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Express F as a unit vector and find the Scalar Projection of F onto OA

  • #1
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Homework Statement



Express the 5.2-kN force F as a vector in terms of the unit vectors i, j, and k. Determine the scalar projections of F onto the x-axis and onto the line OA.

I have attached an image of the problem.


Homework Equations


Fx = Fcos(θ)
Fy = Fcos(θ)
Fz = Fcos(θ)


The Attempt at a Solution



Firstly I tried to find the unit vector nxy
nxy = sin(4)i + cos(4)j

Then I tried to find the unit vector of F:
nF = cos(41)nxy + sin(41)k

which becomes:

nF = cos(41)[sin(4)i + cos(4)j] + sin(41)k
n = cos(41)sin(4)i + cos(41)cos(4)j + sin(41)k

Then F = F°nF (dot product)
F = 5.2[cos(41)sin(4)i + cos(41)cos(4) + sin(41)k]

F = 0.2738i + 3.9149j + 3.412 k

I checked that this was true by squaring i, j and k, adding them together and then squaring the result.

sqrt(0.2738^2 +3.9149^2 + 3.412^2)) = 5.20 kN

However, it says that only my value for K is correct.

With the projection of F onto OA I used the dot product, whereby:

F = 0.2738i + 3.9149j + 3.412k

n)A = cos(33)i + sin(33)j

F°nOA = (0.2738*cos(33))i + (3.9149*sin(33))j + 3.142k*0

F°nOA = 0.2296 + 2.132
= 2.36187 kN

Unsurprisingly this was wrong too.

Any suggestion for where I'm going wrong?
 

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Answers and Replies

  • #2
Simon Bridge
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2. Homework Equations
Fx = Fcos(θ)
Fy = Fcos(θ)
Fz = Fcos(θ)
... these are not correct - taken as written, they would mean that the components are all the same magnitude.

I don't understand your strategy ... you are given the vector ##\vec{F}## in terms of a magnitude and two angles. If you put ##\theta=41^\circ## (i.e. ##\theta## is the angle ##\vec{F}## makes to the x-y plane) then ##F_{xy}=F\cos(\theta)## would be the projection of ##\vec{F}## onto the x-y plane.

Label the angle that ##\vec{F}_{xy}## makes to the x-axis, ##\phi##, then ##\phi = 53^\circ##

From that and basic trig you should be able to work out the components without all that messing about with unit vectors.
You can pull two important triangles out of the diagram ... one is F-Fz-Fxy, it has a theta in it; and the other is Fxy-Fx-Fy, it has a phi in it.
 
Last edited:
  • #3
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To be honest, I didn't understand what I was doing earlier; I was attempting to imitate the method I saw in an online video. But if I'm understanding you correctly, then my calculations are as follows:

Fxy = F cos(41)
Fxy = 3.924 kN

Fz = Fcos(49)
Fz = 3.4115 kN

My answer for Fz is consistent with the answer I found with my previous method.

Then I think that I can break up Fxy into it's Fx and Fy components as follows:

Fx = F sin(37)
Fx = 2.362 kN

Fy = F cos(37)
Fy = 3.1338 kN

When it asks for the projection of F onto the x-axis, would that be Fx?

But I'm still a little confused about how I should do the projection of F onto OA. Is it just the dot product of F and OA:

F = <2.262, 3.1338, 3.4115>

This next part, is the unit vector of OA but I'm not sure if I can do this.
OA/|OA| = <cos(33), sin(33), 0>

2.362*cos(33) + 3.1388*sin(33) + 0
= 3.6877 kN

Am I making any sense?
 
  • #4
Simon Bridge
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It is good to have confirmation by different methods. Do you see how this way is easier to understand?
Fz = Fcos(49)
OK that works - cos(49)=sin(41).
Fx = F sin(37)
Careful of your notation - it is communication too.
That should be Fx = Fxy.cos(53) = Fxy.sin(37).

I don't know why you are using the complimentary angles throughout but ho hum it all works.

The projection ##p## of ##\vec{u}## onto ##\vec{v}## is the amount of ##\vec{u}## in the direction of ##\vec{v}##... the direction of ##\vec{v}## is ##\vec{v}/v## so: $$p=\frac{\vec{u}\cdot\vec{v}}{v}$$
 
  • #5
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So, am I right to think that the projection of F onto OA is:

p = F°OA/|OA|

But I'm not sure how I find the vector OA. Am I right in saying that OA = cos(33)i + sin(33)j?
 
  • #6
Simon Bridge
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But I'm not sure how I find the vector OA. Am I right in saying that OA = cos(33)i + sin(33)j?
Notice that this is consistent with how you found the x and y components of F?

Since the vector is in 3D you should explicitly include the z component even though it is zero... good discipline for later when you'll be writing ##\vec{u}\cdot\vec{v}=\vec{u}^t\vec{v}##

BTW: it is acceptable to say v=vector(OA); p=F.v/|v| better v=vector(OA); p=v.F/v (see how I used the normal period for a dot product - better if it is boldface - but much better to use the actual symbols from the Ʃ menu or just writing LaTeX).
 

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