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Re: Simple Machines B/Compound Machines C

Posted: April 7th, 2015, 3:38 am
by mjcox2000
Unless I'm doing it wrong (and I don't think I am) no. I think you're messing up somewhere in the beginning.
Are the masses of the pulleys relevant?
7.62 kg? I'm honestly thinking of just giving up and posting another question.
21*0.9^5=12.4 with sig figs

Re: Simple Machines B/Compound Machines C

Posted: April 7th, 2015, 9:41 am
by JonB
Ok then:

If M2 has a mass of 20.0 kg, and for all pulleys the mass is 1.00 kg and the efficiency is 90.0%, solve for the mass of M1 needed to balance the system (with proper significant figures) (assume all ropes are exactly vertical; I'm not sure how them being at angles would affect the system).
29.9kg

Re: Simple Machines B/Compound Machines C

Posted: April 7th, 2015, 10:00 am
by Unome
Ok then:

If M2 has a mass of 20.0 kg, and for all pulleys the mass is 1.00 kg and the efficiency is 90.0%, solve for the mass of M1 needed to balance the system (with proper significant figures) (assume all ropes are exactly vertical; I'm not sure how them being at angles would affect the system).
29.9kg
Hmm... your answers are usually correct so I'm going to just post my work and ask about it; maybe I'm just doing the pulley efficiencies wrong (perhaps it doesn't matter, since everything is stable)
[attachment=0]Untitled.png[/attachment]

Re: Simple Machines B/Compound Machines C

Posted: April 7th, 2015, 10:35 am
by JonB
Ok then:

If M2 has a mass of 20.0 kg, and for all pulleys the mass is 1.00 kg and the efficiency is 90.0%, solve for the mass of M1 needed to balance the system (with proper significant figures) (assume all ropes are exactly vertical; I'm not sure how them being at angles would affect the system).
29.9kg
Hmm... your answers are usually correct so I'm going to just post my work and ask about it; maybe I'm just doing the pulley efficiencies wrong (perhaps it doesn't matter, since everything is stable)
[attachment=0]Untitled.png[/attachment]
Now that I have double-checked my work, I get a new
of 18.1 kg.
Referring to the picture below, on the left side we see that T + .9T = 20 + 1, and T therefore = 11.053.
Then, on the right side where the two separate pulley trains connect, we see that .9^3 T + .9^4 T + 1 = .9 M1, and therefore solving for M1 we arrive at ~18.1kg.
Does this seem right? I do include inefficiency in the pulleys because if you imagine a very rusty inefficient pulley, you could probably hang unequal weights on both sides that differ by a small amount and still have the pulley hang stationary.
[attachment=0]Untitled.png[/attachment]

Re: Simple Machines B/Compound Machines C

Posted: April 8th, 2015, 6:37 am
by Unome
29.9kg
Hmm... your answers are usually correct so I'm going to just post my work and ask about it; maybe I'm just doing the pulley efficiencies wrong (perhaps it doesn't matter, since everything is stable)
[attachment=0]Untitled.png[/attachment]
Now that I have double-checked my work, I get a new
of 18.1 kg.
Referring to the picture below, on the left side we see that T + .9T = 20 + 1, and T therefore = 11.053.
Then, on the right side where the two separate pulley trains connect, we see that .9^3 T + .9^4 T + 1 = .9 M1, and therefore solving for M1 we arrive at ~18.1kg.
Does this seem right? I do include inefficiency in the pulleys because if you imagine a very rusty inefficient pulley, you could probably hang unequal weights on both sides that differ by a small amount and still have the pulley hang stationary.
[attachment=0]Untitled.png[/attachment]
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.

Re: Simple Machines B/Compound Machines C

Posted: April 8th, 2015, 9:49 am
by 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
paintshop.png (6.08 KiB) Viewed 2638 times

Re: Simple Machines B/Compound Machines C

Posted: April 8th, 2015, 1:23 pm
by 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.

Re: Simple Machines B/Compound Machines C

Posted: April 9th, 2015, 1:21 pm
by 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?"

Re: Simple Machines B/Compound Machines C

Posted: April 9th, 2015, 8:47 pm
by jkang
What is the "law of the lever?"
d1F1 = d2F2, where d = distance from the fulcrum and F = force applied

Re: Simple Machines B/Compound Machines C

Posted: April 10th, 2015, 2:35 pm
by 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!
[math]d_1F_1 = d_2F_2[/math]
Your turn!