UTF-8 U+6211 U+662F wrote:Are the masses of the pulleys relevant?Unome wrote:Unless I'm doing it wrong (and I don't think I am) no. I think you're messing up somewhere in the beginning.7.62 kg? I'm honestly thinking of just giving up and posting another question.
21*0.9^5=12.4 with sig figs
Unome wrote: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)JonB wrote:Unome wrote: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
[attachment=0]Untitled.png[/attachment]
Now that I have double-checked my work, I get a newUnome wrote: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)JonB wrote:Unome wrote: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[attachment=0]Untitled.png[/attachment]
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.JonB wrote:Now that I have double-checked my work, I get a newUnome wrote: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)JonB wrote:29.9kg[attachment=0]Untitled.png[/attachment]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]
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.
Ok... I sort of get it. Anyway, the first person that gets here can ask a question.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.
Extremely simple question:Unome wrote:Ok... I sort of get it. Anyway, the first person that gets here can ask a question.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.
d1F1 = d2F2, where d = distance from the fulcrum and F = force appliedUTF-8 U+6211 U+662F wrote:What is the "law of the lever?"
Remember to hide your answer!jkang wrote:d1F1 = d2F2, where d = distance from the fulcrum and F = force appliedUTF-8 U+6211 U+662F wrote:What is the "law of the lever?"
[math]d_1F_1 = d_2F_2[/math]
