Resultant force projected along a line

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Discussion Overview

The discussion revolves around calculating the resultant force at point D in Cartesian form and determining the magnitude of the projected component of this force along the line AD. Participants are addressing a homework problem that involves vector forces and unit vector calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Tim expresses uncertainty about calculating the resultant force at point D, suggesting it might be F1= (-30i) and F2= (-40k), leading to FR= (-30i - 40k) with a magnitude of 50N.
  • Tim proposes a unit vector UAD calculation as UAD = (-0.2/0.245)i + (-0.1/0.245)j + (0.1/0.245)k.
  • One participant points out a sign error in Tim's unit vector UAD and suggests including units in the force expression, providing an alternative unit vector UAD = (1/√6)(2i + j - k).
  • Another participant questions how the alternative unit vector was calculated and provides their own calculations for the resultant force, leading to a discussion about the magnitude and direction of vectors.
  • Tim asks for clarification on the approach to solving the problem to gain a better understanding.
  • A participant notes that Tim's calculations for UAD may be incorrect due to using the wrong reference vector, suggesting a sign error based on the direction from A to D.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct calculation of the unit vector UAD and the resultant force. Multiple competing views and calculations are presented, leading to ongoing debate about the signs and values used in the calculations.

Contextual Notes

There are unresolved issues regarding the assumptions made in the calculations, particularly concerning the directionality of the vectors and the definitions of the points A and D. The calculations rely on specific coordinate values that may not be universally agreed upon.

U1021981
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Homework Statement



(i) Find the resultant force acting at point D and write it in Cartesian form. (Note that CD is parallel to the y-axis, the 40N force acts parallel to the z-axis and the 30N force is parallel to the x-axis.)

(ii) Find the magnitude of the projected component of the resultant force acting along the line AD.

Image of problem is attached.

Homework Equations



(i) Unsure about calculating resultant force
(ii) Find Unit vector of UAD and find the dot product of UAD and FR from (i)

The Attempt at a Solution



(i) Unsure if resultant force at point D is simply F1= (-30i) and F2= (-40k) so FR= (-30i - 40k) (Magnitude 50N) ?

(ii) UAD = (-0.2/0.245)i + (-0.1/0.245)j + (0.1/0.245)k

Then F.UAD = (Fi)(-0.2/0.245) + (Fj)(-0.1/0.245) + (Fk)(0.1/0.245)

Thanks for your help,

Tim
 

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U1021981 said:
(i) Unsure if resultant force at point D is simply F1= (-30i) and F2= (-40k) so FR= (-30i - 40k) (Magnitude 50N) ?

(ii) UAD = (-0.2/0.245)i + (-0.1/0.245)j + (0.1/0.245)k

Then F.UAD = (Fi)(-0.2/0.245) + (Fj)(-0.1/0.245) + (Fk)(0.1/0.245)

You have a sign error in your unit vector UAD, and I would recommend you include the units (Newtons in this case) in your expression for force and use exact numbers in your calculations. I get \frac{1}{\sqrt{6}}(2\mathbf{i}+\mathbf{j}-\mathbf{k}) for UAD.
 
You have a sign error in your unit vector UAD, and I would recommend you include the units (Newtons in this case) in your expression for force and use exact numbers in your calculations. I get (1/√6)(2i+j−k) for UAD.

How did you calculate this?

I have used u = F(Coord)/F(magnitude) = (Fx/F)i + (Fy/F)j + (Fz/F)k

F (aka RDA) = (XA-XD)+(YA-YD)+(ZA-ZD)
= (0-0.2)+(0-0.1)+(0-(-0.1))
= -0.2i - 0.1j + 0.1k
magnitude = √0.06

Which is how I have got:

UDA = (-0.2/0.245)i + (-0.1/0.245)j + (0.1/0.245)k


Could you please outline the approach/steps that you believe I should take to solve the problem so I can get a better understanding.

Thanks for your reply,

Tim
 
U1021981 said:
How did you calculate this?

I have used u = F(Coord)/F(magnitude) = (Fx/F)i + (Fy/F)j + (Fz/F)k

F (aka RDA) = (XA-XD)+(YA-YD)+(ZA-ZD)
= (0-0.2)+(0-0.1)+(0-(-0.1))
= -0.2i - 0.1j + 0.1k
magnitude = √0.06

Which is how I have got:

UDA = (-0.2/0.245)i + (-0.1/0.245)j + (0.1/0.245)kCould you please outline the approach/steps that you believe I should take to solve the problem so I can get a better understanding.

Thanks for your reply,

Tim

Just on top of that, you have the same answers 2/√6 = .2/√.06

except in the drawing it shows that the pipe is +ve 0.2m (where you have -0.2m) and that if the line is in fact A to D then does it not share this property?
 
U1021981 said:
How did you calculate this?

I have used u = F(Coord)/F(magnitude) = (Fx/F)i + (Fy/F)j + (Fz/F)k

F (aka RDA) = (XA-XD)+(YA-YD)+(ZA-ZD)
= (0-0.2)+(0-0.1)+(0-(-0.1))
= -0.2i - 0.1j + 0.1k
magnitude = √0.06

Notice first that \sqrt{0.06} = \frac{1}{10}\sqrt{6}. Next, consider that UAD usually means "the unit vector from A to D". By using RDA as F, instead of RAD, you've calulated the opposite and hence have a sign error.
 

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