Simple Machines B/Compound Machines C

[JonB]

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.

I can best explain this with an example:

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.

paintshop.png (6.08 KiB) Viewed 3936 times

[Unome]

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.

I can best explain this with an example:

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.

paintshop.png

Ok... I sort of get it. Anyway, the first person that gets here can ask a question.

[UTF-8 U+6211 U+662F]

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.

I can best explain this with an example:

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.

paintshop.png

Ok... I sort of get it. Anyway, the first person that gets here can ask a question.

Extremely simple question:
What is the "law of the lever?"

[jkang]

What is the "law of the lever?"

d1F1 = d2F2, where d = distance from the fulcrum and F = force applied

[UTF-8 U+6211 U+662F]

What is the "law of the lever?"

d1F1 = d2F2, where d = distance from the fulcrum and F = force applied

Remember to hide your answer!

Answer (Click to Show Hidden Text)

[math]d_1F_1 = d_2F_2[/math]

Your turn!

[jkang]

Whoops. Sorry about that.
Another pretty simple question: Identify all of the simple machines in a nail clipper (include different classes of levers, if applicable).

[mjcox2000]

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.

[jkang]

Nail clipper diagram.jpg

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.

Correct, your turn!

[mjcox2000]

How do actual mechanical advantage and ideal mechanical advantage differ? When would one want to know IMA instead of AMA, and vice versa? (This wasn't original - it was on our state test - but I thought it was not a bad question.)

[finagle29]

answer (Click to Show Hidden Text)

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)