Two force vectors, F = 8.0 N at an angle of 60° above the x-axis and F = 5.5 N at an angle of 45° below the...?
Two force vectors, F = 8.0 N at an angle of 60° above the x-axis and F = 5.5 N at an angle of 45° below the x-axis are applied to a particle at the origin. What third force F=? would make the net, or resultant, force on the particle zero?
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Force 1
x component y component
Fcos(angle) Fsin(angle)
8cos60 8 sin60
4 6.9282
Force 2
x component y component
Fcos(angle) Fsin(angle)
5.5cos(-45) 5.5sin(-45)
3.88909 3.88909
Now add the x's and the y's
x component y component
7.88909 3.03912
use pythagorean theorem to find resultant
sqrt(7.88909^2 + 3.03912^2)
8.45423
to find angle
tan(angle)=y/x
tan(angle)=3.03912/7.88909
angle=21.0682
The resultant is 8.45423 @ 21.07 degrees
thus to make force on particle be zero make the vector opposite. To do this add 180 to the angle
F=8.45423 @ 181.07 degrees
Sum of the forces = 0 Sum of the forces in the x route = 0 ==> 0 = 8*cos(60) + 5.5*cos(40 5) - F3x F3x = 8*cos(60) + 5.5*cos(40 5) F3x = 7.89 N Sum of the forces in the y route = 0 ==> 0 = 8*sin(60) - 5.5*sin(40 5) -F3y F3y = 8*sin(60) - 5.5*sin(40 5) F3y = 3.04 N F3 = sqrt(F3x^2 + F3y^2) F3 = 8.40 six N theta = tan^-a million(F3y/F3x) theta = 201.a million degress (vector sensible F3y and x are appearing in the different route so as that suggests that's one hundred eighty ranges opposite the importance of F1 + F2.)