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#1
O.k. I know this problem isnt that hard but I cant remember how its done...can anyone help me out?

|U|= 25,theta(lowercase u)=219degrees
|v|=37,theta(lowercase v) = 137degrees
0
18 years ago
#2
did i hear you say it wasnt hard - i must b so stupid!
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#3
doesnt really have anything to do with being smart or not....judging from your picture you look pretty smart
0
18 years ago
#4
|U|= 25,theta(lowercase u)=219degrees
|v|=37,theta(lowercase v) = 137degrees[/QUOTE]

Assuming angles measured clockwise form +ve X axis then Horiz get 25Cos39+37Cos43 acting to left ---(1)
and
37Sin43-25sin39 acting up---(2)

Using Pythag you can get resultant magnitude
sqrt{(1)^2+(2)^2}
Using Trig: Tan^-1 {(1)/(2)} to get angle to left of vertical so bearing is 90 + calc angle above.

OK not a full solution but should be enough to get you your answer
0
18 years ago
#5
(Original post by djwarfield)
O.k. I know this problem isnt that hard but I cant remember how its done...can anyone help me out?

|U|= 25,theta(lowercase u)=219degrees
|v|=37,theta(lowercase v) = 137degrees
u y = 25sin219 = -15.73
u x 25cos219 = -19.43

v y = 37sin137 = 25.23
v x = 37cos137 = -27.06

add em together and |u|+|v| = (-19.43+-27.06, -15.73+25.23) =
translation of (-46.49, 9.50)

or something
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#6
So pretty much you have to find both the x and y axis for each of the equations. so the absolute value of vector U = 25. Considering this, Xu could be understood as absolute value of vector Ucosthetau which would be 25cos219degrees which in turn would be -19.43

and that would be the first equation for the X axis.....how am I doing so far?
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18 years ago
#7
Dont work in angles greater than 90deg

Draw diagram, two lines, one to left and up at angele 43deg
The other to left and down at 35deg

Now resolve Horiz and then Resolve vert

as I aluded to above
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#8
(Original post by Phil_C)
Dont work in angles greater than 90deg

Draw diagram, two lines, one to left and up at angele 43deg
The other to left and down at 35deg

Now resolve Horiz and then Resolve vert

as I aluded to above

I dont quite undrstand where your going with that? Can you eloborate please.
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18 years ago
#9
To keep vector problems a little easier, only work in angles <90 so if a problem, such as yours quotes 219deg it's saying 219deg from some datum, in problems of angles of any magnitude the datum is the +ve x axis abd +ve angle is measured anticlockwise so straight up is 90deg and to teh left would be 180 deg etc.

Drawing all vectors out in a digramatic form with the acute angle between the vector and the x axis (either direction) will simplify matters as now all angles are acute so the Cos and Sin ratios will be +ve no need to remember the CAST diagram or try to work out why you are getting negative values on your calculator when dialing in Sin 219 or Cos143 say.

Make sense?
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