You're carrying a 4.0-mm-long, 22 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 cm from its tip.
How much force must you exert to keep the pole motionless in a horizontal position?

    F = 118.14 N

Explanation:

For this exercise we will use the rotational equilibrium condition

      ∑ τ = I α = 0

Since the pole  is in balance.

Let's fix a reference system located the point where the post rests on the fence, select the counterclockwise rotation as positive

         -F (L -0.35) + W L / 2 = 0

         F = W L / 2 (L-0.35)

         L = 4.0 m

Let's calculate

         F = 22 9.8 2 / 3.65

         F = 118.14 N

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

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