Angle between Fz and S, picture included

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Homework Statement


Hint: W = ~S · ~F .
Consider a force ~F with components Fx =28 N, Fy = 57 N, and Fz = 99 N, as illustrated in the figure below. Work is done when a particle moves up the vertical z-axis a distance of 4 m.
[PLAIN]http://img691.imageshack.us/img691/5594/problem4e.jpg [Broken]
What is the angle θ between the body diagonal which represents the force ~F and the front-left edge of the block which represents both the z-component Fz of the vector ~F and the vector ~S the particle traversed?
Answer in units of ◦

Homework Equations


They give me a hint of W= S *F
but I don't know S or W...


The Attempt at a Solution


F= 28i + 57 j + 99 k
 
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Answers and Replies

  • #2
jhae2.718
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Work for a constant force is the dot product between the force and the displacement,
[tex]W=\vec{F} \cdot \vec{d}[/tex].

What are the two definitions of the dot/scalar product, and how can you use them to find the angle θ?
 
  • #3
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Work for a constant force is the dot product between the force and the displacement,
[tex]W=\vec{F} \cdot \vec{d}[/tex].

What are the two definitions of the dot/scalar product, and how can you use them to find the angle θ?

W= F * d= mag(F delta d) costheta?
so is the distance only in the z direction?
and if so does that mean only Fzk matter?
 
  • #4
jhae2.718
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The problem says "Work is done when a particle moves up the vertical z-axis a distance of 4 m", so what does that imply about the direction of the displacement?

You can relate the dot product in terms of the magnitudes and cos(theta) with the dot product in terms of the components to find cos(theta), and then use arcos(cos(theta)).
 
  • #5
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The problem says "Work is done when a particle moves up the vertical z-axis a distance of 4 m", so what does that imply about the direction of the displacement?

You can relate the dot product in terms of the magnitudes and cos(theta) with the dot product in terms of the components to find cos(theta), and then use arcos(cos(theta)).

So the direction is in the z direction... so does this mean I only need to care about Fzk(which is also S) and F?
 
  • #6
jhae2.718
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Essentially; [tex]W=F_z\hat{k} \cdot S\hat{k}[/tex] in this case. Now, from this, how can you get [tex]\theta[/tex]?

Note that you don't need to take [tex]\vec{F}[/tex] itself in the calculation once you've put it into component form.
 
  • #7
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Essentially; [tex]W=F_z\hat{k} \cdot S\hat{k}[/tex] in this case. Now, from this, how can you get [tex]\theta[/tex]?

Note that you don't need to take [tex]\vec{F}[/tex] itself in the calculation once you've put it into component form.

So, Fzk=99k
W= 99k * S= mag (99k*S)costheta
theta=arccos( 99k*S)/mag (99k*S)

but how do I get S?
 
  • #8
jhae2.718
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S is 4 m.

You're making a mistake. It is the magnitude of [tex]\vec{F}[/tex] in the dot product, so you should have:
[tex]99\,\mathrm{N}\cdot 4\,\mathrm{m}=|\vec{F}| \cdot 4\,\mathrm{m} \cos(\theta)[/tex].

[tex]|\vec{F}|=\sqrt{F_x^2+F_y^2+F_z^2}[/tex].
 
  • #9
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S is 4 m.

You're making a mistake. It is the magnitude of [tex]\vec{F}[/tex] in the dot product, so you should have:
[tex]99\,\mathrm{N}\cdot 4\,\mathrm{m}=|\vec{F}| \cdot 4\,\mathrm{m} \cos(\theta)[/tex].

[tex]|\vec{F}|=\sqrt{F_x^2+F_y^2+F_z^2}[/tex].

Oh, I see now.

Thanks so much!
 

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