Simple Machines B/Compound Machines C

[Unome]

Does anyone know how to solve number 4 on the Captain's Tryouts test for Compound Machines?

Yeah, I spent so much time trying to figure out that one (and I still don't know); I never understood tension beyond what I needed to solve Div B level problems.

[daydreamer0023]

Does anyone know if you are required to simplify your mass ratios while doing device testing? And is it required that you know how to do tension for division B? Thanks to anyone who decides to help.

[Unome]

Does anyone know if you are required to simplify your mass ratios while doing device testing? And is it required that you know how to do tension for division B? Thanks to anyone who decides to help.

Ratio simplification/format is up to the supervisors individually (from the FAQs); most tests won't have anything that requires tension knowledge to solve, but it's possible.

[RontgensWallaby]

Just a question I came across earlier that I'm not quite sure about. It went something like this...
"A box is placed on top of an inclined plane. The slope of the inclined plane is gradually raised until the box begins to fall down the inclined plane. If the friction coefficient is 0.4121 (I think this was it), what is the angle made between the inclined plane and the ground when the box begins to fall?"
I'm not sure if this is solvable, but I did it one way and my answer was 36°. I would really appreciate if anyone could solve and explain it to me or tell me if it is unsolvable.

[sciolyboy123]

What is the average time for you guys going to nationals (Div.B)?

[jkang]

Just a question I came across earlier that I'm not quite sure about. It went something like this...
"A box is placed on top of an inclined plane. The slope of the inclined plane is gradually raised until the box begins to fall down the inclined plane. If the friction coefficient is 0.4121 (I think this was it), what is the angle made between the inclined plane and the ground when the box begins to fall?"
I'm not sure if this is solvable, but I did it one way and my answer was 36°. I would really appreciate if anyone could solve and explain it to me or tell me if it is unsolvable.

A box on an inclined plane is in equilibrium when mu = tan(theta) (where the force of friction and force of gravity cancel each other out). Substituting mu for .4121, you get the angle to be around 22.4°.

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

Just a question I came across earlier that I'm not quite sure about. It went something like this...
"A box is placed on top of an inclined plane. The slope of the inclined plane is gradually raised until the box begins to fall down the inclined plane. If the friction coefficient is 0.4121 (I think this was it), what is the angle made between the inclined plane and the ground when the box begins to fall?"
I'm not sure if this is solvable, but I did it one way and my answer was 36°. I would really appreciate if anyone could solve and explain it to me or tell me if it is unsolvable.

A box on an inclined plane is in equilibrium when mu = tan(theta) (where the force of friction and force of gravity cancel each other out). Substituting mu for .4121, you get the angle to be around 22.4°.

I got 0.3909 rad (in sig figs) which is 22.40 deg, so I support your answer. (just need to put it in sig figs)

[RontgensWallaby]

Just a question I came across earlier that I'm not quite sure about. It went something like this...
"A box is placed on top of an inclined plane. The slope of the inclined plane is gradually raised until the box begins to fall down the inclined plane. If the friction coefficient is 0.4121 (I think this was it), what is the angle made between the inclined plane and the ground when the box begins to fall?"
I'm not sure if this is solvable, but I did it one way and my answer was 36°. I would really appreciate if anyone could solve and explain it to me or tell me if it is unsolvable.

A box on an inclined plane is in equilibrium when mu = tan(theta) (where the force of friction and force of gravity cancel each other out). Substituting mu for .4121, you get the angle to be around 22.4°.

I got 0.3909 rad (in sig figs) which is 22.40 deg, so I support your answer. (just need to put it in sig figs)

Thanks so much, guys. I remember solving it twice and getting something around there the first time but thinking that was too shallow of an angle if that much of the input work was converted to friction.

[Unome]

A box on an inclined plane is in equilibrium when mu = tan(theta) (where the force of friction and force of gravity cancel each other out). Substituting mu for .4121, you get the angle to be around 22.4°.

I got 0.3909 rad (in sig figs) which is 22.40 deg, so I support your answer. (just need to put it in sig figs)

Thanks so much, guys. I remember solving it twice and getting something around there the first time but thinking that was too shallow of an angle if that much of the input work was converted to friction.

Yeah, the same thing happened to us on the test at Dodgen; we spent 10 minutes working out a problem (which now seems sort of easy) only to find out that the answer was 1.

[RontgensWallaby]

Thanks so much, guys. I remember solving it twice and getting something around there the first time but thinking that was too shallow of an angle if that much of the input work was converted to friction.[/quote]
Yeah, the same thing happened to us on the test at Dodgen; we spent 10 minutes working out a problem (which now seems sort of easy) only to find out that the answer was 1.[/quote]

Was it the final question?