Machines B/C

JoeyC · Post by **JoeyC** » September 29th, 2019, 5:42 am

Given that a #7-32-5 (UNF) screw is being turned with a 5 inch long wrench, what is the IMA of the action?

viditpok · Post by **viditpok** » October 9th, 2019, 7:11 pm

JoeyC wrote: ↑September 29th, 2019, 5:42 am Given that a #7-32-5 (UNF) screw is being turned with a 5 inch long wrench, what is the IMA of the action?

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?

JoeyC · Post by **JoeyC** » October 10th, 2019, 5:16 pm

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

viditpok · Post by **viditpok** » October 11th, 2019, 7:05 am

Oh, that makes so much more sense
Thank you so much for the help!

AlfWeg's Userpage · Post by **AlfWeg** » October 13th, 2019, 6:03 pm

Who's next? @JoeyC

JoeyC · Post by **JoeyC** » October 13th, 2019, 6:22 pm

Anyone else I guess?
Alf you can do it if you want.

AlfWeg's Userpage · Post by **AlfWeg** » October 14th, 2019, 3:07 am

JoeyC wrote: ↑October 13th, 2019, 6:22 pm Anyone else I guess?
Alf you can do it if you want.

Naw

Creationist127 · Post by **Creationist127** » October 14th, 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?

Umaroth's Userpage · Post by **Umaroth** » October 14th, 2019, 7:37 am

Creationist127 wrote: ↑October 14th, 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?

mgsinθ = μmgcosθ
μ = sin26/cos26 = tan26
μ = 0.488

F = μmgcosθ
F = 107N

Creationist127 · Post by **Creationist127** » October 14th, 2019, 9:04 am

Umaroth wrote: ↑October 14th, 2019, 7:37 am
Creationist127 wrote: ↑October 14th, 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?
mgsinθ = μmgcosθ
μ = sin26/cos26 = tan26
μ = 0.488

F = μmgcosθ
F = 107N

Correct! Your turn.

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