Consider two vectors where F1 = 53 N, F2 =

81 N, �1 = 320� , and �2 = 13� ,measured from
the positive x axis with the counter-clockwise. What is the magnitude of the resulting vector?

1 is angle 1 and 2 is angle 2

α1 =360⁰-320⁰ =40⁰

F(x) =F1(x)+F2(x)
=F1cosα1 + F2cosα2 =
=53•cos40 +81•cos13 =40.6+78.9=119.5 N

F(y) =F1(y)+F2(y)
= - F1sinα1 + F2sinα2 =
= - 53•sin40 +81•sin13 =
= - 34.07 +18.22 = - 15.85 N
F=sqrt{F(x)²+F(y)²} =
=sqrt{119. 5²+15.85²} =120.5 N

To find the magnitude of the resulting vector, we can use the concept of vector addition.

Step 1: Represent each vector as components in the x and y directions.

For F1:
Magnitude = 53 N
Angle = 320 degrees

Using trigonometry, we can calculate the x and y components of F1:
Fx1 = Magnitude * cos(angle) = 53 N * cos(320 degrees)
Fy1 = Magnitude * sin(angle) = 53 N * sin(320 degrees)

For F2:
Magnitude = 81 N
Angle = 13 degrees

Using trigonometry, we can calculate the x and y components of F2:
Fx2 = Magnitude * cos(angle) = 81 N * cos(13 degrees)
Fy2 = Magnitude * sin(angle) = 81 N * sin(13 degrees)

Step 2: Calculate the x and y components of the resulting vector by adding the corresponding x and y components of F1 and F2.

Rx = Fx1 + Fx2
Ry = Fy1 + Fy2

Step 3: Calculate the magnitude of the resulting vector using the Pythagorean theorem.

Magnitude of resulting vector (R) = sqrt(Rx^2 + Ry^2)

Now, you can substitute the values and calculate the magnitude of the resulting vector.