Question

Vectors
determine the tension in each cable supporting the load

Answer 1

are you given any other values?

Answer 2

no

Answer 3

Remember that the horizontal components are equal and the vertical components add up to gravitational force.

Answer 4

So \[A\cos(50)=B\cos(30)\]\[A\sin(50)+B\sin(30)=F_{gravity}\]

Answer 5

Who equations, two unknown variables.

Answer 6

A and B are the tensions on A and B respectively.

Answer 7

@ifrah34 Still need help?

Answer 8

the answer are Tac=1758.8 and Tbc=1305.4

Answer 9

First of all, the answer needs to be force.

Answer 10

the vertical and horizontal components are not used for this problem

Answer 11

Yes they are!

Answer 12

I tried it your way, I don't have the right answers

Answer 13

Okay, well hold on and let me check.

Answer 14

Did you use degree mode?

Answer 15

yes

Answer 16

is mass in grams?

Answer 17

or kg

Answer 18

http://www.wolframalpha.com/input/?i=A+cos%2850%29+-+B+cos%2830%29+%3D+0%2C+A+sin%2850%29+%2B+B+sin%2830%29+%3D++2000

I get the same answer!!!

Answer 19

@ifrah34 The problem is that you didn't solve the system correctly.

Answer 20

mg is F gravity right?

Answer 21

Yeah, well in the engineering system, lbs is force, not mass.

Answer 22

Mass is measured in slugs.

Answer 23

my teacher used a shortcut

Answer 24

How hard it is the solve a system of equations?

Answer 25

Are you going to admit that I wasn't wrong?

Answer 26

can you use law of sine?

Answer 27

who is supposed to do that...
@ifrah34 are you doing engineering?

Answer 28

@ifrah34 said my way doesn't get him the right answers, but it does get the right answers.

Answer 29

ur way is right, but its long

Answer 30

can't you use law of sine to solve for it?

Answer 31

Look you can make it shorter if you want my getting a general formula.

Answer 32

\[A\cos(\alpha)=B\cos(\beta)\]\[A=B\frac{\cos(\beta)}{\cos(\alpha)}\]So:\[B\frac{\cos(\beta)}{\cos(\alpha)}\sin(\alpha)+B\sin(\beta)=F\]\[B(\tan(\alpha)\cos(\beta)+\sin(\beta)) = F\]\[B=\frac{F}{\tan(\alpha)\cos(\beta)+\sin(\beta)}\]\[B=\frac{F\cos(\alpha)}{\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)}\]And then:\[A=B \frac{\cos(\beta)}{\cos(\alpha)}\]\[A=\frac{F\cos(\alpha)}{\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)}\cdot \frac{\cos(\beta)}{\cos(\alpha)}\]\[A=\frac{F\cos(\beta)}{\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)}\]

Answer 33

thanks wio

Answer 34

Then using trig identities:\[A=\frac{F\cos(\beta)}{\sin(\alpha)\cos(\beta)+sin(\beta)\cos(\alpha)} =\frac{F\cos(\beta)}{\sin(\alpha+\beta)}\]\[B=\frac{F\cos(\alpha)}{\sin(\alpha)\cos(\beta)+sin(\beta)\cos(\alpha)} =\frac{F\cos(\alpha)}{\sin(\alpha+\beta)}\]Oops I meant this, before was mistake.

Answer 35

@ifrah34 I made little mistake.

Answer 36

its ok, I rather do it the long way, less complications

Answer 37

It's fixed now.

Answer 38

Try that! It works. But it only works for this particular type of problem.

Answer 39

if you were to solve it a trig way, would you approach it differently?

Answer 40

Umm, I'm using trig already

Answer 41

nevermind