Two forces F1 and F2 act on a particle. F1 has magnitude 5N and in direction 30° and F2 has magnitude 8N and in direction 90° find the magnitude and direction of their resultant

Answer 1

To find the magnitude and direction of the resultant force, we can add the two forces together using vector addition.

First, let's find the horizontal and vertical components of each force.

The horizontal component of F1 can be found using trigonometry:
F1x = F1 * cos(30°)
F1x = 5N * cos(30°)
F1x ≈ 4.33N

The vertical component of F1 can also be found using trigonometry:
F1y = F1 * sin(30°)
F1y = 5N * sin(30°)
F1y = 2.5N

Since F2 has a direction of 90°, its horizontal component is 0N and its vertical component is equal to the magnitude itself:
F2x = 0N
F2y = 8N

Now, let's add the horizontal and vertical components of the forces to find the total horizontal and vertical components:
Ftotalx = F1x + F2x
Ftotalx = 4.33N + 0N
Ftotalx ≈ 4.33N

Ftotaly = F1y + F2y
Ftotaly = 2.5N + 8N
Ftotaly = 10.5N

To find the magnitude of the resultant force, we can use the Pythagorean theorem:
|Ftotal| = sqrt(Ftotalx^2 + Ftotaly^2)
|Ftotal| = sqrt((4.33N)^2 + (10.5N)^2)
|Ftotal| ≈ sqrt(18.7129N^2 + 110.25N^2)
|Ftotal| ≈ sqrt(129.9629N^2)
|Ftotal| ≈ 11.4N

To find the direction of the resultant force, we can use trigonometry and find the angle formed with the horizontal axis:
θ = atan(Ftotaly / Ftotalx)
θ = atan(10.5N / 4.33N)
θ ≈ 67.5°

Therefore, the magnitude of the resultant force is approximately 11.4N and its direction is approximately 67.5°.