Question

Q3: Force A, 12N acting horizontally to the right, force B, 20N acting. at 140° to force A; force C, 16N acting at 290° to force A. (Ans.: 3.06 kN, -45° to force A) ​

Answer 1

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.