A straight weightless rod $60$cm in length rests in a horizontal position between two pegs. The pegs are placed at a distance of $6$cm apart, one peg being at one end of the rod, and a weight of $2$N is suspended from the other end. What is the pressure on the pegs?
Well, I saw the solution of this question in a book and it was solved using resultant of unlike parallel forces.
This is how it's solved, but i cant understand it!
Let P and Q be the forces on the pegs A and C (force p is in upward direction and force q is in downward direction). $2$N is the resultant of P and Q.
thus P=$(2*53)/6 = 18$N
and Q=$(2*60)/6 = 20$N
My question is: How does the rod remain in horizontal if the resultant is $2$N? Is it that the the forces give resultant of $2$N in the opposite direction of the suspend weight and that's how it's balanced? and why is $2$ unlike force supposed in the pegs?
Can you please explain it to me clearly? Thank you.