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Hi there.
00:01
So for this problem, we have the situation that is shown in here.
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The information that we are given is that the mass of the sky is 60 kilograms and the slope is 30 degrees.
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And there is a force.
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We are going to call this the force of earth, f .a.
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And that magnitude of the force is given.
00:22
That is 10 newtoms.
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We are also given the information about the coefficient of dynamic friction.
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That coefficient, we're going to call just simply as mu and that is equal to 0 .08.
00:35
So the question in here is about what is the resultant force that adds on this kire? so we need to just simply add the net force in here.
00:46
So with that set, let me choose.
00:49
First of all, we're going to set the axis parallel to, yes, parallel to the incline, just like this, then we will have the the x -direction, the x -axis, oh well, let's put it the x -axis, the positive x -axis, and to the right, and the negative and the positive y direction perpendicular to the surface, just like this.
01:22
Okay, so that's what we have.
01:28
Now, the forces that are apting on this, skyer are the normal force always perpendicular to the surface of contact.
01:37
We also have the force, the resistive force of the earth, f .a.
01:44
This opposes the motion because the skyer is moving downward.
01:51
So the direction of that resistive force is in the opposite direction.
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And also we have some frictional force, f .r.
02:00
Because we are given the information about the dynamic frictional, the coefficient of dynamic friction.
02:09
And now the other forces that are opting on this are the weight always points towards the center of the earth, the weight.
02:19
And, well, that's it.
02:21
So we know that this weight meets an angle with respect to the vertical, which corresponds to the same 30 degrees of the incline.
02:29
Now what we can do is to sum all of the forces acting on the, let's start with the y direction.
02:37
Now as you can see from the figure, we will have just simply the normal force minus the weight because points downward.
02:45
And that will be the vertical component of the weight, which is the magnitude of the weight times the cosine of 30 degrees.
02:54
And because this system is not moving in the vertical direction, then we set this equal to zero.
03:01
So from this, we obtained that the normal force is equal to the weight, which you can recall that the weight is defined as the mass times the acceleration due to the gravity, times the cosine of 30 in this case...