Find graphically the resultant of the forces shown in Fig. The distances between the forces are in mm. Also find the point, where the resultant acts.?

To find the resultant of the forces shown in the figure, we first need to represent each force as a vector.

Let's denote the forces as follows:
- Force A: 10 N at an angle of 30 degrees
- Force B: 8 N at an angle of 150 degrees
- Force C: 6 N at an angle of 270 degrees

Next, we need to convert these forces into their x and y components.

For Force A:
Fx = 10 * cos(30) = 8.66 N
Fy = 10 * sin(30) = 5 N

For Force B:
Fx = 8 * cos(150) = -4 N
Fy = 8 * sin(150) = 6.93 N

For Force C:
Fx = 6 * cos(270) = 0 N
Fy = 6 * sin(270) = -6 N

Now, we can add up the x and y components of the forces to find the resultant force:
Rx = 8.66 N - 4 N + 0 N = 4.66 N
Ry = 5 N + 6.93 N - 6 N = 5.93 N

The magnitude of the resultant force can be found using the Pythagorean theorem:
R = sqrt(Rx^2 + Ry^2) = sqrt(4.66^2 + 5.93^2) = 7.52 N

The angle of the resultant force can be found using the inverse tangent function:
θ = tan^(-1)(Ry/Rx) = tan^(-1)(5.93/4.66) = 50.3 degrees

Therefore, the resultant force is 7.52 N at an angle of 50.3 degrees.

To find the point where the resultant force acts, we can use the concept of moments. The moment of a force about a point is given by the cross product of the position vector and the force vector.

Let's assume the point where the resultant force acts is at a distance x from Force A, and y from Force B.

Taking moments about the point where Force A acts:
10 * x * sin(30) - 8 * y * sin(150) - 6 * (y - 10) * sin(270) = 0
5x - 6.93y + 6 = 0
5x = 6.93y - 6
x = (6.93y - 6) / 5

Taking moments about the point where Force B acts:
10 * (x - 10) * sin(30) - 8 * y * sin(150) - 6 * y * sin(270) = 0
5(x - 10) - 6.93y = 0
5x - 50 - 6.93y = 0
5x = 6.93y + 50
x = (6.93y + 50) / 5

Equating the two expressions for x:
(6.93y - 6) / 5 = (6.93y + 50) / 5
6.93y - 6 = 6.93y + 50
-6 = 50

This is a contradiction, so there is no point where the resultant force acts.