MEDAL!!!!!!!!!!!!!
Problem stated below.

- calculusxy

MEDAL!!!!!!!!!!!!!
Problem stated below.

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- calculusxy

I did an experiment today relating to torques. The materials that I had were a meter stick, masking tape, and weights.
We had to use the ruler and place the two weights on each side, to try to make it balance to each other. Also, we had to note how far the tape (holding the weight) was from the center (or 50cm). Here is my first data:
Mass 1: 200g Distance from 50cm: 19cm
Mass 2: 100g Distance from 50cm: 36 cm
Then we had to multiply the mass times the distance for each and were asked to look for a pattern.
Here are my products:
Mass 1: 200g x 19 cm = 3800
Mass 2: 100g x 36 cm = 3600
I don't notice a pattern and need help in finding one.

- calculusxy

@Nnesha

- calculusxy

@ikram002p

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## More answers

- DanJS

looks pretty close

- anonymous

okay so you say your lab involved torque right

- calculusxy

Yes

- DanJS

The real world will have small errors, not ideal... you did it correct

- calculusxy

@DanJS So can my conclusion be like since they are both near of each other, it means that they are almost balanced to each other.

- DanJS

Are you discovering torque moments for the first time here, or do you know already

- DanJS

torque = (force perpendicular to the line of action) *(distance)

- calculusxy

Well it is the second to last day of class, and my teacher didn't want to explain torques to us fully and just let us do the experiment. So we needed to find out pattern with the products.

- DanJS

The rotational moment about a point (here 50cm mark) = distance from that point, times the perpendicular force

- anonymous

here this is what i used to study torque last year
https://www.youtube.com/watch?v=sAMNoRBBjzg
i hope it helps :)

- DanJS

so if you have a force on an angle, you need to break into components first..

- calculusxy

I just need to know if the statement that I wrote earlier would be fine for my conclusion or not.

- DanJS

hah, if your teacher is cramming it in this late, probably wont be much of it on the final

- calculusxy

I already took the final. It's just an extra thing.

- DanJS

oh, is that the only single trial point you did? did you try different distances?

- calculusxy

yes until the weights were balanced (almost)

- DanJS

I would say something like, The distance multiplied by the force seems almost the same... Need to try further experiments with different masses and distances to test if this is true

- DanJS

one test isn't enough

- calculusxy

I did another trial, but with three different weights ( two on one side and one on the other).
Mass 1 : 100g Distance from 50cm: 28cm
Mass 2: 200g Distance from 50cm : 20cm
Mass 3: 300g( this weight was kept on the other side) Distance from 50cm: 20cm

- calculusxy

@DanJS

- DanJS

ok, apply the same reasoning as before...
If a torque is clockwise it is negative, and if it is counter clockwise about a point, it is positive...
The sum of all the torques should add to zero if the thing is balanced and not moving

- DanJS

so 100*28 + 200*20 ideally would be 300*20

- calculusxy

Okay I get it thanks.
I have some other questions, will you be able to help me on them?

- calculusxy

This is an example of a velocity
a. 45 miles per hour
b. 34 meters per second
c. 23 centimeters per year northeast
d. 12 days per mile north
e. 0.1 milliseconds per kilometer
f. all of the above are examples of velocity

- calculusxy

It's C right?

- DanJS

back

- DanJS

yes, a velocity is a vector quantity

- DanJS

has a magnitude and a direction for distance/time

- calculusxy

second one
austin is out for a run. he starts at 98th street and 5th ave and runs 1km north to 110th street in 10 minutes. then, he turns left and runs 0.8km in 9 minutes. then, he turns left and runs down central park west. this takes him 11 min to run down the remaining 1.0km down to 98th street. what is the average velocity?

- calculusxy

@DanJS

- DanJS

sorry, here

- DanJS

First need to draw all those vectors

- DanJS

|dw:1438729327608:dw|

- DanJS

The average velocity will be the resultant of those 3 vectors, the sum of them

- DanJS

|dw:1438729501422:dw|

- DanJS

The average velocity is the total displacement/ total time
Displacement is the start point to the end point, independent of the path

- calculusxy

I got the velocity of 33.33m/s

- calculusxy

I am wrong, right?

- DanJS

since they made a square, the magnitude or R is the same as the top piece

- calculusxy

0.8km/9min = V_avg

- DanJS

|dw:1438729828843:dw|

- DanJS

His net displacement is 0.8 km straight west, the total time is (10 + 9 + 11) min

- DanJS

Average Velocity = Total Disp[lacement / Total Time

- calculusxy

0.8 / 1800 sec

- calculusxy

0.0004

- DanJS

I would just leave it in the units the prob already is in , km/min, since it didn't ask to change it

- calculusxy

Our teacher likes seconds to be used that's why.

- DanJS

0.8km / 30 min

- DanJS

oh, then convert to whatever units you need, but that is the prob

- DanJS

about 0.027 km/min

- DanJS

west

- calculusxy

THANK YOU!!!

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