Question

Four forces act on bolt A as shown; F1 150N, F2 80N, F3 110N and F4 100N. Determine the magnitude and direction of the resultant of the forces of the bolt, A.

Answer 1

Complete Question

The  complete question(reference (chegg)) is shown on the first uploaded image

Answer:

The magnitude of the resultant force is  $F = 199.64 \ N$

The  direction of the resultant force is  $\theta = 4.1075^o$ from the horizontal plane

Explanation:

Generally when resolving force, if the force (F )is moving toward the angle then the resolve force will be  $Fcos(\theta )$ while if the force is  moving away from the angle  then the resolved force is  $Fsin (\theta )$

Now  from the diagram let resolve the forces to their horizontal component

    So

          $\sum F_x = 150 cos(30) + 100cos(15) -80sin (20)$

          $\sum F_x = 199.128 \ N$

Now  resolving these force into their vertical component can be mathematically evaluated as

         $\sum F_{y} = 150 sin(30) - 100sin(15) -110 +80 cos(20)$

         $\sum F_{y} = 14.30$

Now the resultant force is mathematically evaluated as

        $F = \sqrt{F_x^2 + F_y^2}$

substituting values

        $F = \sqrt{199.128^2 + 14.3^2}$

        $F = 199.64 \ N$

The  direction of the resultant force is  evaluated as

       $\theta = tan^{-1}[\frac{F_y}{F_x} ]$

substituting values

       $\theta = tan^{-1}[\frac{ 14.3}{199.128} ]$

       $\theta = 4.1075^o$ from the horizontal plane