How much force must you exert to keep the pole motionless in a horizontal position?
You’re carrying a 3.0-m-long, 23kg 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.
As the pole is not turning SUM(torques) = 0 = L/2*M*g – (L – .35)F; where F is the force I’m hefting the pole. L = 3 m and M = 23 kg.
Solve for F = (LMg/2)/(L – .35) = (3*23*9.8/2)/(3 – .35) = 128 N. ANS.
How much force must you exert to keep the pole motionless in a horizontal position?