3-D vectors forces exerted upon a mast

[late347]

Homework Statement

The thick arrows represent forces exerrted upon the mast
Let $\vec{r_2} = ~~the~~ longer~force~ vector~~$
$\vec{r_1} = ~the ~shorter ~force ~ vector$

correspondinly $\vec{R_2} = cable ~~2$
$\vec{R_1} = cable~~1$

r_1 is in the same direction as R_1
r_2 is in the same direction as R_2
(I could not find the correct latex symbol for this, I want two vertical arrows side-by-side both facing upwards)

The thinner lines parallel to the forces which have endpoints in the xy- plane, those thin lines are the cables which are attached to the ground (xy- plane ostensibly) The cables are attached high up into the top of the mast. There's two cables and two forces in the picture currently.

a) calculate the two forces in the picture and give their coordinate form (using unit vectors)
b) calculate the sum of the two forces

c) a third cable is attached as follows. The tension is 800 N. Its endpoint will be (x, y). Define the force exterted upon the mast from this cable. (using unit vectors)

d) into which point in the xy plane, must the third cable be attached such that the resultant force from the three cable forces will be in the same direction as the mast ( the last part was a little bit unclear, but using common sense it would seem the resultant force needs to point down into the ground, and be parallel to the mast ?)The c) part was a little bit unclear to me what is being asked exactly.

Homework Equations

The Attempt at a Solution

a)

$\vec{r_2} = 272.91 \hat{i} + 682.22 \hat{j} - 818.62 \hat{k}$
$\vec{r_1} = - 471.5 \hat{j} - 707.24 \hat{k}$

b) $\vec{F_{sum}} = \vec{r_1}+ \vec{r_2} = 272.91 \hat{i} + 210.725 \hat{j} - 1525.86 \hat{k}$c) I was a little bit unconviced of finding a definite solution for the endpoint of the third cable on the xy plane.

c1) It seems as though infinite number of points exist as solutions ? The length of the third cable is ostensible finite length, but it is unknown length also.

c2) We only know the force exerted by the third cable upon the mast. Not the length of the third cable.

c3) If the cable were of fixed length such that the third cable is called C

c4) C is attached to the top of the mast, and C is a taut line, somewhat analogous to a stick. C will draw out a circle in the xy plane. All those points in the circumference of the circle are correct solutions for point (x,y)

c5) the radius of the circle in xy plane will be fixed value, but we don't know the radius because we don't know the length of the cable C.

 

[Andrew Mason]

The thinner lines parallel to the forces which have endpoints in the xy- plane, those thin lines are the cables which are attached to the ground (xy- plane ostensibly) The cables are attached high up into the top of the mast. There's two cables and two forces in the picture currently.

a) calculate the two forces in the picture and give their coordinate form (using unit vectors)
b) calculate the sum of the two forces

...3. The Attempt at a Solution

a)
$\vec{r_2} = 272.91 \hat{i} + 682.22 \hat{j} - 818.62 \hat{k}$
$\vec{r_1} = - 471.5 \hat{j} - 707.24 \hat{k}$

Can you show how you got these results?

For $\vec{r_1}$ which is in the y-z plane, the sin of the angle it makes to the z axis is arctan 20/30. So the y component should have length |850 *cos (arctan (20/30))|
$\vec{r_2}$ appears to be correct for the y component.

AM

 

[late347]

small r vectors are force vectors, where as capital R vectors are really just the lengths of the cable which goes from the top of the mast to the ground.(i.e. xy plane)
My interpretation was that the arrows are the force vectors because they are "floating around" with no endpoint.

Where as the cables are attached on both ends, one endpoint in the xy plane, and the other endpoint at the mast.

Of course each force is essentially parallel to the way that the cable is set up. The force of the cable drawing at the mast (force effecting upon the mast), is in the same direction, as the cable from the mast, into the ground.

my work looks so hideous in my notebook on paper so I will abbreviate what I did and give reasoning for it

in my notation decimal separator is a comma, I think...
1.)The capital R lengths can be thought of as vectors.
2.) Then we calculate length of capital R's
3.) then we calculate unit vector for the capital R this will crucially be the same unit vector as it would be for the force vector r_1 for example
4.) multiply the unit vector with the length of the force vector r_1 (R1 and r1 have the same unit vector and R2 and r2 have the same unit vector)
5.) this procedure gives usthe complete force vectors r_1 and r_2 (I think...)
$|| \vec{ R_1} || = \sqrt{(-30)^2+(-20)^2}=10\sqrt{13}$

$\vec{R_1}= -20\ \hat{j} -30 \hat{k}$

$\vec{R_1^{0}} = \frac{\vec{R_1}}{||\vec{R_1}||} = \frac{-20}{10\sqrt{13}} \hat{j} + \frac{-30}{10\sqrt{13}} \hat{k}$

because r_1 and R_1 are vectors in the same direction, and unit vector for R_1 was calculated... I think it is sufficienct to multiply the unit vector by the length of force vector r_1

length of the force vector r_1 is the same as the magnitude 850N
$\vec{r_1}= 850 N * \vec{R_1^0} = -471,5 \hat{j}- 707,24 \hat{k}$

When I use my texas instruments nspire super-calculator, it does calculate the norm from that vector r_1 as correctly being 850Nbelow just for copy pasting into latex code
\vec{} \hat{}$\vec{R_2} = 10\hat{i} +25\hat{j} -30\hat{k}$

$|| \vec{R_2} ||= \sqrt{ 10^2+25^2+(-30)^2 } = 40,3113$

$\vec{R_2^0} = \frac{10}{40,3113} \hat{i} + \frac{25}{40,3113} \hat{j}+ \frac{-30}{40,3113} \hat{k}$

$\vec{r_2} = \vec{R_2^0} * 1100N \leftrightarrow \vec{r_2} = 272,91 \hat{i} + 682,22 \hat{j} -818,62 \hat{k}$

when I use my texas instruments nspire super calculater the norm from that vector comes out at roughly 1100 NWhen I inputted all four vectors
R_1, r_1, R_2, r_2 into my calculator and I calculated the unit vector from each vector.
The unit vector for the pair R1 and r1 was the same
the unit vector for the pair R2 and r2 was the same
There may have been some rounding error but there you go...

