Two forces F1 and F2 act on a particle. F1 has a magnitude of 6N and in direction 045°, and F2 has a magnitude of 7N and in direction 154°. Find the magnitude and direction of their resultant

Answer 1

To find the magnitude and direction of the resultant force, we can use the concept of vector addition.

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

F1:
Magnitude: 6N
Direction: 045°

Horizontal component F1x:
F1x = 6N * cos(45°) ≈ 4.24N

Vertical component F1y:
F1y = 6N * sin(45°) ≈ 4.24N

F2:
Magnitude: 7N
Direction: 154°

Horizontal component F2x:
F2x = 7N * cos(154°) ≈ -5.64N

Vertical component F2y:
F2y = 7N * sin(154°) ≈ 4.89N

Now, let's find the sum of the horizontal and vertical components for the resultant force.

Sum of horizontal components:
Fresx = F1x + F2x ≈ 4.24N + (-5.64N) ≈ -1.4N

Sum of vertical components:
Fresy = F1y + F2y ≈ 4.24N + 4.89N ≈ 9.13N

Next, let's find the magnitude and direction of the resultant force.

Magnitude of the resultant force:
Fres = sqrt(Fresx^2 + Fresy^2) ≈ sqrt((-1.4N)^2 + (9.13N)^2) ≈ 9.32N

Direction of the resultant force:
tanθ = Fresy / Fresx
θ = arctan(Fresy / Fresx) ≈ arctan(9.13N / -1.4N) ≈ -80.6°

The magnitude of the resultant force is approximately 9.32N, and its direction is approximately -80.6°.