Machines B/C
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JoeyC
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Re: Machines B/C
Given that a #7-32-5 (UNF) screw is being turned with a 5 inch long wrench, what is the IMA of the action?
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viditpok
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Re: Machines B/C
I don't understand the #7-32-5 thing, as when I was looking it up, most results showed 7/32-5/8" or something close to it. Could you please describe what it means?
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JoeyC
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Re: Machines B/C
This can help
https://www.hunker.com/12420308/how-to-read-screw-sizes
The #7-32-5 is a unified screw read out, and works like this-
diameter - number of threads per inch - length of screw
Because it's in unified, all units are in inches, and importantly the diameter has an octothorpe (#) in front of it.
This means that the diameter is not 7 inches, but dictated by this formula,
.06+.013*n, where n is the number. So the diameter is .151 inches, there are 32 threads per inch, meaning pitch (important for screw IMA) is .03125, and the screw is 5 inches long.
Using this information, I can use the Screw IMA equation ; IMA = 2pir/p, or IMA = circumference of total effort arm over pitch, the effort arm being 5 inches because my wrench, effort arm, is 5 inches long, and pitch being 1/32, or .03125.
IMA = 1005.30964915
https://www.hunker.com/12420308/how-to-read-screw-sizes
The #7-32-5 is a unified screw read out, and works like this-
diameter - number of threads per inch - length of screw
Because it's in unified, all units are in inches, and importantly the diameter has an octothorpe (#) in front of it.
This means that the diameter is not 7 inches, but dictated by this formula,
.06+.013*n, where n is the number. So the diameter is .151 inches, there are 32 threads per inch, meaning pitch (important for screw IMA) is .03125, and the screw is 5 inches long.
Using this information, I can use the Screw IMA equation ; IMA = 2pir/p, or IMA = circumference of total effort arm over pitch, the effort arm being 5 inches because my wrench, effort arm, is 5 inches long, and pitch being 1/32, or .03125.
IMA = 1005.30964915
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viditpok
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Re: Machines B/C
Oh, that makes so much more sense
Thank you so much for the help!
Thank you so much for the help!
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JoeyC
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Re: Machines B/C
Anyone else I guess?
Alf you can do it if you want.
Alf you can do it if you want.
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Creationist127
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Re: Machines B/C
I'll just go.
There is an inclined plane, with a 26.0* angle of incline. On top of it is a 25.0 kg box, on the cusp of moving. What is the coefficient of static friction between the ramp and the box, and what is the force of friction?
There is an inclined plane, with a 26.0* angle of incline. On top of it is a 25.0 kg box, on the cusp of moving. What is the coefficient of static friction between the ramp and the box, and what is the force of friction?
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Umaroth
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Re: Machines B/C
mgsinθ = μmgcosθCreationist127 wrote: ↑Mon Oct 14, 2019 7:05 am I'll just go.
There is an inclined plane, with a 26.0* angle of incline. On top of it is a 25.0 kg box, on the cusp of moving. What is the coefficient of static friction between the ramp and the box, and what is the force of friction?
μ = sin26/cos26 = tan26
μ = 0.488
F = μmgcosθ
F = 107N
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Creationist127
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Re: Machines B/C
Correct! Your turn.Umaroth wrote: ↑Mon Oct 14, 2019 7:37 ammgsinθ = μmgcosθCreationist127 wrote: ↑Mon Oct 14, 2019 7:05 am I'll just go.
There is an inclined plane, with a 26.0* angle of incline. On top of it is a 25.0 kg box, on the cusp of moving. What is the coefficient of static friction between the ramp and the box, and what is the force of friction?
μ = sin26/cos26 = tan26
μ = 0.488
F = μmgcosθ
F = 107N
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Can I request that we delete 2020 from our memories and do it over again?