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Angle of a Vector 
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#1
Sep2010, 12:44 PM

P: 34

1. The problem statement, all variables and given/known data
two forces, 406N at 17 degrees and 256N at 26 degrees are applied to a a 3400kg car. Find resultant of these two forces and the direction of the resultant force between 180 and 180 degrees. 2. Relevant equations 3. The attempt at a solution 406cos(17) = 388.26, 256cos(26) = 230.09, 230.09 = 388.26 = 618.35N arctan(230.09/388.26) = 30.65 degrees 


#3
Sep2010, 04:36 PM

P: 34

because it's a car and it doesn't move on the yaxis. I'm still not sure how to get the angle though.



#4
Sep2010, 04:41 PM

PF Gold
P: 1,236

Angle of a Vector



#5
Sep2110, 10:18 AM

PF Gold
P: 1,236

Does your answer make sense that the highest forces pulling at 17 degrees somehow generates a force vector that is 30 degress? Your answer is obviously somewhere between (17+26)/2=21.5 degree spread between the two forces. You have 1.58 times the force at 17 degrees than at 26 degrees, so you should be slightly higher off a midpoint mark (1721.5 = 4.5 deg). So your answer should be above 0 degree mark at least, but not at 30 degrees! Your feasible region is thus between (4.5, 17) degrees
Look at it this way.. if 17 degree force has (406/256)= 1.5859375 times more weight than the negative force, then your resultant should be around 1.5859375*17 + 1*(26) ~ 0.96 degrees. Or more closely to the answer now, 1*17 + 0.60591138*(26) ~ 0.60 degrees You need to calculate all resultant forces and follow rules of trigonometry. 


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