Question

A force F1 of magnitude 6.00 units acts at the origin in a
direction 30.0° above the positive x axis. A second force
F2 of magnitude 5.00 units acts at the origin in the direction
of the positive y axis. Find graphically the magnitude
and direction of the resultant force F1 + F2 .

Answer 1

@terenzreignz @Eyad

Answer 2

Physics? Free Body Diagram time :D

Answer 3

Ok ......

Answer 4

So, now, the way to get the resultant vector is to add their respective components :)

Answer 5

Now my vectors are becoming simpler now

Answer 6

So, you understand from here?
If you're given the magnitude and direction (angle) of a vector, you need to break it down into its x and y components...
\[\huge \vec v= x\vec i + y\vec j\]

Answer 7

In this case,
\[\huge x = |\vec v|\cos (\theta) \\\huge y = |\vec v|\sin(\theta)\]

Answer 8

So, for F1
\[\huge \vec{V1}=[6\cos(\theta)]\vec i + [6\sin(\theta)]\vec j\]

NOTE : the i and j here are the horizontal and vertical components of the vectors respectively.

Answer 9

ok

Answer 10

\[\huge \vec{V1}=[6\cos(30^o)]\vec i + [6\sin(30^o)]\vec j\]

Answer 11

ok

Answer 12

Sorry, not V1, but F1
\[\huge \vec{F1}=\left[3\sqrt3\right]\vec i + \left[3\right]\vec j\]

Answer 13

Now, if you do F2, the magnitude is 5, and the direction (angle) is 90 degrees...
\[\huge \vec{F2}=[5\cos(90^o)]\vec i +[5\sin(90^o)]\vec j\]\[\huge \vec{F2}=[0]\vec i + [5]\vec j\]

Answer 14

So...
\[\Large \vec{F1}+\vec{F2}=(3\sqrt3) \ \vec{i}+3\vec j + 0\vec i+ 5\vec j\]\[\huge \vec{F1}+\vec{F2}= \left(3\sqrt 3\right)\vec i +8\vec j\]

Answer 15

\[\huge \vec{F1}+\vec{F2}=\left<3\sqrt3 \ , \ 8\right>\]
Now, find magnitude and direction :)

Answer 16

Thank now i can solve it :)

Answer 17

Yes, and now, (hopefully) you can now solve similar problems :D