anonymous
  • anonymous
how do you find the resultant of a vector quantity using scale drawings?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
It is probably best to use the head-to-tail method here, since we don't have to deal with angles. If we can determine the x and y-components of each vector, we can easily find the length of the resultant vector. Let vector 1 have components \(x_1 ~{\rm and}~ y_1\) and vector 2 have components \(x_2~{\rm and}~y_2\). The length of the resultant is\[L_R = \sqrt{~(x_1+x_2)^2 +(y_1+y_2)^2}\] We can find the angle of the resultant as\[\tan \theta = {y_1 + y_2 \over x_1 + x_2}\]
anonymous
  • anonymous
just move the scale parallel to it make head tale comibnation and go by the simple law of addition of vector
apoorvk
  • apoorvk
|dw:1332707319496:dw| use the parallelogram law of vectors as above.

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anonymous
  • anonymous
|dw:1332727526572:dw| im suppose to find the resultant in m/s
anonymous
  • anonymous
in meter/sec ??? hv u writen the correct question?
anonymous
  • anonymous
ohh no sorry ....... the y value is meant to be 3n and the x as 4 ;; question says let 2cm rep 1n but i dont understand how im suppose to find the resultant based on the information given
anonymous
  • anonymous
answer is 25N if u have given me force 15 N 20 N
anonymous
  • anonymous
ok ok look
anonymous
  • anonymous
im getting 5n if i use pythagorus' therem....because you square the resultant.....
anonymous
  • anonymous
ok tell me how did u use pythagorus theorem steps
anonymous
  • anonymous
umm R2=a2+b2 .......3squared + 4squared.....which is 9+16 >>>25 nd you square that
anonymous
  • anonymous
|dw:1332708622689:dw|
anonymous
  • anonymous
are u gettin the poin that u can move y=2 back and then draw m as a resultant??
anonymous
  • anonymous
|dw:1332708732145:dw|
anonymous
  • anonymous
ohhhh i see...thank you
anonymous
  • anonymous
:-)
anonymous
  • anonymous
u can move vector parallel to itself ok?? and in an vector addition u need to add head to tail
anonymous
  • anonymous
thanks again :)
anonymous
  • anonymous
ok tc

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