Question
Consider two vectors F~
1 with magnitude 44 N
inclined at 163
◦
and F~
2 with magnitude 72 N
inclined at 236
◦
, measured from the positive
x axis with counterclockwise positive.
What is the magnitude of the resultant vector F~
1 + F~
2 ? Draw the vectors to scale on a
graph to determine the answer.
1. 130 N
2. 98 N
3. 123 N
4. 95 N
5. 52 N
6. 72 N
What is the direction of this resultant vector (between the limits of −180◦
and +180◦
from the positive x-axis) ?
1. 103◦
2. 6◦
3. 23◦
4. 37◦
5. −38◦
6. −150◦
1 with magnitude 44 N
inclined at 163
◦
and F~
2 with magnitude 72 N
inclined at 236
◦
, measured from the positive
x axis with counterclockwise positive.
What is the magnitude of the resultant vector F~
1 + F~
2 ? Draw the vectors to scale on a
graph to determine the answer.
1. 130 N
2. 98 N
3. 123 N
4. 95 N
5. 52 N
6. 72 N
What is the direction of this resultant vector (between the limits of −180◦
and +180◦
from the positive x-axis) ?
1. 103◦
2. 6◦
3. 23◦
4. 37◦
5. −38◦
6. −150◦
Answers
R = 44N@163o + 72N@236o.
X =Hor.= 44*cos163 + 72*cos236=-82.3 N.
Y =Ver.= 44*sin163 + 72*sin236=-46.8 N.
1. R^2=X^2 + Y^2=(-82.3)^2 + (-46.8)^2 =8963.53.
R = 95 N. = Resultant force.
tanAr = Y/X = -46.8/-82.3 = 0.56865.
Ar = 29.62o = Reference angle.
A = 29.62 + 180 = 210o,CCW. = -150o,CW.
X =Hor.= 44*cos163 + 72*cos236=-82.3 N.
Y =Ver.= 44*sin163 + 72*sin236=-46.8 N.
1. R^2=X^2 + Y^2=(-82.3)^2 + (-46.8)^2 =8963.53.
R = 95 N. = Resultant force.
tanAr = Y/X = -46.8/-82.3 = 0.56865.
Ar = 29.62o = Reference angle.
A = 29.62 + 180 = 210o,CCW. = -150o,CW.
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