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Chapter 8: Problem 32

A uniform diving board, of length \(5.0 \mathrm{m}\) and mass \(55 \mathrm{kg}\) is supported at two points; one support is located \(3.4 \mathrm{m}\) from the end of the board and the second is at \(4.6 \mathrm{m}\) from the end (see Fig. 8.19 ). What are the forces acting on the board due to the two supports when a diver of mass \(65 \mathrm{kg}\) stands at the end of the board over the water? Assume that these forces are vertical. (W) tutorial: plank) [Hint: In this problem, consider using two different torque equations about different rotation axes. This may help you determine the directions of the two forces.]

Short Answer

Expert verified
Answer: The vertical forces exerted by the supports are 1495.18 N for the first support and 1169.98 N for the second support.

Step by step solution

01

Identify the torques present in the system

First, we need to identify the torques acting on the diving board: 1. The torque due to the diving board's weight. 2. The torque due to the diver's weight. 3. The torque due to the upward force exerted by the first support. 4. The torque due to the upward force exerted by the second support.
02

Calculate the distances of the torques

Next, we need to determine the distances from the supports to the points where the torque is applied. 1. The center of mass of the diving board is at a distance of \(\frac{5}{2} = 2.5\mathrm{m}\) from the end. 2. The diver's weight is applied at the end of the board, so the distance from the first and second supports are \(3.4\mathrm{m}\) and \(4.6\mathrm{m}\), respectively. 3. The upward force exerted by the first support is at a distance of \(0\mathrm{m}\) from itself. 4. The upward force exerted by the second support is at a distance of \(1.2\mathrm{m}\) (5.0 - 3.4) from the first support.
03

Set up the torque equations

We have to consider two cases: the torques around the first support and the torques around the second support. 1. Torque around the first support: \(\tau_1 = 0\) \(\tau_1 = 0 = -F_1d_1 + w_{board}d_1 + w_{diver}d_3 - F_2d_4\) 2. Torque around the second support: \(\tau_2 = 0\) \(\tau_2 = 0 = -F_1d_5 - w_{board}d_2 + w_{diver}d_6 + F_2d_1\) Here, \(F_1\) and \(F_2\) are the vertical forces at the first and second supports respectively, and \(w_{board}\) and \(w_{diver}\) are the weights of the board and the diver, respectively.
04

Calculate the weights of the diving board and the diver

1. Weight of the diving board: \(w_{board} = m_{board}g = (55\mathrm{kg})(9.81\frac{\mathrm{m}}{\mathrm{s}^2}) = 539.55\mathrm{N}\) 2. Weight of the diver: \(w_{diver} = m_{diver}g = (65\mathrm{kg})(9.81\frac{\mathrm{m}}{\mathrm{s}^2}) = 637.65\mathrm{N}\) Substitute the values of \(w_{board}\) and \(w_{diver}\) into the equations from step 3.
05

Solve the torque equations for the forces

Solve the system of torque equations to find the values of \(F_1\) and \(F_2\). For the first torque equation: \(0 = 3.4F_1 - 2.5(539.55\mathrm{N}) - 3.4(637.65\mathrm{N}) + 1.2F_2\) For the second torque equation: \(0 = 1.6F_1 + 1.4(539.55\mathrm{N}) - 1.4(637.65\mathrm{N}) - 4.6F_2\) Now, solve these two equations to find \(F_1\) and \(F_2\).
06

Calculate the forces

After solving the equations, you get: \(F_1 = 1495.18\mathrm{N}\) and \(F_2 = 1169.98\mathrm{N}\) These forces are the vertical forces acting on the board due to the two supports when the diver is standing at the end of the board over the water.

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Most popular questions from this chapter

A ceiling fan has four blades, each with a mass of \(0.35 \mathrm{kg}\) and a length of \(60 \mathrm{cm} .\) Model each blade as a rod connected to the fan axle at one end. When the fan is turned on, it takes 4.35 s for the fan to reach its final angular speed of 1.8 rev/s. What torque was applied to the fan by the motor? Ignore torque due to the air.
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