Question

Determine the components of reaction at the fixed support

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The components of reaction at the fixed support are  

    $A_{(x)} = 400 \ N$ ,  $A_{(y)} = -500 \ N$ ,  $A_{(z)} = 600 \ N$ ,  $M_x = 1225 \ N\cdot m$ , $M_y = 750 \ N\cdot m$ ,  $M_z = 0 \ N\cdot m$

Explanation:

Looking at  the diagram uploaded we see that there are two  forces acting along the x-axis on the fixed support    

   These force are  400 N  and  $A_{(x)}$ [ i.e the reactive force of  400 N  ]

Hence the sum of forces along the x axis is mathematically represented as

        $A_{(x)} - 400 = 0$

=>     $A_{(x)} = 400 \ N$

Looking at  the diagram uploaded we see that there are two  forces acting along the y-axis on the fixed support  

   These force are  500 N  and  $A_{(y)}$ [ i.e the force acting along the same direction with 500 N   ]

Hence the sum of forces along the x axis is mathematically represented as

        $A_{(y)} + 500 = 0$

=>     $A_{(y)} = -500 \ N$

Looking at  the diagram uploaded we see that there are two  forces acting along the z-axis on the fixed support  

       These force are  600 N  and  $A_{(z)}$ [ i.e the reactive force of  600 N  ]

Hence the sum of forces along the x axis is mathematically represented as

        $A_{(z)} - 600 = 0$

=>     $A_{(z)} = 600 \ N$

Generally taking moment about A along the x-axis we have that

    $\sum M_x = M_x - 500 (0.75 + 0.5) + 600 ( 1 ) = 0$

=>   $M_x = 1225 \ N\cdot m$

Generally taking moment about A along the y-axis we have that

    $\sum M_y = M_y - 400 (0.75 ) + 600 ( 0.75 ) = 0$

=>   $M_y = 750 \ N\cdot m$

Generally taking moment about A along the z-axis we have that

    $\sum M_z = M_z = 0$

=>   $M_z = 0 \ N\cdot m$