You're carrying a 4.0-m-long, 25 {\rm kg} pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 {\rm cm} from its tip.How much force must you exert to keep the pole motionless in a horizontal position?

The point on the fence where one end of the pole is placed is taken to be the pivot about which the moments due to the weight of the pole and the applied force could be taken.

We assume that the pole is uniform such that the weight w acts exactly at the centre. i.e 2m from either end.

we apply the principle of moments which states that for any system to be in equilibrium, the clockwise moment at any point must be equal to the anti-clockwise moment at the same point.

Also it should be noted that the force applied 35cm (0.35m) from the tip should be upwards in order to counterbalance the weight which is already acting downwards.

The following equation therefore holds by principle of moments;

[tex]F*3.65=W*2.................... (1)\\[/tex]

Also; w = mg.

Given; m = 25kg and the value of g is assumed to be [tex]9.8m/s^2[/tex]

therefore; w = 25 x 9.8 =245N

Substituting into equation (1), we obtain the following;

[tex]F*3.65=245*2\\\\[/tex]

F = 490/3.6

F = 134.25N