A particle undergoes two displacements, measured from the positive x -axis, with counter-clockwise positive. The first has a magnitude of 11 m and makes an angle of 67◦ with the positive x axis. The resultant displacement has a magnitude of 73 m directed at an angle of 105degrees from the positive x-axis.

Find the Magnitude and angleof the 2nd displacement

Question

A particle undergoes two displacements, measured from the positive x -axis, with counter-clockwise positive. The first has a magnitude of 11 m and makes an angle of 67◦ with the positive x axis. The resultant displacement has a magnitude of 73 m directed at an angle of 105degrees from the positive x-axis.

Find the Magnitude and angleof the 2nd displacement

anonymous · Answer

I found the magnitude.
I am trying to interpret the angle

anonymous · Answer

$$\theta=\tan^{-1} (\frac{ -3.07 }{ -6.19 })=26.421$$

anonymous · Answer

since this Bx<0 and By<0 I assume this is in the 3rd quad, thus would I just do  26.421-180 to find the angle??

anonymous · Answer

@ganeshie8 can you help possibly?

anonymous · Answer

THanks for coming :)

jim_thompson5910 · Answer

hmm this is certainly a tough one, but it's doable

anonymous · Answer

I have solved the majority of it, just trying to do the trig part of it

jim_thompson5910 · Answer

did you draw out a parallelogram?

anonymous · Answer

no, I just used the laws of sines and cosines

anonymous · Answer

for the magnitude

anonymous · Answer

if both x and y are negative it is in the 3rd quad?

jim_thompson5910 · Answer

where does it say they're negative?

anonymous · Answer

i figured that out in part 1 finding the magnitude.  deltax -6.19 and deltay is -3.07

jim_thompson5910 · Answer

ok one sec

anonymous · Answer

Rcos(r)-Acos(a) = Bx
Rsin(r)-Asin(a) = By
thus b = sqrt(by^2+bx^2)

anonymous · Answer

my issue is the angle.. is it negative?  do i need to add or subtract to the value I got from arctan(deltay/deltax) ?

jim_thompson5910 · Answer

drawing it out, one sec

anonymous · Answer

awesome, thanks

anonymous · Answer

just curious, why a parallelogram?  I drew it myself and now I can see thats its a negative angle (I had part of it drawn out earlier)

jim_thompson5910 · Answer

you draw a parallelogram because the two initial vectors (ie forces) pull and create a resultant vector which is really the diagonal of the parallelogram

so you need a parallelogram to figure out the magnitude of the diagonal and the angle of the diagonal

jim_thompson5910 · Answer

so it's like this

|dw:1378771093709:dw|

jim_thompson5910 · Answer

use the parallelogram law

|dw:1378771136642:dw|

jim_thompson5910 · Answer

resultant is this diagonal here

|dw:1378771170948:dw|

anonymous · Answer

ahh very nice explanation!

anonymous · Answer

that is the best physics explanation ive heard in 3 weeks of lecture :)

anonymous · Answer

theta = -153.58

jim_thompson5910 · Answer

lol I'm glad to hear that

jim_thompson5910 · Answer

atm I'm trying to accurately draw it so I can have it to check

anonymous · Answer

Thank you Jim.  I entered it on the utexas physics website and it accepted the angle.

jim_thompson5910 · Answer

oh you figured it all out?

anonymous · Answer

yep :)

jim_thompson5910 · Answer

ok that's great

anonymous · Answer

I was almost there.. but every answer u answer wrong they mark you points off... so i was making triple sure.. and ur explanation sealed the deal

jim_thompson5910 · Answer

I'm glad it did