Two force 1oN and 20N are inclined at angle 60degree . find the magnitude and direction of the force
To find the magnitude of the resulting force, we can use the Law of Cosines:
F^2 = (10N)^2 + (20N)^2 - 2(10N)(20N)cos(60°)
F^2 = 100N^2 + 400N^2 - 400N^2
F = 20N
So the magnitude of the resulting force is 20N.
To find the direction of the force, we can use the Law of Sines:
sin(60°) / F = sin(A) / 10N
sin(A) = (10N sin(60°)) / F
sin(A) = (10N sin(60°)) / 20N
sin(A) = 0.5
A = 30.0°
So the direction of the force is 30° above the 10N force (as measured from the positive x-axis).
To find the magnitude and direction of the resulting force, we can use the concept of vector addition.
Step 1: Resolve the forces into their horizontal and vertical components.
- Force of 10N:
- Horizontal component = 10N * cos(60°) = 10N * 0.5 = 5N
- Vertical component = 10N * sin(60°) = 10N * √(3)/2 ≈ 8.66N
- Force of 20N:
- Horizontal component = 20N * cos(60°) = 20N * 0.5 = 10N
- Vertical component = 20N * sin(60°) = 20N * √(3)/2 ≈ 17.32N
Step 2: Add the horizontal and vertical components separately.
- Horizontal component: 5N + 10N = 15N
- Vertical component: 8.66N + 17.32N ≈ 25.98N
Step 3: Use the Pythagorean theorem to find the magnitude of the resulting force.
Magnitude of the resulting force = √((15N)^2 + (25.98N)^2) ≈ √(225N^2 + 675.12N^2) ≈ √900.12N^2 ≈ 30N
Step 4: Use inverse tangent function to find the direction of the resulting force.
Direction of the resulting force = tan^(-1)(vertical component / horizontal component) ≈ tan^(-1)(25.98N / 15N) ≈ tan^(-1)(1.732) ≈ 60°
Therefore, the magnitude of the resulting force is 30N and its direction is 60°.