## Simple Machines B/Compound Machines C

Unome
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### Re: Simple Machines B/Compound Machines C

blindmewithscience wrote:
JonB wrote:
mjcox2000 wrote:In the image of the pulley system shown below, where there are 2 concentric circles over 2 intersecting ropes, those ropes are tied together at that point; triangles in the image symbolize fixed anchor points. How much downwards effort force must be applied to the arrow (on the rightmost rope) to lift the 10kg weight, as illustrated? Assume that this pulley system is an ideal machine, and g=9.81 m/s^2.
Complex pulley system.jpg
From team member (not myself as the coach):
`9.81N`
Would you be able to explain this solution? Really complex pulley systems are one of the things I'm currently having trouble with.
Same; neither me or my partner could get this one (well, he said he could, but he didn't have the time)
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hscmom
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### Re: Simple Machines B/Compound Machines C

The complex pulley system would be so much easier to explain using a whiteboard and some markers. I feel like I am doing WIDI here...

Print out or draw the diagram and label the pulleys are A-F starting at the left and the extra ropes (with the double-circle attachments) 1-3, again starting at the left. For now, ignore the 10 kg; we'll come back to it. Let's find the IMA by imagining we are pulling at the lower right arrow with one unit of force. When we get to the mass hanging in the lower left, we'll have our IMA figured out. Label the forces on the ropes and lines (let's call the non-rope lines lines, OK?)

Start by labeling the down-right arrow with a "1" since we're pulling with one unit of force. The next thing you encounter is the right knot of rope 3. Label it n (which is a value between 0 and 1 - we don't need to know what it is) and the vertical rope going into pulley on the right F is 1-n.

So, the line coming out and to the left of pulley F is also 1-n, as is the force on either side of pulley E.

Then above and left of pulley E our line intersects rope 3. So, we add the forces: n + 1 - n = 1. Yup, sort of boring so far. Rope 3 and (fixed) pulleys E and F have gained you nothing in terms of force. You put in one unit of force and just above the left side of rope 3, we're back to 1 unit of force again. So, you could really lop off pulleys E and F and rope 3, save yourself some time and money and still have the same IMA.

So, label the line to the right of pulley D and immediately above the knot of rope 3 "1" because that's how much force is on the right of pulley D. And, it's also the same force on the left of pulley D going down to C.

Rope 2 supports (moveable) pulley D. It has to counterbalance the two lines, left and right, of pulley D. Both of those lines have a force of 1 so 1 + 1 = 2. So, label rope 2 with a 2.

Now, the two lines of (moveable) pulley C have a force of 1, so write "1" on either side of pulley C. Pulley C is held in place (ok, it moves) with line 1 which will have a force of 2 on it because 1 + 1 = 2 (which are the forces on both sides of pulley C). Label rope 1 with "2"

Now, look at the line to the left of pulley C, going up to (fixed) pulley B. It's got a force of 1, but rope 2, which carries a force of 2, joins it just below pulley B. So, the force going into pulley B is 3 on the right. And that's balanced out by 3 on the left. Label them. If you feel like labeling the support stuck to the ceiling, it's 6 (3+3). Rope 1 is tied on to the rope coming down from the left of pulley B. So, add those forces. We have a force of 3 from pulley B and a force of 2 from rope 1. They add up to 5, so just below where rope 1 is tied to that line, write 5. That's right above pulley A which supports the mass. It's got to be balanced on the other side, so the leftmost line of the drawing (to the left of pulley A and supported at the ceiling) also has a force of 5. Add up the 5 on the right and the 5 on the left and you get that the force supporting the mass is 10 times the force that you input. So, the IMA of this ideal system is 10.

But the question asked for force input. So, if you have an IMA of 10 and the mass is 10 kg (see, I said we'd get back to it), then you need a force sufficient to lift a 1 kg force input at the right arrow. We know that 9.81 N of force will lift a kilogram, so that's what we need because the IMA means this pulley system will lift 10 kg when we enter 9.81N of force.

Whew.

Again, would've been better (and shorter) with a white board. Unome, you really give some great problems! Keep it up.
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Unome
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### Re: Simple Machines B/Compound Machines C

hscmom wrote:The complex pulley system would be so much easier to explain using a whiteboard and some markers. I feel like I am doing WIDI here...

Print out or draw the diagram and label the pulleys are A-F starting at the left and the extra ropes (with the double-circle attachments) 1-3, again starting at the left. For now, ignore the 10 kg; we'll come back to it. Let's find the IMA by imagining we are pulling at the lower right arrow with one unit of force. When we get to the mass hanging in the lower left, we'll have our IMA figured out. Label the forces on the ropes and lines (let's call the non-rope lines lines, OK?)

