Question

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

Answer 1

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

Answer 2

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

Answer 3

change to vectors and calculate...

Answer 4

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

Answer 5

we add the vectors right?

Answer 6

|M|<cos(t),sin(t)> for each vector

Answer 7

52<cos(100),sin(100)> = a

82<cos(190),sin(190)> = b

111<cos(310),sin(310)> = c

Answer 8

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...

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

our new vector <x,y> has a magnitude of sqrt(x^2 + y^2)

and a direction of tan^-1(y/x) + pi

Answer 13

so the direction is 72.156?

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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)?

Answer 18

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

Answer 19

ok

Answer 20

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

Answer 21

90.86?

Answer 22

200.985 is what I get

Answer 23

but let me recheck my fingers lol

Answer 24

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

Answer 25

tan^-1 spits out 20.9854

Answer 26

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

Answer 27

(52 * cos(100)) + (82 * cos(190)) + (111 * cos(310)) =

-8.03584413 according to google

Answer 28

so thats our x coord

Answer 29

(52 * sin(100)) + (82 * sin(190)) + (111 * sin(310)) =

149.937061 google calculator again

Answer 30

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

Answer 31

((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

Answer 32

ohhh I see

Answer 33

-86.93218027

Answer 34

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

Answer 35

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

Answer 36

so if I add the 180 I get 93.06781973

Answer 37

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

Answer 38

Nope I can show you my answer choices...

Answer 39

are your numbers right in the problem?

Answer 40

no typos?

Answer 41

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

Answer 42

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

Answer 43

ok

Answer 44

almost done :)

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

99.99?

Answer 51

im working and mags v1.3 :)

Answer 52

i know.......

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

say what?

Answer 58

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

Answer 59

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

Answer 60

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!

Answer 61

woohoo....i got a 249 degrees

Answer 62

its B

Answer 63

Ha ha thank you!