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Mathdino wrote:

Here's a IMA question (Click to Show Hidden Text)

Suppose that the pulley in the below image is connected to something weighing 10 kg, and I need to pull on it with 9 N of force to hold it up. What's the efficiency? [img]http://02.edu-cdn.com/files/89501_89600/89569/file_89569.jpg[/img]

IMA = (Force out of the machine)/(Force into the machine). 
So for this scenario, the Force into the machine would be 9N, I think. And the Force out of the machine would be 10kg * 9.8m/s/s = 98N. 
Therefore, the IMA = 98N/9N = 10.89? 
I feel like I need to use the visual somehow, but this is what I ended up doing... :?

lchs wrote:

Mathdino wrote:

Here's a IMA question (Click to Show Hidden Text)

Suppose that the pulley in the below image is connected to something weighing 10 kg, and I need to pull on it with 9 N of force to hold it up. What's the efficiency? [img]http://02.edu-cdn.com/files/89501_89600/89569/file_89569.jpg[/img]

I'm not so sure about this but I'll give it a shot...

Answer (Click to Show Hidden Text)

IMA = (Force out of the machine)/(Force into the machine). 
So for this scenario, the Force into the machine would be 9N, I think. And the Force out of the machine would be 10kg * 9.8m/s/s = 98N. 
Therefore, the IMA = 98N/9N = 10.89? 
I feel like I need to use the visual somehow, but this is what I ended up doing... :?

1.36

chinesesushi wrote:

lchs wrote: I'm not so sure about this but I'll give it a shot...

Answer (Click to Show Hidden Text)

IMA = (Force out of the machine)/(Force into the machine). 
So for this scenario, the Force into the machine would be 9N, I think. And the Force out of the machine would be 10kg * 9.8m/s/s = 98N. 
Therefore, the IMA = 98N/9N = 10.89? 
I feel like I need to use the visual somehow, but this is what I ended up doing... :?

But I believe Mathdino is asking for the efficiency in the problem, not the IMA?

idk though... (Click to Show Hidden Text)

1.36

The [u]AMA[/u] is the (force out)/(force in), not IMA; lchs got the AMA correctly. Efficiency is AMA/IMA. chinesesushi almost got it, but I assume you got the IMA as 8 instead of 16. Notice that there are 4 class 2 pulleys and 2 class 1 pulleys, and 1*1*2*2*2*2=16. Efficiency is 10.89/16, or about 68.06%.
Edit: A simpler way to explain it is that the IMA of the pulley on the right is 4 (count the rope sections) and the IMA of the one on the right is 4 (count). 4*4=16.

Mathdino wrote:

chinesesushi wrote: But I believe Mathdino is asking for the efficiency in the problem, not the IMA?

idk though... (Click to Show Hidden Text)

1.36

Almost (Click to Show Hidden Text)

The [u]AMA[/u] is the (force out)/(force in), not IMA; lchs got the AMA correctly. Efficiency is AMA/IMA. chinesesushi almost got it, but I assume you got the IMA as 8 instead of 16. Notice that there are 4 class 2 pulleys and 2 class 1 pulleys, and 1*1*2*2*2*2=16. Efficiency is 10.36/16, or about 64.75%.
Edit: A simpler way to explain it is that the IMA of the pulley on the right is 4 (count the rope sections) and the IMA of the one on the right is 4 (count). 4*4=16.

Your turn, chinesesushi.

[img]http://www.gweaver.net/techhigh/projects/Rube%20Goldberg/Rube%20Goldberg%20Websites/Period%205/Rube%20Website%20Kyle,%20Jared,%20Sam,%20Nicole,%20Arlene/Images/animatedfourpulley.gif[/img] What is the IMA of the pulley system and how much force is required to lift it the appropriate amount. The pulley is in an ideal state.

Sciolapedia wrote:

Sciolapedia's Answer (Click to Show Hidden Text)

Im new at this event, but here is what I think

IMA of a pulley is number of supporting strings or input d / out put d which is 4.

Ama=Ima in a perfect world so . . .                                200/x=4
x=50 Newtons?

I think :?

Close. The IMA is 4. You can do it either the way your did it or by D(in)/D(out) which is 4 m/1 m = an IMA of 4. However, the reason the answer is not 50 Newtons is because the [b]mass[/b] of the object is 200 kg, not the weight. The weight (or force) would be 200 kg * 9.8 m/s^2 = 1960 N. Therefore, the force needed to support the object would be 490 Newtons.

Sciolapedia wrote:Explain the force distance trade off of work. Use at least one example from the inclined plane family and the pulley family to back up your answer(show how changing input distance and force changes output distance and force in that example).

in order to do the same amount of work with less force, the distance over which the force is applied must be greater. e.g. With an inclined plane, in order to raise the object a certain height with less force, the slope of the inclined plane must be lower, resulting in a greater distance required to push the object to attain the same height. In the pulley family, in order the raise an object the same height with less force, the amount of strings supporting the object must increase thereby increasing the distance required to pull the leading rope to raise the object.