Start by labeling the down-right arrow with a "1" since we're pulling with one unit of force. The next thing you encounter is the right knot of rope 3. Label it n (which is a value between 0 and 1 - we don't need to know what it is) and the vertical rope going into pulley on the right F is 1-n.

So, the line coming out and to the left of pulley F is also 1-n, as is the force on either side of pulley E.

Then above and left of pulley E our line intersects rope 3. So, we add the forces: n + 1 - n = 1. Yup, sort of boring so far. Rope 3 and (fixed) pulleys E and F have gained you nothing in terms of force. You put in one unit of force and just above the left side of rope 3, we're back to 1 unit of force again. So, you could really lop off pulleys E and F and rope 3, save yourself some time and money and still have the same IMA.

So, label the line to the right of pulley D and immediately above the knot of rope 3 "1" because that's how much force is on the right of pulley D. And, it's also the same force on the left of pulley D going down to C.

Rope 2 supports (moveable) pulley D. It has to counterbalance the two lines, left and right, of pulley D. Both of those lines have a force of 1 so 1 + 1 = 2. So, label rope 2 with a 2.

Now, the two lines of (moveable) pulley C have a force of 1, so write "1" on either side of pulley C. Pulley C is held in place (ok, it moves) with line 1 which will have a force of 2 on it because 1 + 1 = 2 (which are the forces on both sides of pulley C). Label rope 1 with "2"

Now, look at the line to the left of pulley C, going up to (fixed) pulley B. It's got a force of 1, but rope 2, which carries a force of 2, joins it just below pulley B. So, the force going into pulley B is 3 on the right. And that's balanced out by 3 on the left. Label them. If you feel like labeling the support stuck to the ceiling, it's 6 (3+3). Rope 1 is tied on to the rope coming down from the left of pulley B. So, add those forces. We have a force of 3 from pulley B and a force of 2 from rope 1. They add up to 5, so just below where rope 1 is tied to that line, write 5. That's right above pulley A which supports the mass. It's got to be balanced on the other side, so the leftmost line of the drawing (to the left of pulley A and supported at the ceiling) also has a force of 5. Add up the 5 on the right and the 5 on the left and you get that the force supporting the mass is 10 times the force that you input. So, the IMA of this ideal system is 10.

But the question asked for force input. So, if you have an IMA of 10 and the mass is 10 kg (see, I said we'd get back to it), then you need a force sufficient to lift a 1 kg force input at the right arrow. We know that 9.81 N of force will lift a kilogram, so that's what we need because the IMA means this pulley system will lift 10 kg when we enter 9.81N of force.

Whew.

Again, would've been better (and shorter) with a white board. Unome, you really give some great problems! Keep it up.
That wasn't my problem... but okay (although now I'll have to think up something original for the next time I give one; all of my other ones were from tests I took)
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mjcox2000
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### Re: Simple Machines B/Compound Machines C

I made that problem based off of what I read in this document: http://efclimbers.net/wp-content/upload ... +Paper.pdf - I found it a really helpful resource for finding the MA of complex pulley systems.
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### Re: Simple Machines B/Compound Machines C

Well,since no one has said anything in a while:

If the parts of the load resting on the inclined plane have the same slope as the inclined plane does at that point, what is force F needed to support the load?
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mjcox2000
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### Re: Simple Machines B/Compound Machines C

Unome wrote:Well,since no one has said anything in a while:

If the parts of the load resting on the inclined plane have the same slope as the inclined plane does at that point, what is force F needed to support the load?
`I don't think the part with slope 2/3 affects the force required, so I'm ignoring that. Then, it becomes a simple matter of calculating the IMA to find friction: hypotenuse is sqrt 109, so IMA is (sqrt 109)/3, so force is 60(9.81)(3/(sqrt 109)) - 60(9.81)(10/(sqrt 109))/4, which evaluates to about 28.19 newtons, according to Wolfram Alpha (or 30 newtons if sig figs are taken into account).`
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Unome
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### Re: Simple Machines B/Compound Machines C

Well, that's how my reasoning went, so I'll take it as correct. Your turn.
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UTF-8 U+6211 U+662F
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### Re: Simple Machines B/Compound Machines C

Question for Simple Machines:
Who identified the pulley?

UTF-8 U+6211 U+662F
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### Re: Simple Machines B/Compound Machines C

UTF-8 U+6211 U+662F wrote:Reviving this thread:
Question for Simple Machines:
Who identified the pulley?
Oops, that's not the right question.

mjcox2000
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### Re: Simple Machines B/Compound Machines C

UTF-8 U+6211 U+662F wrote:
UTF-8 U+6211 U+662F wrote:Reviving this thread:
Question for Simple Machines:
Who identified the pulley?
Oops, that's not the right question.