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0:01
PROFESSOR: Hello, guys. How are you doing? Today we're going to be solving some problems. I'm going to do some practice problems. So you can go and review them at home. Basically, it's going to be about chapter 2. We're going to be working with vectors, just basically vector the composition. The first problem that I want to work with you is this one. And he says, "If the magnitude of the resultant force needs to be 500 newton, directed along the positive y-axis,
0:35
determine the magnitude of the force" at its direction angle, or this angle here. Not the direction angle, this angle here. So the first thing that you have to do, as always, is to learn how to pick up the clues from the problems. And the first clue that we have is the magnitude of the resultant force. But we also know the direction of the resultant force. Because it says that it's going to be detected along the positive y-axis. Meaning the resultant force is going to be acting in this axis.
1:05
And it's positive. So that means that it's going to be pointing up. This problem, basically, the way that it's stated, you can do it in two ways. The first way, you can do it by using trigonometry and using the law of sines and cosines. And this is the first approach that we are going to do. So basically, we're going to do the parallelogram law. What we have to do is build a parallelogram here. This is going to be parallel to the force. And at the same time, we're going
1:36
to have the other one parallel to this one in this direction. That should coincide with the end of the force, if this is made out of a scale. But I know for sure, that the resultant force has to be here. So this is going to be my resultant one, from here to here. And I know the value. Because they are telling me here the value. The value for this one, for this, resultant force is 505 500 newton. I'm not going to put the units but you know it's newtons in this case.
2:08
So once again, remember what we have to look for. What we are looking for here is the value of this force F, and this angle, and this position. What else do we know regarding this problem? Well, we know this angle here, which is 15 degrees, is given with the horizontal axis.
2:34
If this is 15 degrees and this is 90, because x and y are orthogonal-- it doesn't say it but they are orthogonal.
2:48
Usually when the angles, the axes are referred as x, y, it's because they are orthogonal. If they weren't orthogonal, they would use a different type of letters, like u, v, or something like that. So the total angle, if this is 90 degrees here and this is 15 degrees, the total angle between these two vectors, or these two distances, is going to be 90 plus 15. So it's going to be 105 degrees, this angle here. Remember, this angle is 90 degrees and this is 15. 90 plus 15 is going to be 105.
3:23
And the value of the force F, the magnitude, is going to be this side of this triangle, force F. This is going to be force F. Well, if you remember the law of cosines, the law of cosines says that if you have a triangle, you can say that c squared is equal to a squared plus b squared minus 2 times a times b times cosine of the angle between a and b. Well, in this particular example that we are following, this will be-- or these will be a and this will be b.
4:04
And this is the angle between both of them. And what I'm looking for is F. So I can say that F is going to be equal-- I'm going to take just the square root on both sides, here and here, to eliminate the c squared. So F is going to be the square root of a. But a is 700. 700 squared plus b. But b is these other vectors that we have here, the other 400, 500. So it's going to be 500 squared minus 2 times 700 times 500 times cosine of the angle between a b. Cosine of 105.
4:51
Once you solve this, F is going to be equal to 959.78. And remember to use the unit, newton. First part of the problem done. This is the first part of the problem, magnitude. Now the second part of the problem. It says find it's direction, this direction here. This I know. I can find that angle in different ways. But the easiest way, or not the easiest way, the way I'm going to show you right now is finding this angle through this other angle here. I'm going to call this angle alpha.
5:37
And of course, this angle theta is going to be 90 degrees minus alpha. And we are going to apply the other law, which is the law of sines.
