250kg weight, on the left you have 2 pulleys and a 250 kg weight.
Obviously your right hand would be experiencing the full 250kg whereas your left hand would be experiencing only 125kg. If you then tied the ropes together again you would have 125kg difference acting to the right and the right hand weight would fall to the ground, raising the left hand weight by half the distance.
It's actually nothing to do with energy or inertia. The easiest way to see it is to cut the problem in two straight down the middle. Imagine you cut the rope in the middle of the horizontal section and are left holding each end. On the right you have one pulley and a 250kg weight, on the left you have 2 pulleys and a 250 kg weight.
Obviously your right hand would be experiencing the full 250kg whereas your left hand would be experiencing only 125kg. If you then tied the ropes together again you would have 125kg difference acting to the right and the right hand weight would fall to the ground, raising the left hand weight by half the distance.
2. - I am nearly tempted to click it and see
#65
There is of course in reality a small consideration that must be made... for the question we assume the mass of the rope and the pulleys is insignificant. Obviously if you use very heavy pulleys and/or chain instead of rope then you must take this into account!
#69
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The answer is 50, not 51; if you applied a force of 51g N, not only would you be lifting the weight, you'd be accelerating it upwards at a rate of 1/50g, which is more than needed to meet the definition of 'lift'.
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Try it and see - It won't be slightly lit even if you have infrared night vision owl goggles on. The wire in the spur will have no more resistance than the wire leading to the spur (unless it is a lower grade of wire, which isn't specified) so the spur will be able to cope with 100% of the load.
Normally, at this most basic level of electronics, wires would be considered ideal, ie. zero resistance. Therefore, there can be zero potential across the bulb and, therefore, zero current through it.
Consider the wire to have finite resistance, and it's a 3 step process to understand what's going on:
- work out the combined parallel resistance of the wire and the bulb (Rp = 1/ (1/Rw + 1/Rb))
- from that, work out the potential across the wire//bulb combination (Vp = Vo (Rp / (2* Rb + Rp))
- then you can work out the current through the shorted bulb (Ib = Vp / Rb)
There are mathematical simplifications, of course, but this method describes the behaviour of the system without having to resort to misconceptions like current 'choosing' to take a particular route, or a wire 'coping' with a load.
You could also work it out by considering current:
I = Vo / (2* Rb + Rp)
Vp = I * Rp
Therefore, Ib = Vp / Rb
*sigh* it's too early in the morning for this... I do this for a living and my coffee hasn't even soaked in yet
#72
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It's nothing to do with 'coping' with the load, though. That's a concept used by electricians worrying about heat, fire and long runs of cable - it has no real meaning here unless you're specifically told that the current is high or the wire is very thin.
Normally, at this most basic level of electronics, wires would be considered ideal, ie. zero resistance. Therefore, there can be zero potential across the bulb and, therefore, zero current through it.
Consider the wire to have finite resistance, and it's a 3 step process to understand what's going on:
- work out the combined parallel resistance of the wire and the bulb (Rp = 1/ (1/Rw + 1/Rb))
- from that, work out the potential across the wire//bulb combination (Vp = Vo (Rp / (2* Rb + Rp))
- then you can work out the current through the shorted bulb (Ib = Vp / Rb)
There are mathematical simplifications, of course, but this method describes the behaviour of the system without having to resort to misconceptions like current 'choosing' to take a particular route, or a wire 'coping' with a load.
You could also work it out by considering current:
I = Vo / (2* Rb + Rp)
Vp = I * Rp
Therefore, Ib = Vp / Rb
*sigh* it's too early in the morning for this... I do this for a living and my coffee hasn't even soaked in yet
Normally, at this most basic level of electronics, wires would be considered ideal, ie. zero resistance. Therefore, there can be zero potential across the bulb and, therefore, zero current through it.
Consider the wire to have finite resistance, and it's a 3 step process to understand what's going on:
- work out the combined parallel resistance of the wire and the bulb (Rp = 1/ (1/Rw + 1/Rb))
- from that, work out the potential across the wire//bulb combination (Vp = Vo (Rp / (2* Rb + Rp))
- then you can work out the current through the shorted bulb (Ib = Vp / Rb)
There are mathematical simplifications, of course, but this method describes the behaviour of the system without having to resort to misconceptions like current 'choosing' to take a particular route, or a wire 'coping' with a load.
