Machines B/C

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Justin72835
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Re: Machines B/C

Post by Justin72835 » September 11th, 2019, 2:36 pm

Things2do wrote:
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...
Turns out I missed the first part about raising one side heh. Anyway, looking at your work, you should be dividing by 1.216, not multiplying. When you divide by that you get 6125 N, which is half of my original answer.

If you're interested in the formula: F1*d1=F2*d2
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Re: Machines B/C

Post by Things2do » September 11th, 2019, 7:56 pm

Justin72835 wrote:
September 11th, 2019, 2:36 pm
Things2do wrote:
September 10th, 2019, 9:46 pm
Justin72835 wrote:
September 8th, 2019, 9:11 pm


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...
Turns out I missed the first part about raising one side heh. Anyway, looking at your work, you should be dividing by 1.216, not multiplying. When you divide by that you get 6125 N, which is half of my original answer.

If you're interested in the formula: F1*d1=F2*d2
I figured that I was supposed to divide somewhere in there... Anyways, it's your turn.
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Re: Machines B/C

Post by Justin72835 » September 14th, 2019, 4:12 am

What is the IMA of an inclined plane with an angle of 35 degrees?
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Re: Machines B/C

Post by Creationist127 » September 14th, 2019, 7:25 am

Justin72835 wrote:
September 14th, 2019, 4:12 am
What is the IMA of an inclined plane with an angle of 35 degrees?
Would it be 1.7?
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Re: Machines B/C

Post by Justin72835 » September 14th, 2019, 9:31 am

Creationist127 wrote:
September 14th, 2019, 7:25 am
Would it be 1.7?
Yup, your turn!
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Re: Machines B/C

Post by Creationist127 » September 17th, 2019, 9:04 am

A wedge is 1 inch thick and has a 25* angle at its narrow edge. It is placed with the thin edge in a crack in a block of wood. If 5 newtons of force is exerted on the end, how much force will the wedge split the block with?
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Re: Machines B/C

Post by amk578 » September 17th, 2019, 9:52 am

Creationist127 wrote:
September 17th, 2019, 9:04 am
A wedge is 1 inch thick and has a 25* angle at its narrow edge. It is placed with the thin edge in a crack in a block of wood. If 5 newtons of force is exerted on the end, how much force will the wedge split the block with?
11.25 N or 10 N w/ sig figs
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Re: Machines B/C

Post by Creationist127 » September 17th, 2019, 10:26 am

amk578 wrote:
September 17th, 2019, 9:52 am
Creationist127 wrote:
September 17th, 2019, 9:04 am
A wedge is 1 inch thick and has a 25* angle at its narrow edge. It is placed with the thin edge in a crack in a block of wood. If 5 newtons of force is exerted on the end, how much force will the wedge split the block with?
11.25 N or 10 N w/ sig figs
Correct, and sorry about the sig figs. Your turn!
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Re: Machines B/C

Post by amk578 » September 18th, 2019, 5:04 am

Which of the following screws has the highest mechanical advantage?

a) a screw with a (1/2) inch radius with 10 threads per inch
b) a screw with a (1/2) inch radius with 12 threads per inch
c) a screw with a (1/4) inch radius and 10 threads per inch
d) a screw with a (1/4) inch radius and 12 threads per inch
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Re: Machines B/C

Post by CPScienceDude » September 18th, 2019, 7:21 am

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