Machines B/C

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Machines B/C

Post by Creationist127 » September 6th, 2019, 6:09 am

Let's start easy. If a class A lever is in static equilibrium, with a 500 kilogram mass on one side and a 30 kilogram effort force on the other, what is the actual mechanical advantage of the lever?
Remember to use "spoiler" to hide your answers!

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[spoiler]I'm not telling the answer[/spoiler]
I'm not telling the answer
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Re: Machines B/C

Post by JoeyC » September 6th, 2019, 8:39 am

16.67
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Re: Machines B/C

Post by Creationist127 » September 6th, 2019, 10:18 am

JoeyC wrote:
September 6th, 2019, 8:39 am
16.67
Correct! Your turn.
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Re: Machines B/C

Post by JoeyC » September 6th, 2019, 11:19 am

Identify the following gear and list 2 advantages as well as 2 disadvantages of using it.
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Re: Machines B/C

Post by MadCow2357 » September 6th, 2019, 4:26 pm

JoeyC wrote:
September 6th, 2019, 11:19 am
Identify the following gear and list 2 advantages as well as 2 disadvantages of using it.
Image
Can't tell if there is a groove in the middle, but I'm going to say no. That's a Herringbone gear, a special type of double helical gear.

Two advantages include the efficient transfer of torque and smooth motion, and the fact that they don't have the disadvantage of axial thrust like single helical gears do.

One disadvantage is the manufacturing difficulty, and consequently, their cost. Another would be their lack of ability to transfer the axis of rotation like a bevel gear would have (idk if the terminology is correct, and I'm kinda guessing on this last disadvantage).

EDIT: Fixed spoiler thing
Last edited by MadCow2357 on September 7th, 2019, 10:27 am, edited 2 times in total.
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Re: Machines B/C

Post by JoeyC » September 6th, 2019, 5:33 pm

Yeah, except the last disadvantage could include their relative bulkiness compared to regular spur gears causing more force to be needed for acceleration Also remember to use the

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[spoiler][/spoiler]
tag next time, but other than that yes, correct! :)
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Re: Machines B/C

Post by MadCow2357 » September 7th, 2019, 10:28 am

Someone else go for me I can't think of a question lol
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Re: Machines B/C

Post by Things2do » September 7th, 2019, 9:05 pm

MadCow2357 wrote:
September 7th, 2019, 10:28 am
Someone else go for me I can't think of a question lol
You wish to raise one side of a 5,000 kg car to change a flat tire. You have a strong 2x4 to use as a lever, and a 15 cm high block for a fulcrum. The fulcrum is 304,000 µm from the load, and the lever is 1.52 × 10^-6 Mm long. How much force is needed to lift the car? Give your answer in kg.

Assume that the jack point is in the middle of the side, and that you need to lift half of the car's mass.

Sorry if that's not allowed by the rules or is overly complicated. I ain't read the rules that closely yet.

Hint: Notall of the values given are needed. It should be a simple formula.
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Re: Machines B/C

Post by Justin72835 » September 8th, 2019, 9:11 pm

Things2do wrote:
September 7th, 2019, 9:05 pm
You wish to raise one side of a 5,000 kg car to change a flat tire. You have a strong 2x4 to use as a lever, and a 15 cm high block for a fulcrum. The fulcrum is 304,000 µm from the load, and the lever is 1.52 × 10^-6 Mm long. How much force is needed to lift the car? Give your answer in kg.

Assume that the jack point is in the middle of the side, and that you need to lift half of the car's mass.

Sorry if that's not allowed by the rules or is overly complicated. I ain't read the rules that closely yet.

Hint: Notall of the values given are needed. It should be a simple formula.
Before solving, I just want to note that force is measured in Newtons, so I'll give my answer in both kilograms and Newtons :D.

1250 kg or 12250 N
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Re: Machines B/C

Post by Things2do » September 10th, 2019, 9:46 pm

Justin72835 wrote:
September 8th, 2019, 9:11 pm
Things2do wrote:
September 7th, 2019, 9:05 pm
You wish to raise one side of a 5,000 kg car to change a flat tire. You have a strong 2x4 to use as a lever, and a 15 cm high block for a fulcrum. The fulcrum is 304,000 µm from the load, and the lever is 1.52 × 10^-6 Mm long. How much force is needed to lift the car? Give your answer in kg.

Assume that the jack point is in the middle of the side, and that you need to lift half of the car's mass.

Sorry if that's not allowed by the rules or is overly complicated. I ain't read the rules that closely yet.

Hint: Not all of the values given are needed. It should be a simple formula.
Before solving, I just want to note that force is measured in Newtons, so I'll give my answer in both kilograms and Newtons :D.

1250 kg or 12250 N
The kg/N thing was my error... I wanted the answer in Newtons.
If you'd've showed your work in some form, I wouldn't've hadta compute it myself...
I got this:
(0.5×5000kg)(9.80566m/s^2)(0.304m)(1.216m)=9061.998746 Newtons.

9060 N
{I don't round until I get my final answer, so I allow for a little fluctuation...}

Notes: You may have made one or more of the following errors. Without your work, it's hard to tell which.
Mm is Megameters, which converts to 1.52m. But, I gave the total length of the lever here, not the force-fulcrum distance. With that subtracted, you get 1.216m.
µm is micrometers, which converts to 1.304m.
My answer seems a little high, but with made-up values... I tried running it through online calculators, but I wound up with 5 answers from 3 different calculators, without changing anything and with the correct units...

Edit: I might not have the right formula...
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