00:01
Hi there, we are given a meter stick here with two objects suspended on it.
00:06
We are to determine which among the following configurations or arrangements here will result to the meter stick being balanced.
00:18
So we'll take m1 or object 1 as having 0 .50 kilogram mass so that it has a weight of 4 .9 neutrons and the second mass is twice as mass.
00:30
As if that's the first one.
00:32
So its weight is also twice the weight of the first one.
00:37
Written in green box here are the things or the principles that we will be using in answering the main question.
00:46
So for a system to remain in balanced, there has to be, basically we'll use the second condition for equilibrium, which says that at any point, the total torque on the system must be and we have the definition of the magnitude of torque here, which is the product of the applied force and its moment arm.
01:08
And if the torque causes the object to rotate counterclockwise, then we'll take it as positive torque and negative if otherwise.
01:17
So let's start with configuration a.
01:21
Since the fulcrum is at the midpoint, then that is also where we will put our pivot.
01:27
So let's have our pivot here so that all moment arm measurements will be with respect to the middle of this uniform meter state.
01:38
So in other words, weight one here has a moment arm of 0 .50 meters.
01:46
So from the point of application to the pivot, this is now the moment arm or the distance that is perpendicular to the axis of rotation.
01:57
So that makes a 90 degree angle with that.
02:00
So this is moment arm of w1, which is, of course, half a meter.
02:07
And the same thing is true with the weight number two.
02:11
So it's also half a meter away from the pivot.
02:15
So this is moment arm two.
02:17
By symmetry, since they both have the same moment arm, but weight two is more massive is greater than weight.
02:28
Weight 1, then it's already obvious that the clockwise rotation that will be caused by weight 2 will be greater than the counterclockwise rotation by weight 1.
02:40
So we're saying here that the torque due to weight 2 is greater than the torque due to weight 1.
02:48
Therefore, the summation of the torque here is not 0.
02:53
So this configuration will not result to a balanced stick.
02:59
So this is not our answer.
03:02
Let's go to letter b.
03:04
Weight 1 is suspended at 10 centimeters from the left end.
03:08
So from here going to the axis of rotation, this will be the moment arm 1, which is 0 .40 meters.
03:17
And then this one, which is placed at the 70 centimeter mark.
03:23
We'll have a moment arm of, with respect to this, so the moment arm of two is 0 .20 meter.
03:33
Let's determine the total torque and see if it's equal to 0.
03:38
Okay.
03:39
So we have here a counterclockwise rotation, positive torque, clockwise rotation, negative torque.
03:47
So the total torque with respect to the center would be positive torque due to one, negative torque due to 2.
03:55
So this would just be weight 1, moment arm 1, minus weight 2, moment arm 2.
04:01
So this will be our equation for the rest of the configurations here.
04:07
So we have 4 .9 times 0 .4 minus 9 .8 newtons times 0 .2.
04:17
So the total torque here in configuration c is 0.
04:23
Therefore, this is this system will be balanced...