5:53
And remember, the law sines, what it says is, if you have a triangle-- I'm going to put it here-- If you have a triangle like this, a-- and this angle is a. This side is b, and this angle is b. And then this is the angle c, and this is the side c. In the same way that you have c squared equals a squared, b squared minus 2 time ab cosine alpha. Well, in this case, alpha will be this angle here, c for the law of cosines. The law of sines established that a divided by the sine of A
6:41
has to be equal to b divided by this sign of A. And it has to be equal to c divided by the sine of C. So the side divided by the sine of the opposite angle has to be equal. It's the same ratio for every side of the triangle. So if I'm going to apply the law of sines for this one. I know this value here because this distance from here to here is going to be 700. How do I know that? Because it's 700 here. And I know this one because this one is 500. And I know also this one because this one here I
7:15
just calculated like 959. Also, I know this angle here. This here is 105. What else do I know? If this angle here is 105, that means that this angle here is also 105 because these are two parallel lines cut by a straight line. Or you can say alternate [INAUDIBLE] a triangle. There are two parallel lines cut by one straight line. So this is 105. This is alpha. If I apply the law of sines, I can say perfectly that the sine of alpha, sign of these angles divided by 700--
8:03
which is basically the same formulation. I'm just flipping it, both sides-- has to be equal to the sine of 105 divided by this side. And side I just calculated. And it's 959.78. From here, we can solve for alpha. And you can say that alpha is going to be equal to the inverse, or 700 times 959.78. Multiply by sine of 105. And alpha is going to come to be 44.79 degrees. But this is not the answer because alpha will be this angle here. The angle theta is the one that we are looking for.
9:06
And that angle is going to be 90 degrees minus alpha because alpha and theta are complementary. So 90 degrees minus 44.79 degrees. And theta is going to be equal to 45.2 degrees. That's the first way of approaching this problem. As you can see, it's a really easy problem.
9:35
It's a very easy problem. But I don't know why a lot of people struggle with the law of sines and cosines. I don't know if the physics background that you need or geometry, trigonometry background is not sufficient for some students. But if this is the case, and you don't even remember what we're discussing here, I'm going to ask you, please, to go on and review that part. OK. Let's check the second approach, which is more like a vector approach in this problem.
10:11
What we are going to do, we're going to use the components of the forces. And we're going to use also the principles of equilibrium.
10:28
And basically, if this is the same problem and we know the resultant force is going to be directed, and it's going to be 500-- and I like this approach more because, don't get me wrong. It's not that I don't like math or trigonometry. But I like more physics. I like to see what is happening. I like to see what happens in the real world. So I know that if this is 500 newton and it's acting vertical, going up, it's because the horizontal force, the resultant force in x is 0. There's nothing next.
11:06
The resultant force is only vertical. So then I can come and decompose this in the corresponding components. The horizontal component of this 700 force, which is this one, is going to be 700 multiplied by cosine of 15.
11:27
And also, the horizontal component of this force F, which will be acting here, is going to be F cosine of the angle. And I know the resulting force in x is 0. Or I can say the summation of these two is 0. So this is going to be then F cosine theta. Positive because it's going in the positive direction of x minus-- remember, negative, it's acting in the negative direction, pointing to the left-- 700 times cosine of 15 degrees. And that's going to be equal to 0.
12:08
Or I can say also that F cosine theta equals 700 times cosine 15 degrees. And thus, I'm going to call that equation one. Now I'm going to do the same thing with the vertical component. The vertical component of this force here, F, acting in this direction, I can say that the summation-- I didn't copy this here before-- but the summation of forces in x is going to be equal to 0. And now I'm going to say that the summation of forces in y is going to be equal to the resultant force, which is 500.
12:48
Why? Because the problem states that, it's telling you that 500 directed along the positive y-axis.
12:58
So if I know that, then I'm going to do it. This component here of F, this one, this one, once again, is going to be F times sine of the angle. And this other component of this other force here is going to be F, which is 700 multiplied by sine of 15.
13:27
And this is 500. 500. So what I have is F sine of the angle minus 700 sine 15. It's really important that you keep in mind and keep track of the signs. Negative because it's acting to the opposite side of the positive side, in this case, downwards. That has to be equal to 500, which is the resultant force. And from here, I can solve and say F sine of the angle is going to be equal to 500 plus 700 sine of 15 degrees. And I'm going to call this my equation number two. This is a number.
