Question

A 65 kg-mass person stands at the end of a diving board, 1.5 m from the board's pivot point. Determine the torque the person is exerting on the board with respect to the pivot point. Show your work.

Answer 1

The torque exerted by a force F at a distance r from the pivot point is given by the formula:

τ = F x r x sin(θ)

where θ is the angle between the force vector and the vector from the pivot point to the point where the force is applied.

In this case, the person's weight is the force being exerted on the board, and its magnitude is:

F = m x g = 65 kg x 9.8 m/s^2 = 637 N

The distance from the pivot point to the person is r = 1.5 m. Since the person is standing vertically, the angle between the weight vector and the vector from the pivot point to the person is 90 degrees, so sin(θ) = 1. Substituting the values into the torque formula, we get:

τ = 637 N x 1.5 m x 1 = 955.5 Nm

Therefore, the person is exerting a torque of 955.5 Nm on the diving board with respect to the pivot point.

Answer 2

Step-by-step explanation:

To calculate the torque exerted by the person on the diving board, we need to know the force exerted and the lever arm.

The force exerted by the person is the weight of their body, which can be calculated as:

F = mg

F = 65 kg x 9.81 m/s^2

F = 637.65 N

note: The acceleration of gravity "g" is therefore the result of gravitation (gravitational attraction) between the Earth and other celestial bodies, and of the centrifugal acceleration, due to the movement of the earth's rotation and its average global value is 9.81 ms -2.

The lever arm is the distance from the person to the pivot point, which is given as 1.5 m.

The torque (τ) can then be calculated as:

τ = F x d

τ = 637.65 N x 1.5 m

τ = 956.47 Nm

Therefore, the torque exerted by the person on the diving board with respect to the pivot point is 956.47 Nm.