four forces 8N,6n,2n and 4n act on a point O in the directions north,east,south and west respectively,find the magnitude of their result and the direction of their resultant force.
net north = 8 - 2 = 6 Newtons
net east = 6-4 = 2 Newtons
|F| = sqrt (36 + 4) = 2 sqrt 10
tan angle n of e = 6/2 = 3
angle n of e = 71.6 degrees north of east
or
90 - 71.6 = 18.4 degrees east of north (compass direction)
To find the resultant force, you can use vector addition. Since each force is given in a specific direction, you can break down each force into its horizontal and vertical components.
Let's assign the directions as follows:
- North as positive y-direction
- East as positive x-direction
- South as negative y-direction
- West as negative x-direction
Let's denote the four forces as:
- Force 1 (8N) in the north direction (y-component: 8N, x-component: 0N)
- Force 2 (6N) in the east direction (y-component: 0N, x-component: 6N)
- Force 3 (2N) in the south direction (y-component: -2N, x-component: 0N)
- Force 4 (4N) in the west direction (y-component: 0N, x-component: -4N)
To find the resultant force, we can simply add up the x and y components of the four forces separately.
Summing up the y-components: 8N - 2N = 6N (north minus south)
Summing up the x-components: 6N - 4N = 2N (east minus west)
Using these components, we can now find the magnitude and direction of the resultant force using the Pythagorean theorem and trigonometry.
Magnitude (R) = sqrt((sum of squares of x-components) + (sum of squares of y-components))
= sqrt((2N)^2 + (6N)^2)
= sqrt(4N^2 + 36N^2)
= sqrt(40N^2)
= 2N * sqrt(10)
To find the direction (θ) of the resultant force, we can use the inverse tangent (tan^-1) function:
θ = tan^-1[(sum of y-components) / (sum of x-components)]
= tan^-1[6N / 2N]
= tan^-1(3)
Converting θ to degrees:
θ (in degrees) = tan^-1(3) * (180° / π)
≈ 71.57°
Therefore, the magnitude of the resultant force is 2N * sqrt(10), and the direction is approximately 71.57° counterclockwise from the positive x-axis (east).