Question

Joe, a pole-vaulter, is holding a vaulting pole parallel to the ground. The pole is 5 m long. Joe 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 light (its mass is only 2.5 kg), the forces that Joe must exert on the pole to maintain it in this position are large. How large are they? (Assume that Joe 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

Joe exerts a larger force on the pole when he holds it with his right hand closer to the center of gravity (cg) of the pole and his left hand farther away. The total force exerted by Joe is equal to the weight of the pole, which can be calculated using the equation: Total force = weight of the pole = mass of the pole x acceleration due to gravity. Therefore, Joe exerts a total force of 24.5 N on the pole.

Step-by-step explanation:

Joe exerts a larger force on the pole when he holds it with his right hand closer to the center of gravity (cg) of the pole and his left hand farther away.

This is because the force exerted on the pole is larger when it is farther away from the cg. In this case, the right hand exerts a downward force to counterbalance the weight of the pole, while the left hand exerts an upward force.

The total force exerted by Joe is equal to the weight of the pole, which can be calculated using the equation:

Total force = weight of the pole = mass of the pole x acceleration due to gravity

Using the given information that the mass of the pole is 2.5 kg and the acceleration due to gravity is 9.8 m/s^2, we can calculate the weight of the pole:

Weight of the pole = 2.5 kg x 9.8 m/s^2 = 24.5 N

Therefore, Joe exerts a total force of 24.5 N on the pole.