0:00
Hi there.
00:01
So for this problem we need to calculate the magnitude of the torque and the direction of this and on each of these cases that are shown in this figure.
00:14
So we are given the magnitude for the vector force that is equal to 10 newtoms.
00:26
And we are also given the magnitude of these or the length of the road, that is the magnitude also of the vector art.
00:40
Well, i'm going to call this better the length.
00:45
So the length and has a length of four meters.
00:52
Now, with this information, we need to calculate the torque.
00:56
So we know that the torque, it is a vector, and it is defined as the cross -product between the position bettor, from the point of rotation to the point of where the force is applied, cross the force that is being applied.
01:15
Now we can obtain the magnitude of this vector, and we know that the cross product in this case will be given by the product between the magnitudes of the position vector and the force vector times the sign of the angle that that is the angle that meets the position vector with the force vector.
01:46
Now, for the case a, we will have the following.
01:52
As you can see from the picture, we have a vector force.
01:57
Well, this is the vector that i'm going to try in red.
02:02
That is the vector of position from zero, where zero is the point.
02:09
Of rotation and in here we have a force that is f we can translate that to upward it doesn't change the problem so in here we have that the angle between these two is 90 degrees so we can find first the magnitude of the torque in this case we can find first the magnitude so we will find that this is the magnitude of d, as you can see in this case, the magnitude of earth corresponds to the length of the rod.
03:03
So we can substitute that in there, so it will be four meters.
03:10
And the magnitude of the force, that is 10 newtons, and the sign of the angle that makes that position vector with the 4.
03:24
Vector that is 90 degrees.
03:26
And we know that the sign of 90 degrees is just one.
03:30
So we obtain from this that the magnitude of the torque in this case is 40 newtons per meter.
03:39
And now to find the direction of this, we know that we can obtain that from the crosse project.
03:50
So for the crosse per audit, we use the right -hand rule.
03:55
So we put our hand in the vector r because it is r cross f.
04:06
So we put our vector f and we go with our palm to the other vector f.
04:13
So we obtain that our thumb must indicate the direction of this.
04:20
So if we put that this is our a axis for this, we will have eds, and we'll have a ceta axis, you will find that the direction of that vector is perpendicular to both, so it will be in this direction, it will be in the ceta direction.
04:41
Now, for part b of this problem, we have the following situation, we have that.
04:47
This is the vector, the vector art, and this is the force that is being applied to that.
05:01
So as you can see, we are already given the angle between the force.
05:11
This angle in here is the one that we want.
05:16
And that angle corresponds to 120 degrees.
05:22
So again, we use the same formula as before.
05:27
The magnitude of the torque is equal to the product between the magnitudes of the position better times the magnitude for the vector force times the sign of the angle that in this case is 120 degrees.
05:43
So we substitute again all of those values.
05:46
We know that the position vector has a magnitude of 4 meters.
05:51
The force vector has a magnitude of 10 mutants and the sign of 120 degrees.
06:02
So plotting this into the calculator, we obtained that the magnitude 4...