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Posted: **April 27th, 2015, 5:58 pm**

plaid suit guy2 wrote: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.

Posted: **April 27th, 2015, 6:07 pm**

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.

Posted: **April 27th, 2015, 7:32 pm**

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 4071 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 4071 times

Posted: **April 28th, 2015, 12:42 pm**

plaid suit guy2 wrote: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]

Posted: **April 28th, 2015, 1:10 pm**

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

Posted: **April 28th, 2015, 1:19 pm**

plaid suit guy2 wrote: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)

Posted: **April 28th, 2015, 1:20 pm**

There are three equations, each with only one variable

Posted: **April 28th, 2015, 1:38 pm**

plaid suit guy2 wrote:There are three equations, each with only one variable

I just give up.

Posted: **April 28th, 2015, 1:42 pm**

You sure?
The first problem is this

Try the second question.

Posted: **April 28th, 2015, 2:00 pm**

plaid suit guy2 wrote:You sure?
The first problem is this
divcanswer.png
Try the second question.

Now I get it. I forgot the $\Sigma F = 0$ 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?

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