Consider two vectors F~
1 and F~
2 with magnitude
F1 = 61 N and F2 = 63 N
and where θ1 = 98 ◦
and θ2 = 50 ◦The
angles are measured from the positive x axis
with the counter-clockwise angular direction
as positive.
What is the magnitude of the resultant vector F where F = F1+F2
F = 61[98o] + 63[50o].
X = 61*Cos98 + 63*Cos50 = -8.49 + 40.50
= 32.0 N.
Y = 61*sin98 + 63*sin50 = 60.41 + 48.26
= 108.7 N.
F = sqrt(X^2+Y^2).
To find the magnitude of the resultant vector F, we can use the law of cosines.
The law of cosines states that for a triangle with sides a, b, and c and the angle opposite to side c is C, the following equation holds true:
c^2 = a^2 + b^2 - 2abcos(C)
In this case, we have a triangle formed by the vectors F1, F2, and the resultant vector F, where F1 and F2 are the known sides, and the included angle between F1 and F is 180 degrees.
Since we know the magnitudes of F1 (61 N) and F2 (63 N), we can use the law of cosines to find the magnitude of F.
Using the equation:
F^2 = F1^2 + F2^2 - 2(F1)(F2)cos(180)
Substituting the given values:
F^2 = (61)^2 + (63)^2 - 2(61)(63)cos(180)
Simplifying further:
F^2 = 3721 + 3969 - 2(61)(63)(-1)
F^2 = 3721 + 3969 + 2(61)(63)
F^2 = 3721 + 3969 + 7686
F^2 = 15376
Taking the square root of both sides:
F = sqrt(15376)
After calculating, we find that F is approximately equal to 124 N. Hence, the magnitude of the resultant vector F is approximately 124 N.