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## How much Force must you Exert?

You’re carrying a 3.8-m-long, 23 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?

Given Variables:

Length: 3.8 m

Mass (m): 23 kg

r1= 35 cm

First:
The force exerted by you and the force exerted by the fence would equal to the weight of thepole.

Force exerted by you = F1

Force exerted by Fence = F2

Weight of Pole = mass x gravity(9.8m/s2)
F1 +F2 = mg

= 23kg x 9.8 m/s2 = 225.4 N

equation derived:

F1 +F2 = 225.4 N

*all the torques sum to zero

To find the center of mass, divide the length of the pole

3.8/2= 1.9m

Find the distance from  F1 to the center of mass i23s

1.9m-0.35 m = 1.55m

the torque equation is:

1.55F1 = 1.9 F2

solve for F2

F2 = 0.815 F1

plug into the first equation: F1 +F2 = 225.4 N

F1+0.82 F1 = 225.4N

Treat F1 as variables which they are

(1)F1+(0.82 )F1 = 225.4N

1.82F1= 225.4N

F1=225.4/1.83N

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