14:18
Sine 15 times 700, that's a number. What I'm going to do now is I'm going to divide equation two, turn by turn, by equation one. And then I'm going to get F sine of the angle equal-- remember, this is a number-- 500 plus 700 sine 15 divided by F cosine theta, which is the second equation. And the second term is 700 cosine 15 degrees. Once you solve this, this and this cancel out. And then you have sine divided by cosine is tangent of the angle. And that's going to be equal to 1.0074.
15:17
And before you say anything, I know the number of decimal places and significant figures here doesn't match the other one. But I don't care. I'm just showing a problem. If you want to run out the result at the end, you will give the appropriate amount of significant figures. And then I can say that alpha is equal to the inverse tangent of 1.0074. And no alpha, I mean theta. And theta equals 45.2 degrees, which is the same threshold that we got before here, in the other procedure.
15:55
Now the other thing that we have to do is calculate the force. And for doing that, I just have to plug this either into equation two or into equation one. Equation one is a little bit easier, not that it's that much easier, but I'm going to plug in 1. Plug in 1. And then we are going to have that F cosine of 45.2 has to be equal to 700 cosine 15, which is equation one. And then F is going to be equal to 959.78 newton. There you go. Two approaches for solving the same problem.
16:55
Now I'm going to try to a do another problem in the same recording. I hope that it's not too long. I was going to do several recordings for several problems. But I'm going to do a new recording [INAUDIBLE].. A new problem in the same recording. Let's try to solve now this other problem.
17:22
This is a problem that usually has a lot of difficulties. And the students always have issues trying to solve these type of problems. Why? Because students are used to say, find the projection of the vectors, projection. And one thing is the projection and the other thing are the components.
17:51
What happens is that the projection matches the components if the axes are perpendicular. For example, let me explain my point. This is y and this is x. And these axes are perpendicular. If I have a vector like that and I want to find the components of that vector, there there's no problem with that. Easy. Why? Because the components of this vector-- this is vector I don't know, F, the force F-- then this force F is going to have a projection. Now the other thing is that this is like the shadow
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that that vector projects over the axis. And like that, this is the projection vector. This is the projection of the force F over the x-axis. And this is the projection of the force F over the y-axis. You see? It's like the shadow of that vector. And of course, if this angle is alpha, x is going to be equal to F cosine alpha. And this one, which is the same side of Fy, then is going to be F sine of alpha. And this is what you are used to doing. Now what happens in a situation like this?
19:18
What happens in a situation where you have one axis like that and the other axis like that? And then you say, oh, wait a second. This axis is u and this axis is v. And then you have a force or any other vector, which is going to be acting somewhere here.
19:53
But this is the tricky part. If I want the projection, I'm going to look for the shadow of this over this axis. And if this angle is alpha, let's say, the projection is going to be this distance. And guess what? If this is F, the projection of F over the axis u is going to be F cosine alpha. This is a projection but this is not a component. In order to find the components, I have to go and I have to go parallel to the axis, parallel to the axis.
20:25
So if I go parallel to this axis here, and I look for this line. And this line is parallel to this one. And in this other case, I'm going to make another line parallel to this other axis here also. Then the projections of that vector-- I mean the components of that vector are going to be this and this.
21:02
You see? This is the component. This is Fu and this is Fv. And as you can see, Fu is very different than the projection. Because the projection, I told you, was this one. This is the projection, this is Fu. So if you say, F cosine of alpha, what you're calculating is this, another component. That's why a lot of people have a lot of trouble doing this type of situation. But I don't know why there is trouble. Because you apply the law of sines or law of cosines. Which by the way, the law of sines
21:40
is exactly of cosines, what you do in this case. Differences. The angle here is 90 degrees. That's the only difference. And when you say F cosine, or F sine, in this particular case, this is a particularity of this case. But this is the general case. So what you do is, you say, once again, this side divided by sine of alpha has to be equal to this side by sine of this angle has to be equal to this side divided by sine of the x angle. Now let's say that I'm this case.
