What is the direction of the resultant of the concurrent forces shown from the positive x-axis?        Given:        P =20 kN; Q = 15 kN; R = 5 kN; S = 5 kN        θ = 60°; α = 45°; β = 19°.QUESTION 2ANSWERA.26.91°B.144.43°C.37.76°D.153.09°

To solve this problem, we need to find the resultant of the forces in both the x and y directions.

Step 1: Resolve each force into its x and y components.

For P (20 kN), it's only in the x direction, so: Px = 20 kN Py = 0 kN

For Q (15 kN), it's at an angle of 60 degrees from the x-axis, so: Qx = 15 cos(60) = 7.5 kN Qy = 15 sin(60) = 12.99 kN

For R (5 kN), it's at an angle of 45 degrees from the x-axis, so: Rx = 5 cos(45) = 3.54 kN Ry = 5 sin(45) = 3.54 kN

For S (5 kN), it's at an angle of 19 degrees from the x-axis, so: Sx = 5 cos(19) = 4.67 kN Sy = 5 sin(19) = 1.59 kN

Step 2: Sum up the x and y components of all the forces to get the resultant force in the x and y directions.

Resultant in x direction (Rx) = Px + Qx + Rx + Sx = 20 + 7.5 + 3.54 + 4.67 = 35.71 kN Resultant in y direction (Ry) = Py + Qy + Ry + Sy = 0 + 12.99 + 3.54 + 1.59 = 18.12 kN

Step 3: Find the angle of the resultant force from the positive x-axis using the formula θ = atan(Ry/Rx).

θ = atan(18.12/35.71) = 26.91 degrees

So, the direction of the resultant of the concurrent forces shown from the positive x-axis is 26.91 degrees. Therefore, the answer is A.26.91°.