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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!

AB=$60$cm

AC=$6$cm

CB=$54$cm

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.

P/CB=Q/AB=$2$/AC

P/$54$=Q/$60$=$2/6$

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.

  • 2
    This belongs on physics.stackexchange.com2012-11-16

1 Answers 1