22:24
Oh, you're going to say, but you're missing here sine of the other angle because you have this one here or you have this one here. Yes, but the other angle is 90 degrees. And when I say, F-- look at this case. This is 90 degrees. So if I apply the law of sines here, I'm going to say, F divided by sine of 90 has to be equal to Fy divided by sine of alpha. So Fy, solving for Fy-- I pass to this to this side-- is going to be F sine of alpha divided by sine of 90. But sine of 90 is 1.
23:08
So basically, I have this. Which is what you did before. But that's a particular case of the law of sines when you have a right triangle. This is not what happens in this case. In this case, as you can see, you have two axes, u and v. And those axes are not perpendicular. So because the axes u and v are not perpendicular, then we have to follow the law of sines. Remember, the law of sines, once again, says that a divided by sine of A has to be equal to b divided by sine of A
23:59
has to be equal to c divided by sine of A, where A, B, and C, capital, are the sides. And a, b, and c, small caps, are the angles.
24:12
And then we come to this problem. And we have to try to solve it. First, read the problem. The problem is asking us to resolve, resolve, not project. Resolve the force F1 into components acting along the u and v axes. Only F1, not F2, no resultant, no, nothing else. Just F1. Well, if I want to find the components of F1 with respect to these two axes, what I have to do is come parallel to the axis. I have a line here, which is the axis.
24:48
And I have to put the other line parallel to the u-axis, which is here, like that. And my components are going to be this one up to here and this one up to here. Right there. Now I have to find the rest of my angles in order for me to solve this problem. So how are we going to do that? Well, I know for sure this angle. How do I know this angle? Once again, I have two parallel lines intersected by a straight line. This angle and this angle are the same. So this angle is 30 degrees here.
25:37
That's the first thing that I have there. What else do I know? Let's think about it. What else do you know here?
25:53
Well, if you look closely to the picture, looking at the big picture here, I have the u-axis here and the v-axis here.
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u and v. And this angle here is 70 degrees. And 70 degrees is the same angle here because they're vertical angles, or both by the vertex 70. So this is 70. That means that if this is 70, this here has to be 40. And for supplementary angles-- not supplementary angles-- the summation of all internal in a triangle has to be 180. That means that this internal angle here has to be 110. 110 degrees.
26:45
Now I know three angles. And I know this side. And this side is 300. So I can say immediately that applying this law, 300 divided by the sine of the opposite-- this is 300 and this is the opposite-- sine of 110 has to be equal to this side, which in this case, we're going to say this is F1 u.
27:15
F1 u divided by the sine of the opposite angle, which is 40.
27:26
And from here, I pass these to the other side. And I can calculate F1 u. And F1 u is equal to 205.2 newton. And it's positive because it's acting in the positive direction of the u-axis. Or I can say also that 300 divided by this sine of 110 has to be equal-- now I'm going to the other one, which is this one-- this side is F1 v because it's acting in v. F1 v divided by the sine of the opposite. But the sine of the opposite is 30.
28:14
And then I can calculate F1 v. And that's 159.6 newton. I want you to try now finding the components of the other one, this one, F2. But F2 is the same. I just moved my axis here to the bottom one. And then my components are going to be this one and this one. The problem is not asking for that. But let's say that what if the problem is asking me also for F2 in u and F2 v? Well, if that's the case, I know this angle is 45. I know this side. Can I calculate the rest? Yes.
29:08
This angle here is 70 here.
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How do I know it's 70? Well, I know this is 70 because this one is 110 and this one is 110. And 110 plus 70 is going to be 180 for supplementary angles. So this is 70 here. You can have something like that, something like that, and something like that. That's what you have. This angle here is 70. This angle here I know it because it's a given, 45 degrees. And because this is 70 and this is 45, this is 115. To 180, this has to be 65 degrees. And I know this value here, which is 500.
