Sterling Archer, despite failing repeatedly at pole-vaulting, is determined to master the skill. He is holding a vaulting pole parallel to the ground. The pole is 5 m long. Archer grips the pole with his right hand 10 cm from the top end of the pole and with his left hand 1 m from the top end of the pole. Although the pole is quite light (its mass is only 2.5 kg), the forces that Archer must exert on the pole to maintain it in this position are quite large. How large are they?

Let g = 10m/s2. The pole gravity force would be [tex]F_p = 5*10 = 50 N[/tex]. Suppose the pole mass distribution is uniform, this means the center of mass is at its geometric center, aka 5/2 = 2.5m from the top end.

In order for the system to stay balanced, the total net force and net moment must be 0

[tex]F_L + F_R = F_p = 50 N[/tex] where [tex]F_L, F_R[/tex] are the forces exerted by his left and right hand, respectively.

We will pick his right hand (0.1 m from the top end) as moment pivot point. The total moment generated by pole gravity and his left hand must be 0

Distance (or moment arm length) from pole center to his right hand is [tex]R_p = 2.5 – 0.1 = 2.4 m[/tex]

Moment generated by gravity around his right hand is: [tex]M_p = F_pR_p = 50*2.4 = 120 Nm[/tex] counter-clockwise

Distance (or moment arm length) from left hand to his right hand is [tex]R_L = 1 – 0.1 = 0.9 m[/tex]

Moment generated by force from his left hand around his right hand is: [tex]M_L = F_LR_L = 1.9F_L Nm[/tex] clockwise

Since the total moment must be 0, the clockwised-direction moment = counterclockwised-direction moment:

[tex]0.9F_L = 120[/tex]

[tex]F_L = 120 / 0.9 = 133 N[/tex] upward

Therefore, the force exerted by his right hand must be 50 – 133 = -83 N or 83N downward