## Simple Machines B/Compound Machines C

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

Ok, that's what I got.
(originally I solved incorrectly for the minimum mass and used the total weight of the block as the force it exerted, for some reason)
Every great and deep difficulty bears in itself its own solution. It forces us to change our thinking in order to find it. - Niels Bohr

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

The only other thing was that my minimum mass was 4 grams heavier than yours but that shouldn't be an issue. Probably a result of different intermediate rounding.
Every great and deep difficulty bears in itself its own solution. It forces us to change our thinking in order to find it. - Niels Bohr

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

http://img.sparknotes.com/content/testp ... pulley.gif
A problem I just came up with. Solved it and just want to make sure I'm right since I doubt my coach will know how to solve it (it's not that complicated).
In the diagram from the link, angle θ is 37 degrees and mass m is 15 kg. The coefficient of friction between mass m and the inclined plane is 0.4. Assume the pulley is frictionless. What are the maximum and minimum masses for mass M if the system is in equilibrium?
Just want to make sure you know you don't have to know this. Div B prohibited topics include coefficient of friction.

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

Okay, so as far as I can tell, if the following system is in static equilibrium, the downward force on the fulcrum would be 16.82; I just wanted to check here and see if that makes sense:
Lever 2nd class.png (3.61 KiB) Viewed 3482 times
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UTF-8 U+6211 U+662F
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### Re: Simple Machines B/Compound Machines C

Okay, so as far as I can tell, if the following system is in static equilibrium, the downward force on the fulcrum would be 16.82; I just wanted to check here and see if that makes sense:
Lever 2nd class.png
Strange... I got an upward force of 16.82 N (with sig figs that's 20 N).

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

Okay, so as far as I can tell, if the following system is in static equilibrium, the downward force on the fulcrum would be 16.82; I just wanted to check here and see if that makes sense:
Lever 2nd class.png
Strange... I got an upward force of 16.82 N (with sig figs that's 20 N).
Wouldn't it be downwards since the outside effort force going upwards is less than the load force going down?
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### Re: Simple Machines B/Compound Machines C

Okay, so as far as I can tell, if the following system is in static equilibrium, the downward force on the fulcrum would be 16.82; I just wanted to check here and see if that makes sense:
Lever 2nd class.png
Strange... I got an upward force of 16.82 N (with sig figs that's 20 N).
Wouldn't it be downwards since the outside effort force going upwards is less than the load force going down?
Since $F_{upwards} < F_{downwards}$, then $F_{upwards} + F_{fulcrum} = F_{downwards}$. Furthermore, if $F_{upwards} + F_{fulcrum} = F_{downwards}$, then static equilibrium is achieved. (Think downwards as negative and upwards as positive) $\Sigma F_{upwards} + \Sigma F_{downwards} = 0$

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

Strange... I got an upward force of 16.82 N (with sig figs that's 20 N).
Wouldn't it be downwards since the outside effort force going upwards is less than the load force going down?
Since $F_{upwards} < F_{downwards}$, then $F_{upwards} + F_{fulcrum} = F_{downwards}$. Furthermore, if $F_{upwards} + F_{fulcrum} = F_{downwards}$, then static equilibrium is achieved. (Think downwards as negative and upwards as positive) $\Sigma F_{upwards} + \Sigma F_{downwards} = 0$
The force exerted by the lever is up, but the force on the lever would be down, right?
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### Re: Simple Machines B/Compound Machines C

The force exerted by the lever is up, but the force on the lever would be down, right?
Yes, at least that's how I see it. Oh, okay, I get it.