Simple Machines B/Compound Machines C

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chinesesushi
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Re: Simple Machines B/Compound Machines C

Post by chinesesushi »

Unome wrote:Correct! Your turn. :D
who's correct?
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Re: Simple Machines B/Compound Machines C

Post by chinesesushi »

Unome wrote:Correct! Your turn. :D
What is the mechanical advantage of each of the pulleys. https://www.lhup.edu/~dsimanek/scenario ... ulley1.gif

If f = 10 N and L has a mass of 3 kg what is the AMA. What is the efficiency.
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Re: Simple Machines B/Compound Machines C

Post by Unome »

chinesesushi wrote:
Unome wrote:Correct! Your turn. :D
What is the mechanical advantage of each of the pulleys. https://www.lhup.edu/~dsimanek/scenario ... ulley1.gif

If f = 10 N and L has a mass of 3 kg what is the AMA. What is the efficiency.
Sorry, yes you were the correct one (Unfortunately, I can't also answer the question because the page won't load (it's a problem on my side not, with the image))
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Re: Simple Machines B/Compound Machines C

Post by mjcox2000 »

chinesesushi wrote:
Unome wrote:Correct! Your turn. :D
What is the mechanical advantage of each of the pulleys. https://www.lhup.edu/~dsimanek/scenario ... ulley1.gif

If f = 10 N and L has a mass of 3 kg what is the AMA. What is the efficiency.
A) IMA=4, AMA=3, efficiency=75% (80% with sig figs)
B) IMA=32, AMA=3, efficiency=9.375% (10% with sig figs)
C) IMA=3, AMA=3, efficiency=100%
(these are assuming the gravitational constant is rounded to 10, because otherwise C's efficiency is above 100%)
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Re: Simple Machines B/Compound Machines C

Post by chinesesushi »

a is right.
b is right.
c is wrong (btw why can't it be 9.8. AMA = 9.8*3/10 = 2.94. efficiency = AMA / IMA = 2.94 / 3 < 1)
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Re: Simple Machines B/Compound Machines C

Post by bernard »

mjcox2000 wrote:
chinesesushi wrote:
Unome wrote:Correct! Your turn. :D
What is the mechanical advantage of each of the pulleys. https://www.lhup.edu/~dsimanek/scenario ... ulley1.gif

If f = 10 N and L has a mass of 3 kg what is the AMA. What is the efficiency.
A) IMA=4, AMA=3, efficiency=75% (80% with sig figs)
B) IMA=32, AMA=3, efficiency=9.375% (10% with sig figs)
C) IMA=3, AMA=3, efficiency=100%
(these are assuming the gravitational constant is rounded to 10, because otherwise C's efficiency is above 100%)
mjcox2000, since chinesesushi has checked your answer, feel free to post a question!
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Re: Simple Machines B/Compound Machines C

Post by mjcox2000 »

bernard wrote: mjcox2000, since chinesesushi has checked your answer, feel free to post a question!
Sorry - I forgot. Here's a problem:
Lever 1.jpg
In the above diagram of a class 1 lever, the arm on the right is horizontal and the arm on the left is 60.0 degrees down from horizontal. The fulcrum supports a force of 32.0 newtons. A 20.0 newton force is supported by the 30.0 centimeter arm on the right. Find F, the force on the left side of the lever, and L, the length of the left-side arm, rounded to proper significant figures. Assume the lever is still (no unbalanced forces or moments act on it), the lever arm is massless, the device is frictionless, etc.
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Re: Simple Machines B/Compound Machines C

Post by Unome »

FL cos 60 degrees is the torque on the bent arm, which should be equal to the torque 6 J/rad on the other side (I don't know how gravity affect this) and somehow that 32.0 N force on the fulcrum forms another equation... this isn't really getting me anywhere. Nice problem though!
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Re: Simple Machines B/Compound Machines C

Post by blindmewithscience »

Maybe the 32N force is acting as a "normal force" for the system as a whole? So the mass weighs 32-20=12N? I've also not seen something like this, and I'm not sure if this is correct at all...
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Re: Simple Machines B/Compound Machines C

Post by mjcox2000 »

Unome wrote:
FL cos 60 degrees is the torque on the bent arm, which should be equal to the torque 6 J/rad on the other side (I don't know how gravity affect this) and somehow that 32.0 N force on the fulcrum forms another equation... this isn't really getting me anywhere. Nice problem though!
blindmewithscience wrote:
Maybe the 32N force is acting as a "normal force" for the system as a whole? So the mass weighs 32-20=12N? I've also not seen something like this, and I'm not sure if this is correct at all...
You're both right about what you've gotten so far, and together, you're pretty close to the solution.
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