29:48
So the side of the triangle, this value is going to be F2. Yes, F2 u. And this is going to be F2 v. I do the same thing. I say F2 u divided by this sine of 45 has to be equal to 500 divided by the sine of 70. And even more, I can say also, that it's going to be equal to F2 v divided by a sine of 65. And then I select whatever part-- if I want to calculate this, I select this part. If I want to calculate the other one, I select this part. And then I get that F2 u is equal to 376.24 newton.
30:43
And F2 v is equal to 482.24 newton. Now pay attention to this. If you look closely, even though this is going down-- because this is the other thing that you are used to doing. You are using your mind to say, everything that goes on is negative. No sir. It's not like that. This is not negative because it's coming in the positive direction of the v-axis. Wherever you have the letter, that's a positive direction. So this component is positive. But look at this one. The component of u is negative.
31:22
And guess what? The law of sines is not going to give you the negative. The negative you have to add it. This is the one. If I want to calculate the resultant force, which the problem is not asking me, then I have to add everything that goes in u together. So I'm going to have 205.2 minus 376.24. Remember, this goes in the u direction, plus whatever goes in the v direction. 159.6 plus 482.24. And this goes in the v direction. Whatever this operation is, I didn't calculate it. You can do it.
32:08
Now once again, pay attention. This is going to be negative. And this is going to be positive. Why? Because it's bigger than this. So if these are your axes-- once again, this is u and this v--
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then the u component is going to be negative. And the v component is going to be positive. And it's going to follow the axis all the time. That means that your resultant force is going to be somewhere here.
32:55
How are you going to calculate that resultant force? I don't know. You have this angle here because you have it. You have this value, you have this value. You apply the law of sine again and calculate the resultant. Do not apply Pythagoras, which is the square root of this square plus this square because Pythagoras, once again, applies only to right angles. And this is not a right triangle-- I mean right triangles. And this is not a right triangle. So don't do it like that.
33:23
You can calculate R if you want to by the law of cosines. Because you know this angle here. And then you can say that R is equal to this square plus this square minus 2 times this times this cosine of that angle. And then you can calculate the resultant R in that way. I hope this problem helps you clarify that situation. Let me see. I think I have another problem that I can do. And we have time. Yes, I think we have time. Let me see another problem. Let's try to do-- OK, this problem is nice.
34:03
Let's do this problem.
34:10
This is a nice problem. And also, it allows us to, I don't know, help you at least to read the problem and pick up the clues that are intrinsic in the problem. Let's read the problem. And one of the things that you have to learn is how to read problems. It's says, if the magnitude of the resultant force acting on the eyebolt is 600 newton-- magnitude of the resultant force. Magnitude of the resultant force is 600 newton. Now is the other part, which is also important.
34:56
Because the other part, a lot of people just keep it and don't read this clockwise part here. It's direction measured clockwise. Clockwise means in this direction, from the positive x-axis is 30 degrees. That means that my resultant force is going to be acting in this direction. This is going to be the resultant force. This is going to be 600 newtons. And this angle from here to here because it says clockwise. Otherwise, if you doesn't say clockwise, it says each direction is 30 degrees,
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it has to be counterclockwise. Should be counterclockwise because per default, the direction is measured counterclockwise from the positive direction of the x-axis. But in this particular problem, it's telling you, the problem is telling you that it's measure clockwise from the positive x-axis. So this angle is 30 degrees. Right there. And this is 600. Now we have to find the magnitude of F1 and the angle. How are we going to do this problem? Basically, this problem is solved in the same way
36:12
that we solved the other problem. We are going to decompose every force in components. Then we're going to add x, add y. The summation of the forces in x has to be equal to the component x of the resultant force. And the summation in y has to be equal to the component in y of the resultant force. That's basically what we're going to do. So let's start with the first problem, with the first force. And then the first force is going to be-- let's have a summation of forces in x equals 0.
