Final answer:
To calculate the upward pull by a pole vaulter on a pole, apply the principle of moments using the weight of the pole, the distance between the hands, and the position of the center of gravity relative to the hands. Set the torques exerted by each hand around a pivot point equal, and solve for the unknown force.
Step-by-step explanation:
The student is asking about the calculation of forces exerted by the hands of a pole vaulter on a pole, which involves the concepts of static equilibrium and torques. Assuming that the pole vaulter's left hand is exerting an upward force and the right hand is exerting a downward force, we can use the information provided to find these forces. Given that the total weight of the pole is 26.4 N and the distance between the hands is 0.771 m (77.1 cm), the location of the center of gravity (cg) is crucial for the calculation.
To find the upward pull by the vaulter on the pole (FL) with the cg at a known distance from the vaulter's left hand (in this case, 2.00 m from the left hand as per Figure 9.22), we can apply the principle of moments, where the sum of the clockwise torques equals the sum of the counter-clockwise torques around a pivot point. The forces can be calculated as follows:
- Let FL represent the upward force (left hand) and FR the downward force (right hand).
- The torque produced by the left hand about the right hand: FL × distance between hands = FL × 0.771 m.
- The torque produced by the weight of the pole about the right hand: 26.4 N × (2.00 m - 0.771 m).
- Set the clockwise torque equal to the counter-clockwise torque: FL × 0.771 m = 26.4 N × (2.00 m - 0.771 m).
- Solve for FL.=42.082
Through this process, the student will be able to determine the upward pull by the pole vaulter's left hand on the pole.