00:01
This exercise, we have a beam that's hold in equilibrium, as shown in the picture, by a force that acts upward with an angle of 30 degrees with respect to the line of the beam.
00:18
And the beam itself makes 40 degrees with the horizontal.
00:23
There's also another force, f, of which we don't have any other information.
00:29
And finally, the final force that acts on the beam is the weight force itself, such that mg, the weight force, is equal to 124 newton.
00:44
And it's pointing down.
00:46
Now, we know that the beam is held in equilibrium, meaning that the sum of all the forces is equal to zero.
00:55
So let me write all the forces in a vector component form.
01:02
So for the weight force, the weight is equal to m g, that's just the weight of the beam, times minus j.
01:16
J is in vector in the x direction.
01:18
So i'm setting up a coordinate system such that the y -axis is pointing upwards and the x -axis to the right.
01:25
So the weight force is minus 124 j nons.
01:33
Also, the force of 100 newton's acting with an angle of 30 degrees with the line of the beam, which i'm going to call lowercase f, this is equal to, well, in order to find this force, the first thing we need to do is to find what is the angle that this force does with the horizontal.
01:57
Since the beam does an angle of 40 degrees with a horizontal and the force does an angle 30 degrees with a beam.
02:03
Then the force does an angle of 40 plus 30 degrees with a horizontal in total the angle is 70.
02:12
So the force is pointing to this direction.
02:17
This here is 70.
02:21
The x component of the force is the magnitude of the force, which is 100 newtons, times the cosine of 70, i plus 100 times the sign of 70 j.
02:37
So the force is equal to 34 .2i plus 93 .7 j newton.
02:59
And the final force is that we want to find out.
03:02
So what the exercise asks us to do is to calculate the magnitude and direction of the force, capital f...