37:04
Let's do that. Not equal 0, equals the resultant force in x. This is force 1. Force 1 has this horizontal component. The value for this horizontal component is going to be-- now the axes are perpendicular. So then it's going to be F1 x, which is this one, is going to be F1 multiplied by cosine of the angle. Why cosine? Because it's adjacent. So F1 cosine of the angle to the right. So it's positive because this is going to the right. Now plus the horizontal component
37:43
of this one, which is this one. What is its value? I don't know. I need this angle here. How much is that angle over there? That angle over there, either you say that, oh, wait a second. This is 60 and this is 30. So you can use this one or you can use this one also. If you use this one, then you're going to say, sine of 30. But if you use the other one, which is this one here, you're going to say cosine 60. It's the same thing. Let's use this force, this component here.
38:27
Because I don't want to mess up this drawing more. So this component is going to be F2 sine 30. And it's positive also. Plus F2 sine 30, which is the same cosine 60, remember. Here is the opposite but here it is adjacent to 60. What else? Now I have this one. You know this little triangle here, the component, the horizontal component of F3 is going to be this one. And this little triangle, what it's telling you is the ratio between geometry and force.
39:12
So basically, what it's telling you is that this force, this horizontal component is going to be 3 divided by 5, 3 divided by 5 multiplied by 450. And this is negative. Why? Because it's pointing to the left, to the negative direction. And that's important. Minus 450 times 3 divided by 5. And this F2, remember, I didn't put it. But this F2 is 500 because it's given also. The summation of forces in x is this one. And that has to be equal to the horizontal component of the resultant force.
39:58
And the horizontal component of the resultant force is 600. Let me put it here. This is the resultant force.
40:12
This is the horizontal component of the resultant force. So it's going to be 600 cosine 30.
40:27
Or I can say that F1 cosine of the angle. And I solve all these numbers because those are numbers. I take this, pass it to the other side, negative. I take this, pass it to the other side, positive. What we have there now is this is going to be equal to 539.62 newton. And I call this equation one.
41:08
And then I do the same thing. And I say, summation of forces is in y has to be equal to the resultant component in y. And summation of forces in y is this direction here. So the y component of F1 is going to be the side of the triangle, which is going to be F1 sine of the angle. We're going to have F1 sine of that angle, this one. And now the vertical component of F2 is this one. And this one is, [INAUDIBLE] that, first of all, it's negative. Why negative? Because it's pointing down. Negative.
41:44
Negative what? Negative 500 multiplied by cosine 30.
41:51
And the vertical component of this one is also this side. And this side is 4 this time. You see? 4 divided by 5. And it's negative, 4 divided by 5. Because that's what the little triangle is telling us. 4 divided by 5 multiplied by 450. And that has to be equal to the summation of forces in y. Summation of forces in y-- I'm going to copy it in this side because I ran out of space here. The decimation of forces in y-- the resultant force in y is this one over here.
42:26
And this one over here is negative, first of all. Why negative? Same reason. It's acting downwards. So negative. Remember this is the component. Negative 600 sine 30 is going to be equal to that. From here, I get this, pass it to this side, positive. This, pass it to that side, positive. And I get that F1 sine of the angle has to be equal to 493.01 newton. Second equation.
43:05
You can solve this whatever way you want to. But it's easier for me to divide 2 by 1. And if I do that, then I'm going to have F1 sine theta equals 493.01. And the other equation is going to be F1 cosine theta equals 539.62. And these and these cancel out. And then if you go here, you have that tangent of the angle is going to be equal to this divided by this, which is 0.91. And then you take the inverse tangent of this. And the angle is going to be 42.42 degrees.
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And this is the value of the angle. Once we have the angle, we can plug it into here or plug it into here. And we can calculate F1, solving for here. Let's say that we plug it into this one. Then F1 is going to be equal to 493.01. This is going to pass to the other side, dividing sine of 42.42. And then F1 is 730.92 newton. There you go. Another problem that you can go and check and practice on your own. And I hope that these problems help you a little bit with the material.