I may have made a gross error somewhere of course... It would not be the first time for me I suppose

indeed the F_sum vector was wrongly calculated by myself it seeems

with the calculator it comes out at

$F_sum = \vec{r_1} + \vec{r_2} = 272,91 i + 210,73 j -1525,86k$

 

[late347]

part C

give the form of the force vector $\vec{r_3}$ when its magnitude is 800 N

the cable R_3 is parallel to the force vector r_3

cable R_3 is of the following form

$R_3 = x*\hat{i} + y* \hat{j} - 30\hat{k}$

length of the cable is unknown and was not part of any information that was given to us.
the cable's length is of the form as follows
$|| \vec{R_3} || = \sqrt{x^2+y^2+900}$

the unit vector will be of the following form
$\vec{R_3^0} = \frac{x}{\sqrt{x^2+y^2+900}}\hat{i} + \frac{y}{\sqrt{x^2+y^2+900}}\hat{j} + \frac{-30}{\sqrt{x^2+y^2+900}}\hat{k}$
the vector $\vec{r_3}$ is of the following form

$\vec{r_3} = \vec{R_3^0} * 800 N$
$\leftrightarrow \vec{r_3} = \frac{800x}{ \sqrt{x^2+y^2+900}} \hat{i} + \frac{800y}{ \sqrt{x^2+y^2+900}} \hat{j} - \frac{24000}{ \sqrt{x^2+y^2+900}} \hat{k}$I believe that should qualify for an answer for part C because not much was given as information, such that a clear singular solution could be found.part D

where in the xy plane should the cable R_3 be attached such that the resulting force of the three forces will be parallel to the mast itself.

Firstly the resultant force is the $F_{sum} = \vec{r_1} +\vec{r_2} +\vec{r_3}$we already know that $\vec{r_1} +\vec{r_2} = 272,91 \hat{i} + 210,72 \hat{j} - 1525,86 \hat{k}$

hence
$F_{sum} = (272,91 + \frac{800x}{ \sqrt{x^2+y^2+900}}) \hat{i} + (210,725 + \frac{800y}{ \sqrt{x^2+y^2+900}}) \hat{j} + (-1525,86 - \frac{24000}{ \sqrt{x^2+y^2+900}}) \hat{k}$we also know that F_{sum} vector will be of the form
$0* \hat{i} + 0*\hat{j} + n * \hat{k}$ where n is scalar and belongs to real numbers

Then a very complicated looking system of equations can be built using the fact that coefficients for $\hat{i}$ and $\hat{j}$ are both zero.
So, we can set the the thing inside the reqular brackes (coefficient) for $\hat{i}$ = 0

System of equations:
eq1 = $210,725 + \frac{800y}{\sqrt{x^2+y^2+900}} = 0$
eq2 = $272,91 + \frac{800x}{\sqrt{x^2+y^2+900}} = 0$

putting those into my trusty and brand-new TI nspire CX CAS calculator, we get lots of possible solutions for x,and y

the solution which seems to work is as follows
x= -11,3416
y= -8,7573When you input those x and y values into the equation for vector $\vec{r_3}$

and calculate the vector sum $\vec{F_{sum}}$with $\vec{r_3}+ \vec{r_2}+ \vec{r_1}$\hat{} \vec{} \frac{}{} \sqrt{x^2+y^2+900}
the resulting vector sum is about
$-0,000263 \hat{i} - 0,004858 \hat{j} -2247,74 \hat{k}$

which looks quite ok with all the rounding error and such. The i- coefficient and j-coefficient are almost zero but not quite

 

[late347]

A simpler solution would be welcome if anybody has got this figured ouit... I can ask the teacher in a couple of days if this was incorrect or correct...

 

[haruspex]

A simpler solution would be welcome if anybody has got this figured ouit... I can ask the teacher in a couple of days if this was incorrect or correct...

It all looks right to me.
It is not hard to solve eq1 and eq2 manually. Rearrange each into the form R₃²=. You then have a pair of linear simultaneous equations in the unknowns x² and y².

 

[late347]

It all looks right to me.
It is not hard to solve eq1 and eq2 manually. Rearrange each into the form R₃²=. You then have a pair of linear simultaneous equations in the unknowns x² and y².

please elaborate if you have the time

my brain is about to freeze at this point with this exercise.
I suppose with some more effort I could solve the system of Equations manually, but... I think even my textbook used an equations solver in some of the worked examples.

I was reading my textbook and the most sensible thing at first seemed to be to work with the force vector coefficients

that is to say form the system of Eqs by using the force vector i, j and k coefficients
we can certainly set up

( sum of the three i-coefficients) = 0
and
(sum of the three j-coefficients) = 0
because those equations were directly justifiable from the point of view of the vector sum of the forces upon the mast. The vector sum should only have k-component and others should be zero.

in my notation force vectors were small case r letters. Capital R was for cables themselves.

 

[haruspex]

please elaborate if you have the time

Your eqs 1 and 2 are of the form:
x=A√[x²+y²+900)] and y=B√[x²+y²+900)].
Writing X=x² and Y=y², we have
X=A²[X+Y+900], Y=B²[X+Y+900].
That is just a pair of linear simultaneous equations, easily solved.