Question

20. 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? (Assume that Archer exerts only vertical—up or down—forces on the horizontal pole and that the center of gravity of the pole is located at the center of its length.)

Answer 1

Answer:

F1=40.9 N

F2=65.4 N

Explanation:

the diagram is shown in the picture. The torque of forces in A=0

$-F_{1} DA-2.5gBA+F_{2}CA=0\\ -4.9F_{1} +4F_{2} =2.5*9.8\\ -4.9F_{1} +4F_{2} =61.25$

Along vertical is:

$F_{1} -F_{2} +2.5g=0\\F_{1} -F_{2}=-24.5$

solving F1 in the first equation and replacing that value in the second equation we can calculate the values ​​of F1 and F2:

F1=40.9 N

F2=65.4 N