# Resultant force projected along a line

## 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)

Tim

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gabbagabbahey
Homework Helper
Gold Member
(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.

Tim

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.

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?

gabbagabbahey
Homework Helper
Gold Member
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.