44:48
That's my intention, at least. That's why I'm doing it. Last problem for today. I didn't think I was going to do so many problems. But I'm going to do it. Let's check this problem. And this problem is interesting because there are several ways that you can represent a vector. You can represent a vector using the magnitude and the cosine directors, or the unit vector. You can represent a vector by the cosine directors. And you can represent a vector or you
45:17
can decompose a vector using the projection angles. And you have to be able to identify which one is which. In this particular case, we have two. Now let's start with F1. F1 is this one here. Well, let's start by reading the problem. The problem says that the bracket is subject two forces shown. Express each force in Cartesian vector form. And then determine the resultant force. Find the magnitude and coordinate direction angles of the resultant force. That's basically what we have to do.
45:46
Let's start. Let's start with the angle F1. This is a rope here, and this is a rope here acting in that bracket. Both of them are going to be applying a tensile force in this direction. And we have to calculate the resultant. In order to do that, the easiest way-- all the time is find a Cartesian coordinate. And even more now that we are in 3D. For the Cartesian components, express the forces in Cartesian notation at, whatever goes in i with whatever goes in i. Add all the j's together.
46:23
And add all the k's together. And that's going to be the force. Once we have the force, find the magnitude. How do we find the magnitude. The square root of this summation of the squares of each one of the components. And how do we find the direction. By finding the cosine directors. And I explained that in class how to do it. But anyway, let's do it again. So F1 is this force here. The first thing that we know about F1 is that if you look at this angle here and this angle here,
46:53
are projection angles. How do I know they are projection angles? Because they are octagonal. You see. So the first thing that I'm going to do is I'm going to project this angle, this force, over the plane xy.
47:10
So the projection of F1, let's say, over the xy plane, it will be this side. And that side is going to be F1 multiplied by cosine of this angle. And this angle is 35. How do I know it's 35? It's given. So the projection of the force F1 over the xy plane is 250 multiplied by cosine of 35 degrees. Perfect. I have this.
47:46
And now I can take this to the y-axis and to the x-axis because I have this other triangle here. So this force I know is this one. If I want to find the x component, which is this one, this is the x component, remember, F1 x is going to be equal to the projection F1 over the plane xy multiplied by sine of 25. This is the angle. This is 25, now over the plane. And this is the opposite. So it's going to be sine of 25 degrees. And F1 y, y which is the projection over this axis here y.
48:28
It's going to be cosine 25. So it's going to be F1 xy cosine 25. But if the projection over the plane xy is this one, so if I want to calculate F1 x, I just plug this into here. And that's going to be 250 times cosine 35 times sine 25. And F1 y is going to be 250 cosine 35 cosine 25 degrees. This value is 86.55. Remember our forces are in newton. And this value here is 185.60 newton also. Wait a second. Why two components only if we are x, y, z? Oh look.
49:27
This is the z component, not the force, direct. That's the easiest one to see. Why? Because this angle here is a vertical angle like that. And it's 35 degrees. So this one is going to be F1.
49:41
F1, which is 250 multiplied by sine of 35 degrees.
49:52
Attention here.
49:56
The z component is pointing downwards, negative. It's pointing in the opposite direction to the letter, negative. The other two were pointing in the same direction of the letters. So be careful with this one. And that's negative 143.39 newton. What we got so far is F1 as a vector. And F1 as a vector is 86.55i plus 185.60j minus 143.39k. And that's our force F1 in Cartesian representation or Cartesian notation. Now if we go to F2, F2-- I'm going to keep working here.
50:57
If we go for F2 now, look at the angles that we have for F2. For F2, we have angles with respect to the axis. Look at these angles. These angles are inclined with respect to the axis. And for these, you have the cosine director angles, or the director angles, alpha, beta, and gamma. And you have to remember the definition of alpha, beta, and gamma. Alpha is the angle between the x-axis and the force, which in this case is clearly stated as 120. So alpha is 120.
