Angle of a Vector


by delfam
Tags: angle, vector
delfam
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#1
Sep20-10, 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
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cronxeh
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#2
Sep20-10, 01:26 PM
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What about the y component?
delfam
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#3
Sep20-10, 04:36 PM
P: 34
because it's a car and it doesn't move on the y-axis. I'm still not sure how to get the angle though.

cronxeh
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#4
Sep20-10, 04:41 PM
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Angle of a Vector


Quote Quote by delfam View Post
because it's a car and it doesn't move on the y-axis. I'm still not sure how to get the angle though.
The car has nothing to do with anything. And changing -26 degrees to 26 was a nice trick, but you need to follow the rules because the next step requires the actual angle.
cronxeh
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#5
Sep21-10, 10:18 AM
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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 (17-21.5 = -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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