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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 man is doing push-ups. He has a mass of \(68 \mathrm{kg}\) and his center of gravity is located at a horizontal distance of \(0.70 \mathrm{m}\) from his palms and \(1.00 \mathrm{m}\) from his feet. Find the forces exerted by the floor on his palms and feet.
A person is doing leg lifts with 3.0 -kg ankle weights. She is sitting in a chair with her legs bent at a right angle initially. The quadriceps muscles are attached to the patella via a tendon; the patella is connected to the tibia by the patella tendon, which attaches to bone \(10.0 \mathrm{cm}\) below the knee joint. Assume that the tendon pulls at an angle of \(20.0^{\circ}\) with respect to the lower leg, regardless of the position of the lower leg. The lower leg has a mass of \(5.0 \mathrm{kg}\) and its center of gravity is $22 \mathrm{cm}\( below the knee. The ankle weight is \)41 \mathrm{cm}$ from the knee. If the person lifts one leg, find the force exerted by the patella tendon to hold the leg at an angle of (a) \(30.0^{\circ}\) and (b) \(90.0^{\circ}\) with respect to the vertical.
A Ferris wheel rotates because a motor exerts a torque on the wheel. The radius of the London Eye, a huge observation wheel on the banks of the Thames, is \(67.5 \mathrm{m}\) and its mass is \(1.90 \times 10^{6} \mathrm{kg} .\) The cruising angular speed of the wheel is $3.50 \times 10^{-3} \mathrm{rad} / \mathrm{s} .$ (a) How much work does the motor need to do to bring the stationary wheel up to cruising speed? [Hint: Treat the wheel as a hoop.] (b) What is the torque (assumed constant) the motor needs to provide to the wheel if it takes 20.0 s to reach the cruising angular speed?
A hollow cylinder, a uniform solid sphere, and a uniform solid cylinder all have the same mass \(m .\) The three objects are rolling on a horizontal surface with identical transnational speeds \(v .\) Find their total kinetic energies in terms of \(m\) and \(v\) and order them from smallest to largest.
Find the force exerted by the biceps muscle in holding a 1-L milk carton (weight \(9.9 \mathrm{N}\) ) with the forearm parallel to the floor. Assume that the hand is \(35.0 \mathrm{cm}\) from the elbow and that the upper arm is $30.0 \mathrm{cm}$ long. The elbow is bent at a right angle and one tendon of the biceps is attached to the forearm at a position \(5.00 \mathrm{cm}\) from the elbow, while the other tendon is attached at \(30.0 \mathrm{cm}\) from the elbow. The weight of the forearm and empty hand is \(18.0 \mathrm{N}\) and the center of gravity of the forearm is at a distance of \(16.5 \mathrm{cm}\) from the elbow.
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