Simple Machines B/Compound Machines C

RontgensWallaby · Post by **RontgensWallaby** » Wed Apr 22, 2015 2:11 pm

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)

RontgensWallaby · Post by **RontgensWallaby** » Wed Apr 22, 2015 2:13 pm

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.

Post by **UTF-8 U+6211 U+662F** » Wed Apr 22, 2015 3:31 pm

RontgensWallaby wrote: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.

Userpage · Post by **Unome** » Wed May 20, 2015 2:35 pm

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

Post by **UTF-8 U+6211 U+662F** » Wed May 20, 2015 2:45 pm

Unome wrote: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).

Userpage · Post by **Unome** » Wed May 20, 2015 3:23 pm

UTF-8 U+6211 U+662F wrote:
Unome wrote: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?

Post by **UTF-8 U+6211 U+662F** » Wed May 20, 2015 3:37 pm

Unome wrote:
UTF-8 U+6211 U+662F wrote:
Unome wrote: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) $\Sigma F_{upwards} + \Sigma F_{downwards} = 0$

Userpage · Post by **Unome** » Wed May 20, 2015 3:44 pm

UTF-8 U+6211 U+662F wrote:
Unome wrote:
UTF-8 U+6211 U+662F wrote: 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?

Post by **UTF-8 U+6211 U+662F** » Wed May 20, 2015 5:00 pm

Unome wrote: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.

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Simple Machines B/Compound Machines C

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