00:01
Okay, so you've said, so we've got a bar, and from a pivot at a distance 30 centimeters, we have a force f1 equals 0 .1g, where g is newton's gravitational acceleration constant.
00:19
Then we've got a force 0 .45 meters away with magnitude f2 equals 0 .05 times g.
00:34
And then finally, we've got an additional mass that you've determined experimentally, and that's given by its mass m3g, and it's a distance d3 meters away from the pivot.
00:55
You found m3 equals 0 .02, and d3 equals 0 .144.
01:02
We want to find out what the condition of rotational equilibrium is.
01:07
So that means that the anticlockwise moment equals the clockwise moment.
01:12
And the anticlockwise moment is just given by f1 times the distance from the pivot 0 .3, or equivalently 0 .1g times 0 .3.
01:21
And the clockwise moment is given by f3 times its distance from the pivot d3, plus f2, which is 0 .05g, times its distance from the pivot 0 .45.
01:34
And so the condition of equilibrium is simply that f3d3 is equal to 0 .1g times 0 .3, minus 0 .05g times 0 .45, which is 0 .0736 newton meters.
02:03
So we want, the next bit is to calculate the counterclockwise moment.
02:09
So the counterclockwise moment is just given by f1 times 0 .1g...