newtonian mechanics - Why there is an extra componet of force in this Equilibrium state?

I have a question about the forces on a body at equilibrium. Look to this image:

here i draw a man standing with a pen on earth. And I think that I draw all possible forces on it. They are as follows:

1) Force $F_1$ due to the weight of pen.

2) Force $F_2$ acting on earth by pen accourding to newton's third law.

3) Force $F_3$ due to friction between hand and pen to hold it.

4) Force $F_4$ equal and opposite to $F_3$ acting on man.

(Note that magnitude of $F_1=F_2=F_3=F_4$ ).

5) weight of man $W$ acting downward.

6) Since $F_4$ acting on the man downward, the normal reaction is $W+F_4$ acting upward.

I think that I mention every forces here. Due to the fact that the body in equilibrium, all forces must cancel each other. But there comes an extra component equal to the weight of pen. Where is my mistake?

Answer

Due to the fact that the body in equilibrium, all forces must cancel each other.

Which forces must cancel each other when a body is in equilibrium?

When a body is in equilibrium, resultant of forces acting on it must be zero. In current question, we have three bodies those are in equilibrium (man, pen and the earth). So, net force exerted on each of them must be zero.

For this purpose, you should draw free body diagram for each body. For example, free body diagram for man is like figure below: (Let's assume that man has held the pen on his palm because if you want to consider to friction, you should consider to the normal forces between hand and pen also) The forces are as follows:

1. 
   
   

   $F_1$ due to the weight of the man.

   
   


2. 
   

   $F_2$ due to the normal reaction of pen.

   
   


3. 
   

   $F_3$ due to the normal reaction of the earth.