51:36
Beta is the angle between the y-axis and the force. And that is 45 degrees. Now look at this. 45 and 60. Oh, wait a second, 45 and 60, should it be 90 degrees here? No it shouldn't be because those are not in the same plane. This 45 and this 60 are inclined angles. So that's one of the indications also, that you are dealing with cosine director angles and not with a projection angle. Now gamma is 60. How do I know? Because it's given. But even if I don't know one of them,
52:17
remember, that I said in class and I stated that you have a relation, direct relation between the three angles. Those are not independent, alpha, beta, and gamma. They are not independent. They always have to be, always, always-- it has to happen that cosine is square alpha plus cosine is square beta plus cosine is square gamma has to be equal to 1. So in case I don't have one of them, I can calculate them by using this expression. And try with these angles.
52:51
I say try with these and these and calculate gamma. And you're going to see that it is true what I'm saying. So F2 as a vector I'm going to use the cosine directors. And they are called cosine directors because the cosine of this, the cosine of this, and the cosine of this are going to give me the unit vector. And I multiply the unit vector by the magnitude. And the magnitude is 400. So 400 is the magnitude multiplied by the unit vector, which is going
53:21
to be cosine 120i plus cosine 45j plus cosine 60k. This is my cosine director. Remember, this is in newton also. So F2 is going to be equal to negative 200i plus 282.84j plus 200k.
53:52
That's F2.
53:55
That's my F2. Now what do I have to do? Add these with this. So I'm going to put F1 here. Remember, this is as a vector, an arrow on top of the--
54:10
or the vector notation. So F1 was 86.55i. The other one in j is 185.60j. And the other one is negative 43.39k. Now I have these two. Remember this is newton, these are newton. And I'm going to add them in order to calculate the resultant force. My resultant force is going to be i together. So this negative 113.45 in i. And this is plus here. Even if you don't have the j, please, put 0 here in case you didn't have it. And put it on top of the j. Because I know you're going to miss
55:00
and you're going to add this with this if you are in a rush in a test, or something. So be organized when you solve the problem. Now when you do this plus this, this is going to be 468.44j plus 56.61k. This is going to be my resultant force as a vector.
55:26
Yeah, resultant force FR. I should be calling this FR not R. And now they are asking me for the magnitude and coordinate direction angles of the resultant force. Their magnitude is really easy. Because the magnitude of the resultant force is just the square root of 113.45 squared plus 468.44 squared plus 56.61 squared. And then you're going to say, oh, wait a second. How come you didn't put these negative? So what. I don't care about the negative.
56:01
Remember, the negative sign is going to be a square. So it's going to be positive no matter what. So then when I do that, the magnitude of the resultant force is going to be equal to 485.30 newton. That's my value for the magnitude of the resultant force. And the other part that is asking in the problem is, coordinate direction angles. Well, the coordinate direction angles, if you remember from class, cosine alpha is going to be equal to what? It's going to be equal to the component of i
56:38
of the force divided by the magnitude of the resultant force. That's cosine alpha. So the alpha is going to be the inverse of this. Alpha is going to be the inverse cosine of this divided by this. But remember, this is negative. Now you have to put the negative.
57:06
Beta is going to be also this, 468.44 divided by the magnitude of the force.
57:20
And remember, this is not the value of beta. This is an inverse cosine of this. And gamma is going to be the inverse cosine of this value, 56.61 divided by the magnitude of that force. Which gives us alpha equal to 103.52 degrees, beta equal to 15.15 degrees, and gamma equal to 83.30 degrees. And our results for the problem are first, the force as vector, then the magnitude of the force, and then the angle, the director angles for the force. Well, you have four problems.
58:20
I solved two of them in different ways. I hope that you enjoyed them. I hope that they clarify a lot. Practice practice practice. Do not procrastinate. And you will be successful in this class. I'm going to prepare another one with chapter 3, probably. And I'm going to try to keep doing this for every chapter. But at least for this chapter, you have some problems that you can practice. See you next class.
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