The sum of two forces, one having a magni-
tude of 2 N acting due west and other having
a magnitude of 8 N acting due north is?
Think hypotenuse.
If you want the direction, think arctan 0.25
To find the sum of two forces, we need to use vector addition. In this case, we have one force acting due west and the other acting due north.
Step 1: Represent the forces as vectors:
The force acting due west can be represented as a vector in the negative x-direction, with a magnitude of 2 N. Let's call it F1 = -2 N î.
The force acting due north can be represented as a vector in the positive y-direction, with a magnitude of 8 N. Let's call it F2 = 8 N ĵ.
Step 2: Perform vector addition:
To find the sum of these two forces, we need to add their respective x and y components. The x-component of the sum will give the net force in the west-east direction, and the y-component will give the net force in the north-south direction.
Adding the x-components:
The x-component of F1 is simply -2 N, as it is acting entirely in the west direction.
The x-component of F2 is zero, as it has no component in the x-direction.
Adding the y-components:
The y-component of F1 is zero, as it has no component in the y-direction.
The y-component of F2 is 8 N, as it is acting entirely in the north direction.
Therefore, the sum of the two forces is:
Sum of forces = (-2 N, 8 N)
In terms of magnitude and direction, the magnitude of the sum of forces can be found using the Pythagorean theorem:
Magnitude of the sum of forces = sqrt((-2 N)^2 + (8 N)^2) = sqrt(4 N^2 + 64 N^2) = sqrt(68 N^2) = 8.246 N.
The direction of the sum of forces can be found using trigonometry:
Direction = arctan((8 N) / (-2 N)) = -75.96 degrees (measured counter-clockwise from the positive x-axis).
Therefore, the sum of the two forces, one having a magnitude of 2 N acting due west and the other having a magnitude of 8 N acting due north, is approximately 8.246 N at an angle of -75.96 degrees from the positive x-axis.