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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