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.

Question

anonymous · Answer

Change the forces into Cartesian coordinates, and add them using vector addition.
You can then change back if you want.

anonymous · Answer

Changing to Cartesian from Polar of magnitude $r$ and angle $theta$: $$
x = r\cos\theta\
y = r\sin\theta
$$Changing from Polar to Cartesian of horizontal magnitude $x$ and vertical magnitude $y$: $$
r = \sqrt{x^2+y^2} \
\theta  = \tan^{-1}(y/x)
$$

anonymous · Answer

x=200cos60 ? and y= 100sin170?

anonymous · Answer

$$
F_1 = 200\cos60^\circ\hat{\imath} + 200\sin60^\circ\hat{\jmath}\
F_2 = 100\cos170^\circ\hat{\imath} + 100\sin170^\circ\hat{\jmath}
$$

anonymous · Answer

$$
F_3 = F_1+F_2 
$$

anonymous · Answer

Is F3 my  my direction or magnitude? and F3=100+100 -98.48-98.48?

anonymous · Answer

.... You really don't understand how vectors work?

anonymous · Answer

$F_3$ is a vector.

anonymous · Answer

So are $F_1$ and $F_2$... do you know how to add vectors?

anonymous · Answer

If i had <1,5> + <3,8> = <4,13> is that what you mean?

anonymous · Answer

Yeah... That's it.  Find $F_1$ and $F_2$ first.

anonymous · Answer

I dont understand why  I have to do F1=200cos60∘ı^+200sin60∘ȷ^ instead of just F1=200cos60∘ı^ and also what is ∘ı^? Is it just supposed to be degrees?

anonymous · Answer

It's vector notation...

anonymous · Answer

$<3,2> = 3\hat{\imath}+2\hat{\jmath}$

anonymous · Answer

Nevermind lol! That was a silly question, i didnt look at the fact that cos for the first and then sin :)

anonymous · Answer

F1=200cos60,200sin60
  =100,173.2
F2=100cos170,100sin170
   =-98.48,17.36

anonymous · Answer

F3=1.52,190.56

anonymous · Answer

Is this the end result?

anonymous · Answer

For the direction using θ=tan−1(y/x) which numbers would i plug in for x and y?

anonymous · Answer

$$\tan^{-1} (170/60)  or \tan^{-1} (60/170)$$

anonymous · Answer

Ummm, well in this case $x =1.52$ and $y=190.56$

anonymous · Answer

$$
\theta = \tan^{-1}(190.56/1.52)
$$

anonymous · Answer

$$
r = \sqrt{(1.52)^2+(190.56)^2}
$$