Finding projected force along line

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In summary, the conversation involved finding the projected component of a force along a specific axis and using unit vectors to calculate the projection. The individuals also discussed the confusion surrounding a diagram and clarified the concept of components being vectors. They also shared their work and calculations to check for any potential errors. Overall, the conversation was focused on solving the mathematical problem at hand and seeking clarification when needed.
  • #1
Saladsamurai
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So I have found the unit vector [itex]u_{ab}[/itex] now I am stuck as to what to do.

I need to find the projected component of the 80 N force along the line axis AB of the pipe.

I have the coordinates of B and A as well. But I am a little lost now.

Any hints are appreciated.

Thanks,
Casey

Also I am having trouble reading the diagram...I have A at [itex]r_A=6i+7j-2k[/itex] meters...does that look right to you?
 
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  • #2
Would it be the dot product [itex]\vec{F}\cdot\vec{r}_{ab}[/itex]?
 
  • #3
Nobody?...
 
  • #4
Ok, your diagram is really confusing, because it looks like the force is pointed along one of the isometric lines. The beginning point of the force is A, and the final point is 0i+0j-12k. Yes, dot product.
 
  • #5
If AD is the vector F, then CD = 0i + 0j - 12k and CA = 6i + 7j - 2k. the AD = -6i -7j -10k
Find unit vector along AD and force F in vector form. Then projection = F.rA
 
  • #6
Dick said:
Ok, your diagram is really confusing, because it looks like the force is pointed along one of the isometric lines. The beginning point of the force is A, and the final point is 0i+0j-12k. Yes, dot product.

Dick! Thank you! I did not even see that! It DOES LOOK like it is parellel to the x-axis, but you are right, the tip lies at -12...damn. I'd been staring at that all night.

Thanks again,
Casey
 
  • #7
So now if [itex]\vec{r}_{AD}=-6i-7j-10k[/itex] then [itex]\vec{u}_{AD}=\frac{\vec{r}_{AD}}{|r|_{AD}}=-.441i-.515j-.735k[/itex] so [itex]\vec{F}=-35.3i-41.2j-58.8k[/itex].

Now the component of [itex]\vec{F}[/itex] projected on [itex]\vec{r}_{AB}[/itex] should be equal to [itex](\vec{F}\cdot\vec{u}_{AB})*\vec{u}_{AB}[/itex]=

(a scalar)* a unit vector= a Vector?

I am just confused because I thought that components were not vectors.

Someone smack me,
 
  • #8
Is this just a semantic question? If a vector is say, i+2j+3k, I don't see anything terribly wrong with saying either "the y component is 2" or "the y component is 2j". They both seem to convey the same information to me.
 
  • #9
It seems to me that components have to be vectors, since they must add up to the vector itself.
 
  • #10
TMM said:
It seems to me that components have to be vectors, since they must add up to the vector itself.

Sure. The given question could also be answered by just giving the magnitude of the "projection of the force". That's why I said "semantics".
 
  • #11
Dick said:
Is this just a semantic question? If a vector is say, i+2j+3k, I don't see anything terribly wrong with saying either "the y component is 2" or "the y component is 2j". They both seem to convey the same information to me.
Right I realized this after some sleep.

But, my problem is mathematical. The answer says 31.1 N and I cannot for he life of me get that out of the numbers I have for F (above post) and [itex]\vec{u}_{AB}=\frac
{-6i-3j+2k}{7}=-.857i-.429j+.288k[/itex]

Well, my Professor verified my vectors are correct via e-mail...so I must be doing something wrong in the procedure.
 
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  • #12
arrrgghh...I'll go through it again and scan in my work. Maybe someone can spot it...
 
  • #13
Saladsamurai said:
arrrgghh...I'll go through it again and scan in my work. Maybe someone can spot it...

Your u_AB looks correct. What do you get for u_DA? Because then you just want 80*u_AB.u_DA.
 
  • #14
Ah, I see you've already posted that. So just do the calc. What do you get for u_AB.u_DA?
 
  • #15
I think I got it. I did [itex]\vec{F}_{AD}\cdot\vec{u}_{AB}=30.97 N[/itex] which is very close to the 31.1 N I am looking for. I am sure that is from round-off.

Dick, thank you. It's always a pleasure...at least for me!
 
  • #16
You did a bad round off on the last component of u_AB. You're welcome!
 

1. What is projected force along a line?

Projected force along a line is a method used in physics to determine the force acting on an object in a specific direction. It takes into account the angle between the force vector and the line of projection, and calculates the component of the force that acts along that line.

2. How is projected force along a line calculated?

The projected force along a line is calculated using trigonometric functions, specifically the cosine function. The formula is Fp = F * cos(θ), where Fp is the projected force, F is the total force, and θ is the angle between the force vector and the line of projection.

3. Why is projected force along a line important?

Projected force along a line is important because it allows us to determine the specific force acting on an object in a particular direction. This is useful in many real-world applications, such as determining the force needed to push a car up a hill or calculating the force acting on a bridge from the wind.

4. Can projected force along a line be negative?

Yes, projected force along a line can be negative. This occurs when the angle between the force vector and the line of projection is greater than 90 degrees. In this case, the projected force acts in the opposite direction of the line of projection, resulting in a negative value.

5. How does projected force along a line relate to work done?

Projected force along a line is directly related to the work done on an object. This is because work is equal to the force applied in a certain direction multiplied by the distance moved in that direction. So, the projected force is the component of the force that actually does work on the object in that specific direction.

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