Question

Two forces with magnitudes of 200 and 100 pounds act on an object at angles of 60° and 170° respectively. Find the direction and magnitude of the resultant force. Round to two decimal places in all intermediate steps and in your final answer.

Answer 1

Oh I see this same question was asked somewhere else already.
Were you able to figure it out? :)

Answer 2

Take Summation of Forces Horizontal and vertical

EFX = 200cos60 - 100cos(180-170) = 1.519224699
EFY = 200sin60 100sin(180-170) = 190.5678785

Resultant/Magnitude
R2 = (ΣFx)² (ΣFy)²
R = 190.5759541

THETA¸ =arctan [EFY2 /EFX2]
THETA¸ =89.54324728

Answer 3

is tht right?

Answer 4

Mmm so sloppy :) Lemme work it out real quick heh.
\[\Large \vec{F_1} \quad=\quad \left<\color{orangered}{200\cos60}, \color{royalblue}{200\sin60}\right>\]\[\Large \vec{F_2} \quad=\quad \left<\color{orangered}{100\cos170}, \quad\color{royalblue}{100\sin170}\right>\]

\[\Large \vec F \quad=\quad \vec F_1+ \vec F_2\]\[\large \vec F \quad=\quad \left<\color{orangered}{200\cos60+100\cos170},\quad\color{royalblue}{200\sin60+100\sin170}\right>\]
Which simplifies to,

Answer 5

Mmm ya looks like you're on the right track so far!\[\large \vec F \quad=\quad \left<\color{orangered}{1.519},\quad\color{royalblue}{190.57}\right>\]

Answer 6

And the magnitude of the resultant vector F will be,\[\Large |\vec F| \quad=\quad \sqrt{1.519^2+190.57^2}\]
\[\Large |\vec F| \quad=\quad 190.576\]

Answer 7

Ya looks fine :O

Answer 8

It would be a good idea to make sure you understand these notes though! :) lol

Answer 9

thank you!, i just wanted to make sure i had it correctly