- #1

- 13

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter mrpburke
- Start date

- #1

- 13

- 0

- #2

- 13

- 0

Anybody???

I found the answer...68.2kN but I have no idea how to get that.

I found the answer...68.2kN but I have no idea how to get that.

- #3

jtbell

Mentor

- 15,803

- 4,034

[tex]{\vec F}_{net} = {\vec F}_A + {\vec F}_B + {\vec F}_C[/tex]

Write down the corresponding x- and y- component equations, using the magnitudes and angles of the forces. For example, the x-component of [itex]{\vec F}_A[/itex] is [itex] |{\vec F}_A} | \cos \theta_A[/itex]. You're given that the three forces have the same magnitude, so use [itex] |{\vec F}_A} | [/itex] for all of them, but of course the angles are different. And [itex]{\vec F}_{net}[/itex] has its own magnitude (which you're given) and angle (which is unknown).

Once you have those two equations, you should have enough information to solve for [itex] |{\vec F}_A} | [/itex].

- #4

- 52

- 0

Then you could write one equation saying 200= the sum of all the components

and a few equations equating the magnitudes.

If that didn't help you, say so, and I'll try to explain it more explicitly.

- #5

jtbell

Mentor

- 15,803

- 4,034

jtbell said:Write down the corresponding x- and y- component equations, using the magnitudes and angles of the forces.

I'd better point out that whenever you have a problem involving magnitudes and angles of forces (or any other vectors, for that matter), this is usually a good way to start.

- #6

- 13

- 0

Could someone work that out a bit more. I'm just not seeing it:(

- #7

FredGarvin

Science Advisor

- 5,067

- 9

- #8

- 13

- 0

I have A, B, and C broken down to x and y components. Now what?

- #9

- 13

- 0

I keep getting 47.09kN for the magnitude of Fa. I think the answer in the book is wrong.

- #10

jtbell

Mentor

- 15,803

- 4,034

Well, I don't think anybody here has psychic "remote vision," so we can't tell you where your mistake is unless you show us your work!

My calculation agrees with the book's answer of 68.2 kN, by the way.

My calculation agrees with the book's answer of 68.2 kN, by the way.

Last edited:

- #11

- 13

- 0

How are you getting that answer?

- #12

jtbell

Mentor

- 15,803

- 4,034

- #13

- 13

- 0

I'm workin on it.

- #14

- 13

- 0

200cos@ = |Fa|1.88

200sin@ = |Fa|2.45

????

when that's worked out it comes to 64.7 which isnt quite right.

- #15

- 13

- 0

- #16

- 13

- 0

Here's what I have. Somebody PLEASE show me what I'm doing wrong!

- #17

jtbell

Mentor

- 15,803

- 4,034

- #18

- 13

- 0

Lemme work it out neatly and post it up!!!

- #19

jtbell

Mentor

- 15,803

- 4,034

Yeah, just fixing those two angles should do it.

- #20

- 13

- 0

I think that's gonna be it!!!

Thanks!

- #21

- 13

- 0

jtbell said:Yeah, just fixing those two angles should do it.

I've been all over that just never did it quite like that. I dont know why I didnt even think of that.

gahh....I hate it when it's something simple like that

Share: