Forces problem key

[zealous harness]

can anyone help me understand why in the matrix she is no longer multiplying x and y times f1 and f2

like its there when she writes down f1 f2 and f3 but then in the matrix its gone

[hoary ruin]

coefficient matrix

F1, F2 and F3 are vectors which were written in coloumns

x and y refer to the components of that vector

[zealous harness]

i dont get it i thought the variable is x and y

so shouldnt the matrix output x and y

also i thought it was f1 times cos(tan...) plus f2times cos(tan...) 0 not without teh Fs

[hoary ruin]

the rows have variables x and y
the coloumns have the force variables f1, f2 and f3

[zealous harness]

how does that work

could you write it llike a system

[hoary ruin]

imagine like

4f1 + 4f2 + f3
5f1 + 4f2 + 2f3

then we can write the system into a **coefficient **matrix

f1 f2 f3
(4 4 1) x
(5 4 2) y

You know like Ax = b

the vector x would be your (x,y) basically

[zealous harness]

Where is z

[hoary ruin]

lemme use other variables

[zealous harness]

so how do u know wha ti toutputs

cuz theres only 2 spots for answers

it gives x = and y= right

how does she get the answer from that

[hoary ruin]

you know polar coordinate

x = cos(theta)
y = sin(theta)

[hoary ruin]

here take a look

,w {{-cos(arctan(24/7)),cos(arctan(24/10))},{sin(arctan(24/7)),sin(arctan(24/10))}}{{x},{y}}={{0},{15}}

[jolly vector]

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Requested by adoniszgks

[hoary ruin]

,w {{N[-cos(arctan(24/7))],N[cos(arctan(24/10))]},{N[sin(arctan(24/7))],N[sin(arctan(24/10))]}}{{x},{y}}={{0},{15}}

[jolly vector]

Results provided by WolframAlpha

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Requested by adoniszgks

[hoary ruin]

here you see the approximated values

It would have been better if your teacher did a straight line

the output are the forces f1 and f2

[hoary ruin]

I mistakenly wrote (x,y) but it should have been (F1, F2) sorry for the confusion

,w {{N[-cos(arctan(24/7))],N[cos(arctan(24/10))]},{N[sin(arctan(24/7))],N[sin(arctan(24/10))]}}{{F1},{F2}}={{0},{15}}