# how do we solve this question which is related to statics(parallel forces)

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.

-
This belongs on physics.stackexchange.com –  Ross Millikan Nov 16 '12 at 4:55

The pegs need to supply the upward 2N so the resultant is 0. You can use that in the calculation. We need to balance the torques around each peg. The torque around the one at the end from the external torque is 2N*60cm=120N-cm. To balance this, we need an upward force on the other of $\frac {120 N-cm}{6 cm}=20 N$. Then to balance the force we need a downward force on the one on the end of $20N-2N=18N$