Simple Machines B/Compound Machines C

[Unome]

What you're saying sounds like it does this.

Ohh... I should have mentioned which parts were stationary... I intended for both the gears to remain stationary.

[plaid suit guy2]

Okay, so if both gears are stationary, then the big circle will be the 1.5m radius of the eccentric (I have no idea where 1.65 came from). The little circle will be the 2m radius of the dot from the center of it's shaft. Even if the two circles are the same size, it will always do this unless they are coaxial.

[plaid suit guy2]

Double posting, but you are right, it isn't an eccentric, but you are wrong about the radii. The radius of the swing must be added to the radius of the second wheel. Meaning the radius would be seen as 2.5m if the swing is .5m. Yielding .6*5π=9.42m

Okay now that that is done: I have two problems.
The first: The easiest question on the Colorado State Comp-comp test:
Ignore friction. This lever is in static equilibrium with a force Fp applied by the piston at a 30° angle θ to the bar, what is the force applied to the fulcrum Φ?

divCstate.png (11.15 KiB) Viewed 4364 times

The second is more a div B level question. What is the load mass to maintain static equilibrium on the lever? (this is so much lower in quality, sorry)

Div.b lever.png (7.91 KiB) Viewed 4364 times

[UTF-8 U+6211 U+662F]

Double posting, but you are right, it isn't an eccentric, but you are wrong about the radii. The radius of the swing must be added to the radius of the second wheel. Meaning the radius would be seen as 2.5m if the swing is .5m. Yielding .6*5π=9.42m

Okay now that that is done: I have two problems.
The first: The easiest question on the Colorado State Comp-comp test:
Ignore friction. This lever is in static equilibrium with a force Fp applied by the piston at a 30° angle θ to the bar, what is the force applied to the fulcrum Φ?

ANS (Click to Show Hidden Text)

Pulley: [math]\text{IMA} = 4[/math]
[math]20N/4 = 5N[/math]
[math]5N * 15 = 75N[/math]
[math]90 - 30 = 60[/math]
[math]cos(60) * F_p = \frac{F_p}{2}[/math]
[math]\frac{F_p}{2} * 5 = \frac{5F_p}{2}[/math]
[math]F_\phi = 75N - \frac{5F_p}{2} \text{upwards}[/math]

[plaid suit guy2]

This is incorrect, Fp can be found, for one thing.

[UTF-8 U+6211 U+662F]

This is incorrect, Fp can be found, for one thing.

There are two variables though... (and only one equation).

Unless I'm making some stupid mistake (still a little tired after States)

[plaid suit guy2]

There are three equations, each with only one variable

[UTF-8 U+6211 U+662F]

There are three equations, each with only one variable

I just give up.

[plaid suit guy2]

You sure?
The first problem is this

Try the second question.

[UTF-8 U+6211 U+662F]

You sure?
The first problem is this

divcanswer.png

Try the second question.

Now I get it. I forgot the part...

Assuming the lever has a constant density (Click to Show Hidden Text)

4 kg

except... shouldn't the weight of the lever arm increase as the arm goes further away from the fulcrum?