00:01
Hi there, so for this problem, we have the situation that is shown in this figure.
00:06
Now there is an uniform being of mass mb, so that is given.
00:15
We also told that the length of this is held, and this is supported on the left by the hinge attached to the wall, and to the other side is attached to a cable with an angle teta with the horizontal, as is shown in here.
00:49
And there is a package of mass mp, this one right here.
00:58
And it's positioned on this bin at a distance adds from the left end, adds in here.
01:08
And we are told that the sum of the masses of the beam and the package is given so the sum of this two it's equal to 61 .22 kilograms and we're also given a graph of the tension in the cable as a function of the cable as a function of of the position of the package.
01:45
And that is given as a fraction of s over elt on the, which, where ll is the pin length.
01:57
Now, we are told that the scale of t, this value in here, t, a, is equal to 500 newton.
02:11
So t .a.
02:14
In the graph is 500 newtons.
02:18
And t .b.
02:24
Is 700 newtons.
02:30
Now for this, what we need to obtain is the angle teta.
02:35
For this first part, we need to obtain the angle tita.
02:41
For part b, the mass mb.
02:47
And for parc the mass of the package.
02:53
Now, in order to solve this, we first need to consider all of the torts in this problem.
03:03
Now, we will have three forces that are making torque.
03:09
The tension in the cable, the weight of the package, and the weight of the bar itself.
03:17
Now, we start to start.
03:19
With the tension.
03:21
So the tension is right here.
03:25
It is pointing to the right.
03:27
This is the tension.
03:31
And we know that from the point of rotation to the application of that force, this is the vector, and that vector has a magnitude of l, the length of the bar.
03:44
Now, if we applied, if we applied the right -hand ruled, we will obtain that that is the cross -brother between the position vector and the tension vector, and that will give us a torque that is out of the screen.
04:04
And if we set that to be the positive direction, we will have in here a positive.
04:11
And of course, these times, the times the distance to the point of application between the point of application of force and the point of rotation.
04:23
And this, of course, because we have in here, we are doing the torque with respect to the vertical components.
04:33
So we will have in here that this is times the sign of theta.
04:41
Now, the other force that is making torque is the force in the package.
04:49
This one right here.
04:51
This is the force in, well, the weight of the package.
04:57
And the distance from the point of rotation to that point of application of that force, if we do the right, if we use the right -hand rule, we'd obtain that that torque is inward, the screen, so we will have that that is minus the weight.
05:17
And the distance from the point of rotation to the point of application of the force is the distance that we are given.
05:26
Eads in here...