You could also work it out by considering current:
I = Vo / (2* Rb + Rp)
Vp = I * Rp
Therefore, Ib = Vp / Rb
*sigh* it's too early in the morning for this... I do this for a living and my coffee hasn't even soaked in yet
thats what you meant to say wasnt it fb
#73
74 % and I know which ones I answered wrong before I even pressed continue! D'oh!
I think its sucked in too!! haha. I remember my old cortina didnt have induction like these new fangled cars today, and it made a sucking noise in the carb!!!!!! Ever stuck your hand over the top of a carb????
I think its sucked in too!! haha. I remember my old cortina didnt have induction like these new fangled cars today, and it made a sucking noise in the carb!!!!!! Ever stuck your hand over the top of a carb????
#74
From Wikipedia:
"Suction is the creation of a partial vacuum, or region of low pressure. The pressure gradient between this region and the ambient pressure will propel matter toward the low pressure area. Physicists consider the notion of "suction" to be specious, since vacuums do not innately attract matter. Dust being "sucked" into a vacuum cleaner is actually being pushed in by the higher pressure air on the outside of the cleaner.
Higher pressure of surrounding air can push matter into a vacuum but a vacuum cannot attract matter."
"Suction is the creation of a partial vacuum, or region of low pressure. The pressure gradient between this region and the ambient pressure will propel matter toward the low pressure area. Physicists consider the notion of "suction" to be specious, since vacuums do not innately attract matter. Dust being "sucked" into a vacuum cleaner is actually being pushed in by the higher pressure air on the outside of the cleaner.
Higher pressure of surrounding air can push matter into a vacuum but a vacuum cannot attract matter."
#76
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I still don't get the water going through pipe with two tubes coming off A and B. I just cannot work out why the fluid goes higher in A, when the restriction is in B.
Anyone have a nice scientific explanation, not a gut feeling one
Anyone have a nice scientific explanation, not a gut feeling one
#78
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You are of course correct that 50kg is not a force, I was answering in the style of the question. However, the definition of lift is "raise to a higher level", a force of 50g N would be insufficient to achieve this effect HTH
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I wasn't 100% sure on this one, but I think the explanation is:
- the rate of flow in terms of mass per second is the same all along the pipe (obviously),
- in the restricted section it's moving faster
- therefore, the pressure is less
- therefore, it can't support such a high column of liquid.
Why is the pressure less? It has to be, in order to accelerate the water from the low speed in the wide section of pipe, up to the high speed needed to maintain the mass flow rate in the restricted section. Just imagine the forces acting on a water molecule at the start of the restriction - why should it accelerate? A: there's a pressure difference acting on it from left to right.
- the rate of flow in terms of mass per second is the same all along the pipe (obviously),
- in the restricted section it's moving faster
- therefore, the pressure is less
- therefore, it can't support such a high column of liquid.
Why is the pressure less? It has to be, in order to accelerate the water from the low speed in the wide section of pipe, up to the high speed needed to maintain the mass flow rate in the restricted section. Just imagine the forces acting on a water molecule at the start of the restriction - why should it accelerate? A: there's a pressure difference acting on it from left to right.
#81
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I was thinking the pressure inside the constricted pipe increased due to its narrow width. The pressure will be lower after the constriction right? As the pipe after the constriction is bigger and therefore you don't need as much pressure to cope with the flow rate being supplied.
I applied that principle to the pressure in the pipe where tube A is.
#82
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I was thinking the pressure inside the constricted pipe increased due to its narrow width. The pressure will be lower after the constriction right? As the pipe after the constriction is bigger and therefore you don't need as much pressure to cope with the flow rate being supplied.
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Hmmm, more food for thought. Just had this pipe thing explained by my dad. He took one look at it, before I had even finished explaining, and said
1) If its a static system then A and B are the same height
2) If there is flow, which there is, then the constriction is a red herring. The pressure at the start of the pipe must be the highest in order to get the water to flow. The pressure falls as you travel along the pipe. The rate of fall of pressure is greatest over the restriction, that is all. Therefore as the pressure is highest at the start of the pipe, A is higher.
Seems damn simple if he is right
1) If its a static system then A and B are the same height
2) If there is flow, which there is, then the constriction is a red herring. The pressure at the start of the pipe must be the highest in order to get the water to flow. The pressure falls as you travel along the pipe. The rate of fall of pressure is greatest over the restriction, that is all. Therefore as the pressure is highest at the start of the pipe, A is higher.
Seems damn simple if he is right
#85
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