Three forces with magnitudes of 52 pounds, 82 pounds, and 111 pounds act on an object at angles 100o, 190o, and 310o, respectively, with the positive x-axis. Find the magnitude and direction of the resultant force. Round answers to two decimal places.
A. 33.13 pounds; 69.01o
B. 51.47 pounds; 249.01o
C. 33.13 pounds; 249.01o
D. 73.18 pounds; 280.53o
E. 51.47 pounds; 69.01o

- anonymous

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- anonymous

http://www.slideshare.net/stootypal/concurrent-forces-1653276

- anonymous

Once you have read the whole document there, you will very easily solve this problem

- amistre64

change to vectors and calculate...

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## More answers

- anonymous

Most probably the person is not acquainted to that method, that is why I have provided that link. Hope that will be helpful.

- amistre64

we add the vectors right?

- amistre64

|M| for each vector

- amistre64

52 = a
82 = b
111 = c

- anonymous

right I have that cause that's how my book sets it up... then I'm confused where to go from there cause I think my book skips a few steps explanation wise...

- amistre64

i get a vetor of<-18.4345,-48.06> which is just a point from the origin

- amistre64

when we add all the vectors together from head to tail we get a new vector that points n the direction of the angle and with a new magnitude

- anonymous

ok so that's what the <-18.4345,-48.06> is from....

- amistre64

our new vector has a magnitude of sqrt(x^2 + y^2)
and a direction of tan^-1(y/x) + pi

- anonymous

so the direction is 72.156?

- amistre64

we add +pi because tan^-1 only spits out angles in Q1 and Q4; so we no we need Q3 which will be 180 degrees from the tan^-1

- anonymous

oh.. ok? I'm kinda lost thought I was following you but I don't think I am any more

- amistre64

##### 1 Attachment

- amistre64

teh shaded area is the range that tan^-1 operates in for define it as a function....

- anonymous

when you say so we now we need Q3 which will be 180 degrees from the tan^-1
do I take tan(180) or tan^-1(180)?

- amistre64

our angle is in -x,-y..... which is Q3

- anonymous

ok

- amistre64

tan^-1(our angles x and y) spits us out in Q1... add 180 to that to get our answer

- anonymous

90.86?

- amistre64

200.985 is what I get

- amistre64

but let me recheck my fingers lol

- anonymous

oh wow I'm way off the problem is neither of our answers match up with theirs.... ugh this dang practice test is gonna be the death of me! lol

- amistre64

tan^-1 spits out 20.9854

- amistre64

let me re check the vector addition.... i tend to mess that up alot :)

- amistre64

(52 * cos(100)) + (82 * cos(190)) + (111 * cos(310)) =
-8.03584413 according to google

- amistre64

so thats our x coord

- amistre64

(52 * sin(100)) + (82 * sin(190)) + (111 * sin(310)) =
149.937061 google calculator again

- anonymous

so I take the tan^-1 of those and then add 180?

- amistre64

((52 * sin(100)) + (82 * sin(190)) + (111 * sin(310)))
------------------------------------------------
((52 * cos(100)) + (82 * cos(190)) + (111 * cos(310))) =
-18.6585328 so lets tan^-1 that amount

- anonymous

ohhh I see

- anonymous

-86.93218027

- amistre64

the add 180 depends on where our angle is pointing; if its in Q1 or Q4 we are good; if not then add 180

- amistre64

with sin (+) and cos(-) we are in Q2 so add 180

- anonymous

so if I add the 180 I get 93.06781973

- amistre64

.......thats what I get too. its not telling you any more info is it

- anonymous

Nope I can show you my answer choices...

- amistre64

are your numbers right in the problem?

- amistre64

no typos?

- anonymous

jk i forgot I posted the answer choices with them lol no everything is right I took it right from my practice exam online....

- amistre64

let me write up a quick javascript to help us..

- anonymous

ok

- amistre64

almost done :)

- anonymous

ha ha ok just let me know thanks for helping me out so much!

- amistre64

this is what i get, if its debbugged all the way

##### 1 Attachment

- anonymous

it doesn't show my anything just blank what numbers should I plug in?

- anonymous

just kidding I'll plugg in my original numbers from the problem

- amistre64

...... ha.....ha

- anonymous

99.99?

- amistre64

im working and mags v1.3 :)

- amistre64

i know.......

- anonymous

ha ok....well idk I might just guess at it

- amistre64

....... either somethings missing in the problem like a gravity vector of a fr a friction vector.... or i forgot how to add :)

- anonymous

ha ha no I think something is missing in the problem and it's just not written right on the review.... oh well thanks sorry I took so much of your time

- amistre64

'sok....try PIs way...and see if you ge tthe same results...

- anonymous

say what?

- anonymous

oh Pi's way... wow I'm sorry you probably think i'm retarded or something lol

- amistre64

PI gave a website....you know, instead of conversing with you and such ;)

- anonymous

ha ha ya I read it lol my book gives me the same way we just tried it but I can try his way not sure if I understand it but I'll give it a shot thanks anyways!

- amistre64

woohoo....i got a 249 degrees

- amistre64

its B

- anonymous

Ha ha thank you!

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