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Unome wrote:Correct! Your turn.

who's correct?

Unome wrote:Correct! Your turn.

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.

chinesesushi wrote:

Unome wrote:Correct! Your turn.

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))

chinesesushi wrote:

Unome wrote:Correct! Your turn.

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%)

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)

mjcox2000 wrote:

chinesesushi wrote:

Unome wrote:Correct! Your turn.

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!

bernard wrote: mjcox2000, since chinesesushi has checked your answer, feel free to post a question!

Sorry - I forgot. Here's a problem: 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.

Unome wrote:

I haven't seen a problem this difficult in nearly a year, and I've never seen a problem like this before, so I'll probably get it wrong, but... (Click to Show Hidden Text)

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:

I'm not really sure, but I think that this is the right way to go off of Unome's part (Click to Show Hidden Text)

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.