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Find the Components

  • Thread starter Mosaness
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  • #1
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Homework Statement



The z component of the force F is 80 lb. (a) Express F in terms of components. (b) what
are the angles x, y, and z between F and the positive
coordinate axes?


Homework Equations



The equations that I attempted to use were directional cosines, trignometric identities and that F = (magF)(cosθ)

The Attempt at a Solution



I haven't really been able to get anywhere. I tried various methods:

1) I tried to find the length of A and use that to find F and so forth.

I have been at this for a good hour and the answer is most likely something simple.
 

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Answers and Replies

  • #2
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If you know the vertical component and the vertical angle, you should be able to find the magnitude of the projection of the force on the horizontal plane. And since you know the horizontal angle, you should be able to find its horizontal component. This is all simple trigonometry.
 
  • #3
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If you know the vertical component and the vertical angle, you should be able to find the magnitude of the projection of the force on the horizontal plane. And since you know the horizontal angle, you should be able to find its horizontal component. This is all simple trigonometry.
I know that Fz = 80 lb and that the angle between F and A is 20°. There is a triangle in there: FOA, where O is the origin and is 60°. But I am stuck, and even if it simple trig...I would appreciate some guidance. I have an exam tomorrow and this is the only problem that I didn't understand.
 
  • #4
rl.bhat
Homework Helper
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Angle between OA and F is given. Find FA. From the figure you can find the angle between OA and x and z axis. Find the components of FA on x and z axis. z comoponent of F is given. Find F. Now proceed.
 
  • #5
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Angle between OA and F is given. Find FA. From the figure you can find the angle between OA and x and z axis. Find the components of FA on x and z axis. z comoponent of F is given. Find F. Now proceed.
Okay. The angle between OA is 30°, and the total angle between the x- and the z- axis ought to be 90°. FA = Fcos(20°)?
 
  • #6
rl.bhat
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Okay. The angle between OA is 30°, and the total angle between the x- and the z- axis ought to be 90°. FA = Fcos(20°)?
Correct. Now find the component of FA on z axis. It's value is given. From that find F.
 
  • #7
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I pictured it in my mind's eye and I can see where I was imagining it wrong. My initial image was that the "rope" F was passing THROUGH the x-axis, when in reality, it can be seen as a rope that is not passing through any of the axis' and is instead attached to an imaginary wall. The line A is drawn and the angle between F and A is 20 yes, but the angle between A and the z-axis is 60, so the angle between A and the x axis is then 30.
 
  • #8
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Correct. Now find the component of FA on z axis. It's value is given. From that find F.
I believe the z-component of FA is 80 lb, and if Fz = FAcos(20°), then FA can be found by 80/cos20?
 
  • #9
rl.bhat
Homework Helper
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I pictured it in my mind's eye and I can see where I was imagining it wrong. My initial image was that the "rope" F was passing THROUGH the x-axis, when in reality, it can be seen as a rope that is not passing through any of the axis' and is instead attached to an imaginary wall. The line A is drawn and the angle between F and A is 20 yes, but the angle between A and the z-axis is 60, so the angle between A and the x axis is then 30.
Yes. Now continue.
 
  • #10
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Yes. Now continue.
I am somewhat stuck on this part. My solution manual states that F = (magnitude of F)(cos20sin60 i + sin20j + cos20cos60 k) and I cannot understand how the manual got to this point.
 
  • #11
rl.bhat
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I believe the z-component of FA is 80 lb, and if Fz = FAcos(20°), then FA can be found by 80/cos20?
No. It should be Fz = FAcos60o
 
  • #12
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No. It should be Fz = FAcos20ocos60o
Why would it be Fcos20cos60? If you don't mind explaining? Please help me understand this...
 
  • #13
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Based on the diagram, if you only look at the line A, then Az would be cos60. If you only look at the line F, then Fz would be cos20
 
  • #14
rl.bhat
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Why would it be Fcos20cos60? If you don't mind explaining? Please help me understand this...
Please refer post. 11.
 
  • #15
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Please refer post. 11.
OH! WE FOUND THAT FA WAS COS20, SO IT'S A MATTER OF SIMPLE SUBSTITUTION NO? F = FACOS60, which is the same as Fcos60cos20
 
  • #16
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I can see what happened there. Sort of. Let me try to explain it out. Using the triangle that was formed, we found that FA, which is the x-component of that specific triangle, was Fcos20°. Now, if you look at the diagram again, you can see that there is another triangle that is formed between the z-axis and the line A, and this triangle has an angle of 60. If I wanted to find Fz, it would be FAcos60°, because FA is one of the sides of the triangle, so FZ becomes Fcos20°cos60°. Is this explanation correct?
 
  • #17
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I believe I finally understood this problem. There are TWO triangles in the diagram given. One if a triangle FOA and the other is a triangle that is formed between the line OA and the z-axis.

Now, for the triangle FOA, it doesn't have a z-component to it, but you know that O is 20°, so the line OA will be Fcos20° and the y-component will simply be Fsin20°.

Those are all the components that you are going to obtain from that specific triangle.

Moving on to the second triangle, you already know that OA is Fcos20° and that the angle between OA and the z-axis is 60°. So taking z- as our adjacent side and OA as the hypotentuse, the z-component is FOAcos60°, which becomes Fcos20°cos60°.

There is no y-component to this triangle, so we move on to the x-component. The x-component is the side that is opposite the angle, and it is defined by FOAsin60°, which becomes Fcos20°sin60°.

So F = F[cos20°sin60° i + sin20°j + cos20°cos60°k], where F is the magnitude and the i, j, and k are the directions.

We are given that Fz = 80lb, and using that, we can solve for F.

Solving gives that F is approximately 170 lbs, and using that, we can now find what the x- and y- components are.
 
  • #18
rl.bhat
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You are right.
 
  • #19
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Thank you Mr. Bhatt. I wouldn't have thought of it doing this way. The way I was originally doing it was by just using the Fz component to do it. I do have a question, HOW did you know HOW to do the question that way? I wouldn't have thought of doing it that way really.
 
  • #20
rl.bhat
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Thank you Mr. Bhatt. I wouldn't have thought of it doing this way. The way I was originally doing it was by just using the Fz component to do it. I do have a question, HOW did you know HOW to do the question that way? I wouldn't have thought of doing it that way really.
Well. It is just practice.
 
  • #21
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I kind of feel like if I see another problem of this sort, using this specific problem as a guide, I will be able to solve it. I do thank you for the help, it is much appreciated. My professor didn't explain it too clearly. Thank You Sir :)
 
  • #22
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Although I do have a slight question. For the second triangle, why has the z-axis chosen as the adjacent side?
 
  • #23
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This is how the Cartesian system of coordinates works. Any vector in it can be decomposed in two triangles, whose sides are all related to the magnitude of the vector via their angles, and so they are all interrelated via their angles.

More generally, in a 3D space you need three numbers to define a vector. There are many different ways to define vectors via three numbers, but they may all be expressed via one another.
 
  • #24
92
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This is how the Cartesian system of coordinates works. Any vector in it can be decomposed in two triangles, whose sides are all related to the magnitude of the vector via their angles, and so they are all interrelated via their angles.

More generally, in a 3D space you need three numbers to define a vector. There are many different ways to define vectors via three numbers, but they may all be expressed via one another.
I do understand that part. I figured it out eventually. It just struck me why I didn't use OA as my adjacent.

My reasoning was that when I made the triangle, the line connecting OA to the z-axis made a 90 degree angle.
 

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