Specify force magnitude and angle

AI Thread Summary
The discussion revolves around the decomposition of a force vector F1 into its components and the subsequent calculations for a second force vector F2. Initial attempts to draw a parallelogram for visualization were deemed ineffective, leading to confusion about the correct angles and magnitudes. After correcting the calculations, the components for F1 were established as F1x, F1y, and F1z, with a focus on ensuring the z-component was negative due to its direction below the x-y plane. The calculations for F2 were then discussed, leading to the determination of its magnitude and angles, with a final correction made to the angle for F2x. The conversation emphasizes the importance of accurately determining vector components and their implications in three-dimensional force analysis.
mtjc1x
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Homework Statement


See picture.

Homework Equations

The Attempt at a Solution


I tried drawing a parallelogram, but it didn't help, so I'm stuck. I'd need an idea in the right direction, if possible.
 

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mtjces said:

Homework Statement


See picture.

Homework Equations

The Attempt at a Solution


I've decomposed F1 into F1x F1y and F1z..
F1x = 150 * sqrt 3
F1y = 150
F1z =150 * sqrt 2[

Now I'm stuck../QUOTE]
 
I'm not seeing how you've accomplished your decomposition of F1. The magnitude of the vector comprised by those components would be about 367 N, which is larger than the given magnitude of F1. So you might want to re-think your method (whatever that was...).

The diagram is very helpful because it gives you the angles between F1 and each of the coordinate axes. This should make finding the projections of F1 on each of the axes easy to find.
 
mtjces said:

Homework Statement


See picture.

Homework Equations

The Attempt at a Solution


I tried drawing a parallelogram, but it didn't help, so I'm stuck. I'd need an idea in the right direction, if possible.
I don't know what sort of parallelogram you drew, but this is a 3-D kinda problem.
 
SteamKing said:
I don't know what sort of parallelogram you drew, but this is a 3-D kinda problem.

yes, I was going to draw a parallelepiped to see if it could help me.
 
gneill said:
I'm not seeing how you've accomplished your decomposition of F1. The magnitude of the vector comprised by those components would be about 367 N, which is larger than the given magnitude of F1. So you might want to re-think your method (whatever that was...).

The diagram is very helpful because it gives you the angles between F1 and each of the coordinate axes. This should make finding the projections of F1 on each of the axes easy to find.

F1z = cos 60 * 300N = 150N
F1x = cos 45 * 300N = 150*sqrt2 N
F1y = cos 60 * 300N = 150 N

Would this be correct?
I'm confused about the next step..
 
mtjces said:
F1z = cos 60 * 300N = 150N
F1x = cos 45 * 300N = 150*sqrt2 N
F1y = cos 60 * 300N = 150 N

Would this be correct?
For the Z component the angle was not 60 degrees. What does the diagram show it to be? Do you expect the Z-compoennt to be less than or greater than zero?

Check your value for cosine 45 degrees. A cosine should not be > 1.*

* EDIT: Oops. I didn't process the fact that the 300 N was divided by two giving 150 N, so the cosine being ##\sqrt{2} / 2## is satisfied! Sorry about that!
I'm confused about the next step..
You'll want to write a vector equation that expresses your desired result. Fix your F1 vector first though.
 
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gneill said:
For the Z component the angle was not 60 degrees. What does the diagram show it to be? Do you expect the Z-compoennt to be less than or greater than zero?

Check your value for cosine 45 degrees. A cosine should not be > 1.

You'll want to write a vector equation that expresses your desired result. Fix your F1 vector first though.

F1z = cos 120 * 300N = -0.5 * 300N = -150N (but how does that make sense?)

cos 45 = (√2)/2
so F1x = 300 N * cos 45 = 150 N * √2 = 150√2 N

F1y = cos 60 * 300N = 150 N

Is this right now?
 
mtjces said:
F1z = cos 120 * 300N = -0.5 * 300N = -150N (but how does that make sense?)

cos 45 = (√2)/2
so F1x = 300 N * cos 45 = 150 N * √2 = 150√2 N

F1y = cos 60 * 300N = 150 N

Is this right now?
Yes, that looks right. The z-component is negative because the vector F1 is directed below the x-y plane.
 
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  • #10
gneill said:
Yes, that looks right. The z-component is negative because the vector F1 is directed below the x-y plane.

so the idea with F2 would be to cancel out the F1z and F1x and amplify F1y by 650N?
So I need a
F2z = 150 N
F2x = -150√2 N
F2z = 650 N

Am I going in the right direction?
 
  • #11
Yes, but sort out your component subscripts. You've listed F2z twice. It would help if you put them in standard order, too: x, y, z. Otherwise, I think you're there.
 
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  • #12
gneill said:
Yes, but sort out your component subscripts. You've listed F2z twice. It would help if you put them in standard order, too: x, y, z. Otherwise, I think you're there.

F2x = -150√2 N
F2y = 650 N
F2z = 150 N

How do I calculate F2?
 
  • #13
mtjces said:
F2x = -150√2 N
F2y = 650 N
F2z = 150 N

How do I calculate F2?
You don't know how to calculate the magnitude from the three force components?
 
  • #14
mtjces said:
F2x = -150√2 N
F2y = 650 N
F2z = 150 N

How do I calculate F2?

F2 = √((-150√2)^2 + 150^2 + 650^2) = 700 N

To calculate the angles of F2 I used:

cos x = (F2x/F2) = (-150√2 N / 700 N) ---> x = 72.36°
cos y = (F2y/F2) = (650N / 700 N) ---> y = 21.79°
cos z = (F2z/F2) = (150N / 700 N) ---> z = 77.63°

Correct?
 
  • #15
mtjces said:
F2 = √((-150√2)^2 + 150^2 + 650^2) = 700 N

To calculate the angles of F2 I used:

cos x = (F2x/F2) = (-150√2 N / 700 N) ---> x = 72.36°
cos y = (F2y/F2) = (650N / 700 N) ---> y = 21.79°
cos z = (F2z/F2) = (150N / 700 N) ---> z = 77.63°

Correct?
Sorry, messed up x:
cos x = (F2x/F2) = (-150√2 N / 700 N) ---> x = 107.64
 
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