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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.
looks pretty close
okay so you say your lab involved torque right
The real world will have small errors, not ideal... you did it correct
Are you discovering torque moments for the first time here, or do you know already
torque = (force perpendicular to the line of action) *(distance)
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.
The rotational moment about a point (here 50cm mark) = distance from that point, times the perpendicular force
here this is what i used to study torque last year https://www.youtube.com/watch?v=sAMNoRBBjzg i hope it helps :)
so if you have a force on an angle, you need to break into components first..
I just need to know if the statement that I wrote earlier would be fine for my conclusion or not.
hah, if your teacher is cramming it in this late, probably wont be much of it on the final
I already took the final. It's just an extra thing.
oh, is that the only single trial point you did? did you try different distances?
yes until the weights were balanced (almost)
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
one test isn't enough
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
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
so 100*28 + 200*20 ideally would be 300*20
Okay I get it thanks. I have some other questions, will you be able to help me on them?
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
It's C right?
yes, a velocity is a vector quantity
has a magnitude and a direction for distance/time
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?
First need to draw all those vectors
The average velocity will be the resultant of those 3 vectors, the sum of them
The average velocity is the total displacement/ total time Displacement is the start point to the end point, independent of the path
I got the velocity of 33.33m/s
I am wrong, right?
since they made a square, the magnitude or R is the same as the top piece
0.8km/9min = V_avg
His net displacement is 0.8 km straight west, the total time is (10 + 9 + 11) min
Average Velocity = Total Disp[lacement / Total Time
0.8 / 1800 sec
I would just leave it in the units the prob already is in , km/min, since it didn't ask to change it
Our teacher likes seconds to be used that's why.
0.8km / 30 min
oh, then convert to whatever units you need, but that is the prob
about 0.027 km/min