Question

A \(2.00-\mathrm{m}\) -long diving board of mass \(12.0 \mathrm{~kg}\) is \(3.00 \mathrm{~m}\) above the water. It has two attachments holding it in place. One is located at the very back end of the board, and the other is \(25.0 \mathrm{~cm}\) away from that end. a) Assuming that the board has uniform density, find the forces acting on each attachment (take the downward direction to be positive). b) If a diver of mass \(65.0 \mathrm{~kg}\) is standing on the front end, what are the forces acting on the two attachments?

Short answer

Answer: a) Without the diver, the forces at attachment A and B are 353.16 N and 470.88 N, respectively, acting upward. b) With the diver, the forces at attachment A and B are 2266.71 N and 3022.28 N, respectively, acting upward.

Step-by-step solution

Calculate the gravitational force on the diving board

To calculate the gravitational force acting on the diving board, we use the formula: \(F_g = mg\). Where m is the mass (12.0 kg) and g is the acceleration due to gravity (9.81 m/s²). \(F_g = (12.0 \mathrm{~kg})(9.81 \mathrm{m/s²}) = 117.72 \mathrm{N}\) The gravitational force acting on the diving board is 117.72 N downward.

Locate the center of mass of the diving board

Since the diving board has a uniform density, its center of mass is at the center of the board. The length of the diving board is 2.00 m, so the center of mass is located at 1.00 m from the back end of the board.

Calculate the torques due to the gravitational force and reaction forces

We will calculate the torque with respect to the attachment at the back end of the diving board (let's call it A). The torque due to the gravitational force is \(Torque_g = \vec{F_g} \times \vec{d_g}\), with \(d_g = 1.00 \mathrm{m}\). \(Torque_g = (117.72 \mathrm{N})(1.00 \mathrm{m}) = 117.72 \mathrm{Nm}\) Let \(F_a\) be the reaction force at attachment A and \(F_b\) be the reaction force at the second attachment, located at 0.25 m from A. Then, \(Torque_a = \vec{F_a} \times \vec{d_a} = 0\), as \(d_a = 0\). \(Torque_b = \vec{F_b} \times \vec{d_b} = (F_b)(0.25 \mathrm{m})\) For equilibrium, the sum of torques is 0: \(Torque_g + Torque_a + Torque_b = 0\)

Calculate the forces at the attachments without the diver

By substituting the values from step 3: \(117.72 \mathrm{Nm} + 0 + (F_b)(0.25 \mathrm{m}) = 0\) Solving for \(F_b\): \(F_b = \frac{-117.72 \mathrm{Nm}}{0.25 \mathrm{m}} = -470.88 \mathrm{N}\) The negative sign indicates that the force is upwards. Now, we can calculate the force at attachment A using the equilibrium condition for forces: \(F_a + F_b + F_g = 0\) \(F_a - 470.88 \mathrm{N} + 117.72 \mathrm{N} = 0\) \(F_a = 470.88 \mathrm{N} - 117.72 \mathrm{N} = 353.16 \mathrm{N}\) The forces at attachment A and B without the diver are 353.16 N and 470.88 N, respectively, acting upward.

Calculate the gravitational force on the diving board with diver

With the diver of mass 65.0 kg standing on the front end of the board, the total mass is now (12.0 + 65.0) kg, and the total gravitational force is: \(F_ {g_{total}} = (77.0 \mathrm{~kg})(9.81 \mathrm{m/s²}) = 755.57 \mathrm{N}\)

Calculate the torques due to the gravitational force and reaction forces with the diver

The torque due to the gravitational force acting on the diving board-diver system is: \(Torque_ {g_{total}} = (755.57 \mathrm{N})(1.00 \mathrm{m}) = 755.57 \mathrm{Nm}\) For equilibrium, the sum of torques is again 0: \(Torque_ {g_{total}} + Torque_a + Torque_b = 0\) \(755.57 \mathrm{Nm} + 0 + (F_b')(0.25 \mathrm{m}) = 0\)

Calculate the forces at the attachments with the diver

By substituting the values from step 6: \(F_b' = \frac{-755.57 \mathrm{Nm}}{0.25 \mathrm{m}} = -3022.28 \mathrm{N}\) The negative sign indicates that the force is upwards. Now, we can calculate the force at attachment A using the equilibrium condition for forces: \(F_a' + F_b' + F_{g_{total}} = 0\) \(F_a' - 3022.28 \mathrm{N} + 755.57 \mathrm{N} = 0\) \(F_a' = 3022.28 \mathrm{N} - 755.57 \mathrm{N} = 2266.71 \mathrm{N}\) The forces at attachment A and B with the diver are 2266.71 N and 3022.28 N, respectively, acting upward. So, the answers are: a) Forces acting on the attachments without the diver are 353.16 N and 470.88 N, acting upward. b) Forces acting on the attachments with the diver are 2266.71 N and 3022.28 N, acting upward.