Find the magnitude of the resultant vector of two 5-pound forces that act on an object with an angle of 70° between them.
let
u = (5,0)
v = (5cos70°,5sin70°)
then
u+v = (5(1+cos70°),5sin70°) = (6.71,4.70)
now just get |u+v| and its direction
To find the magnitude of the resultant vector, we can use the law of cosines. The law of cosines states that for any triangle with sides a, b, and c, and angle C opposite to side c:
c^2 = a^2 + b^2 - 2ab*cos(C)
In this case, the two forces are the sides of a triangle, and we want to find the magnitude of the resultant vector, which is the length of the third side of the triangle.
Let's call the magnitude of the resultant vector R. We have two forces of 5 pounds each, which we'll call F1 and F2. The angle between them is 70°.
Using the law of cosines, we can write:
R^2 = F1^2 + F2^2 - 2*F1*F2*cos(70°)
Substituting the values, we get:
R^2 = (5)^2 + (5)^2 - 2*(5)*(5)*cos(70°)
Now, we can calculate the magnitude of the resultant vector:
R = sqrt( (5)^2 + (5)^2 - 2*(5)*(5)*cos(70°) )
Computing this expression, we find:
R ≈ 8.04 pounds
Therefore, the magnitude of the resultant vector is approximately 8.04 pounds.