Question

You need to repair a gate on the farm. The gate weighs 100 kg and pivots as indicated. A small diagonal bar supports the gate and keeps it from falling downward. What force must this diagonal member be able to provide in order to keep the gate from falling

Answer 1

Answer:

The force is  $F = 3920 \ N$

Explanation:

The diagram for this question is shown on the first uploaded image

   From the question we are told that

          The weight of the gate is $G = 100\ kg$

 

The vertical component of F is  $F_y = F\ sin \theta$

   From the diagram , taking moment about the pivot we have  

                $W_g * 2 - F_y * 1 = 0$

Where $W_g$  is the weight of the gate evaluated as

             $W_g = m_g * g = 100 * 9.8 = 980 \ N$

=>        $980 * 2 - Fsin(30) * 1 = 0$

=>         $F = \frac{1960}{sin(30)}$

=>      $F = 3920 \ N$