four forces 8N 6n 2n and 4n act on a point O in the directionsnorth east south and west respectively find the magnitude of their result and the direction of their resultant force
Force total=8N + 6E + 2S +4W= N (8-2) + E(6-4)= you finish
magnitude = sqrt (6^2+2^2)
direction= arctan E/N=arctan 2/6 E of N
forces in opposite directions subtract from each other
so the "effective forces here are ... 6N north and 2N east
magnitude ... √(6^2 + 2^2)
direction ... [arctan(2 / 6)] east of north
To find the magnitude and direction of the resultant force, we can use vector addition.
Step 1: Draw a four-quadrant diagram and label the directions of the forces.
North (8N)
|
West (4N) -- O -- East (2N)
|
South (6N)
Step 2: Convert the forces into vector form.
North: 8N (0, 8)
East: 2N (2, 0)
South: 6N (0, -6)
West: 4N (-4, 0)
Step 3: Add the vectors together.
Resultant Force (R) = (0 + 2 + (-4), 8 + 0 + (-6))
= (-2, 2)
Step 4: Find the magnitude of the resultant force using the Pythagorean Theorem.
Magnitude (|R|) = √((-2)^2 + 2^2)
= √(4 + 4)
= √8
= 2√2
Step 5: Find the direction of the resultant force.
Direction = arctan(2/(-2))
= -45°
Therefore, the magnitude of the resultant force is 2√2, and the direction is 45° West of North.
To find the magnitude and direction of the resultant force, we can use vector addition. The four forces acting at point O can be represented as vectors.
First, let's assign a coordinate system where the positive x-axis points east, the positive y-axis points north, and the point O is at the center.
The forces in the north and south directions can be represented as:
F1 = 8N (north) = 0i + 8j (where i and j are unit vectors in the x and y directions respectively)
F2 = 2N (south) = 0i - 2j
The forces in the east and west directions can be represented as:
F3 = 6N (east) = 6i + 0j
F4 = 4N (west) = -4i + 0j
Now, we can add these vectors together to find the resultant force vector.
Summing the forces in the x-direction (east/west):
Fx = F3x + F4x = 6i - 4i = 2i
Summing the forces in the y-direction (north/south):
Fy = F1y + F2y = 8j - 2j = 6j
Thus, the resultant force vector is given by:
Fresultant = Fx + Fy = 2i + 6j
To find the magnitude of the resultant:
Magnitude = sqrt(Fresultantx^2 + Fresultanty^2) = sqrt(2^2 + 6^2) = sqrt(4 + 36) = sqrt(40) = 2√10 ≈ 6.324 N.
To find the direction of the resultant force:
Direction = atan(Fresultanty / Fresultantx) = atan(6/2) = atan(3) ≈ 71.57°.
Therefore, the magnitude of the resultant force is approximately 6.324 N, and its direction is approximately 71.57° (measured counterclockwise from the positive x-axis).