Answer:
To find the resultant force and its direction, we can use vector addition.
First, let's break down force B and force C into their horizontal and vertical components:
Horizontal component of force B:
Bx = 20N * cos(140°)
Vertical component of force B:
By = 20N * sin(140°)
Horizontal component of force C:
Cx = 16N * cos(290°)
Vertical component of force C:
Cy = 16N * sin(290°)
Now, let's add up the horizontal and vertical components of all the forces:
Horizontal component of resultant force:
Rx = Ax + Bx + Cx
Vertical component of resultant force:
Ry = Ay + By + Cy
To find the magnitude of the resultant force (R), we use the Pythagorean theorem:
R = sqrt(Rx^2 + Ry^2)
To find the direction (θ) of the resultant force, we can use the inverse tangent function:
θ = atan(Ry / Rx)
Plugging in the given values:
Ax = 12N (horizontal component of force A)
Ay = 0N (vertical component of force A)
Bx = 20N * cos(140°)
By = 20N * sin(140°)
Cx = 16N * cos(290°)
Cy = 16N * sin(290°)
Now let's calculate the values:
Bx = 20N * cos(140°) ≈ -11.55 N
By = 20N * sin(140°) ≈ 9.56 N
Cx = 16N * cos(290°) ≈ 13.82 N
Cy = 16N * sin(290°) ≈ -5.45 N
Rx = 12N + (-11.55N) + 13.82N ≈ 14.27 N
Ry = 0N + 9.56N + (-5.45N) ≈ 4.11 N
R = sqrt(14.27^2 + 4.11^2) ≈ 14.98 N
θ = atan(4.11 / 14.27) ≈ -15.58°
The magnitude of the resultant force is approximately 14.98 N, and the direction is approximately -15.58° (or approximately -45° to force A).
Note: The negative sign indicates that the resultant force is in the opposite direction to force A.