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Correct, your turn!mjcox2000 wrote: 2 wedges, 1 second class lever, 2 third class levers. (That dotted line is a pin connected to the bottom 3rd class lever, going through the top one, and connected to the 2nd class lever.) There's a spring keeping the two 3rd class levers apart - that isn't a simple machine, but it's part of the resistance, so I thought I'd mention it. Effort is applied at the long end of the 2nd class lever and the vertex of the two 3rd class levers.
AMA is the actual mechanical advantage delivered by the machine defined as [math]\frac{\mathrm{F_{out}}}{\mathrm{F_{in}}}[/math]. IMA is the theoretical maximum mechanical advantage able to be delivered by a machine in a frictionless environment defined as [math]\frac{\mathrm{d_{in}}}{\mathrm{d_{out}}}[/math]. One may want to know IMA instead of AMA because IMA gives information about the geometry of the simple machine which is useful in doing theoretical studies or replicating a simple machine. One may want to know AMA instead of IMA when doing work with an actual simple machine and its performance is necessary in determining an unknown quantity (energy problems involving a ramp for example)That sounds right to me, you should go ahead and ask the next question.finagle29 wrote:AMA is the actual mechanical advantage delivered by the machine defined as [math]\frac{\mathrm{F_{out}}}{\mathrm{F_{in}}}[/math]. IMA is the theoretical maximum mechanical advantage able to be delivered by a machine in a frictionless environment defined as [math]\frac{\mathrm{d_{in}}}{\mathrm{d_{out}}}[/math]. One may want to know IMA instead of AMA because IMA gives information about the geometry of the simple machine which is useful in doing theoretical studies or replicating a simple machine. One may want to know AMA instead of IMA when doing work with an actual simple machine and its performance is necessary in determining an unknown quantity (energy problems involving a ramp for example)
finagle29 wrote:How are a machine's Q-factor and efficiency related, and how do these relate to whether or not it is self-locking? Additionally, for each type of simple machine, state whether or not, in its normal mode of use, it would be advantageous for it to be self-locking.
Q-factor is pt much how an oscillator maintains its stored energy over time, so higher Q-factor should correlate to higher efficiency (idk if there's a mathematical relationship that establishes a clear correlation), although considering how simple machines aren't really oscillators idk how these actually relate. The self-locking property occurs when efficiency is at 50%, which should correlate to pretty low Q-factor and efficiency. The only machines I think could possibly have an advantage from self-locking are screws, inclined planes, wedges, and gears... Hope that's everything
but aren't pulleys symmetrical? Therefore, you could flip the image and get an efficiency of ~111%.JonB wrote:I can best explain this with an example:Unome wrote: How does efficiency in pulleys work anyway? My logic (which may not make sense since the system is stable) was that as the rope is pulled through the pulley, it loses 10% of its tension, so the right side ropes need more tension than the left side.
In the following image, the left pulley is assumed to be 100% efficient and the right pulley 1% efficient (some arbitrary number showing great inefficiency)
You can clearly see that the left pulley would only remain stationary if the two masses are equal; this isn't the case with the right pulley. For example, if the source of the right pulley's inefficiency was rust (illustrated rather horribly in Paint), then it is reasonable to assume that the right pulley could remain stable as shown, with unequal masses on either side.
That is the basis for my reasoning that pulley inefficiency does carry over into the stationary case.