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Chapter 11: Problem 11

A diving board 3.00 \(\mathrm{m}\) long is supported at a point 1.00 \(\mathrm{m}\) Trom the end, and a diver weighing 500 \(\mathrm{N}\) stands at the lree end (Fig. E11.11). The diving board is of uniform cross section and weighs 280 \(\mathrm{N} .\) Find (a) the force at the support point and (b) the force at the left-hand end.

Short Answer

Expert verified
(a) Force at the support point B is 1560 N. (b) Force at the left-hand end A is -780 N, which means it acts upwards.

Step by step solution

01

Understand the Problem

First, identify the forces acting on the diving board: the force from the diver (500 N), the weight of the diving board (280 N), and the reactions at the two supports, which we'll call \(F_A\) and \(F_B\). The length of the board is 3.00 m, with a support point 1.00 m from one end.
02

Establish a Coordinate System

We'll place the diving board horizontally with the left-hand end at point A and the right-hand end supporting the diver. The support is located 1.00 m from point A, at point B.
03

Calculate Torque About Point A

Sum the torques about point A. Set counterclockwise as positive.\[ \sum \tau_A = 0 = 500 \times 2.00 + 280 \times 1.50 - F_B \times 1.00 \]This equation ensures rotational equilibrium. The distances (2.00 m and 1.50 m) are the lever arms for the diver's weight and the board's weight, respectively.
04

Solve for Force at Point B

Rearrange the torque equation to find \(F_B\):\[ F_B = \frac{500 \times 2.00 + 280 \times 1.50}{1.00} = 1560 \, \text{N} \]
05

Use Equilibrium Equation for Forces

Sum vertical forces to zero for translational equilibrium:\[ \sum F_y = 0 = F_A + F_B - 500 - 280 \]Substitute \(F_B = 1560\) into the equation:\[ F_A + 1560 = 780 \]
06

Solve for Force at Point A

Rearrange to solve for \(F_A\):\[ F_A = 780 - 1560 = -780 \, \text{N} \]The negative sign indicates that the direction of \(F_A\) is opposite to our assumed direction. Hence, it acts upwards.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium in Physics
Equilibrium in physics refers to a state where an object is either at rest or moving with constant velocity because the forces acting on it are balanced. In simpler terms, when all the forces cancel out, there is no change in motion. For objects in equilibrium, this means both translational and rotational motions are in check.
  • **Translational Equilibrium**: This occurs when the sum of forces acting on an object is zero, ensuring no linear acceleration.
  • **Rotational Equilibrium**: This takes place when all torques acting on the object sum up to zero, preventing any change in the object's rotational speed.
In the diving board problem, we focused on rotational equilibrium to find the forces at the support points A and B. The torques created by the diver and the board's weight needed to balance with the support forces to maintain equilibrium. This careful balance highlights the importance of equilibrium in analyzing static structures.
Understanding Rotational Dynamics
Rotational dynamics involves how forces result in rotational motion. Unlike linear dynamics, which deals with straightforward motion, rotational dynamics is concerned with how objects spin due to various torques applied.
Torque is a rotational force, calculated as the product of force and the distance from the pivot point to where the force is applied (lever arm). The equation is given by:\[ \tau = F \cdot r \cdot \sin(\theta) \]where
  • \( \tau \) is torque,
  • \( F \) is the force applied,
  • \( r \) is the lever arm distance from the pivot point, and
  • \( \theta \) is the angle between the force and lever arm direction.
In the diving board example, torques around the pivot (support) point influence rotational dynamics. We sum the torques from different sources (diver and the board's own weight) and solve to maintain rotational equilibrium. Understanding this lays the foundation for analyzing rotating systems, making rotational dynamics a key part of solving such problems.
Conducting Force Analysis
Force analysis involves identifying and computing all the forces acting on an object. This is critical in equilibrium problems to ensure accurate calculations of unknown forces. In essence, it allows us to establish conditions for translational and rotational equilibrium.
Steps to Conduct Force Analysis:
  • **Identify Forces**: Begin by listing all the forces, like gravitational force, applied forces, support reactions, etc. In our problem, these were the diver's weight, board weight, and supports.
  • **Choose a Reference Point**: Establish a coordinate system to apply force and torque equations accurately. This sets the stage for easier calculations.
  • **Apply Equilibrium Equations**: Use \( \sum F = 0 \) for translational equilibrium and \( \sum \tau = 0 \) for rotational equilibrium to calculate unknown forces.
    In the exercise, using these principles helped find forces at the support points.
By conducting force analysis, one can understand both the balance required for structures and the factors contributing to movements, serving as a basis for problem-solving in engineering and physics.

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Most popular questions from this chapter

BIO Supporting a Broken Leg. A therapist tells a 74 -kg patient with a broken leg that he must have his leg in a cast suspended horizontally. For minimum discomfort, the leg should be supported by a vertical strap attached at the center of mass of the leg-cast system. (Fig. P11.55). In order to comply with these instructions, the patient consults a table of typical mass distributions and finds that both upper legs (thighs) together typically account for 21.5\(\%\) of body weight and the center of mass of each thigh is 18.0 cm from the hip joint. The patient also reads that the two lower legs (including the feet) are 14.0\(\%\) of body weight, with a center of mass 69.0 \(\mathrm{cm}\) from the hip joint. The cast has a mass of \(5.50 \mathrm{kg},\) and its center of mass is 78.0 \(\mathrm{cm}\) from the hip joint. How far from the hip joint should the supporting strap be attached to the cast?

Bulk Modulus of an Ideal Gas. The equation of state (the equation relating pressure, volume, and temperature) for an ideal gas is \(p V=n R T,\) where \(n\) and \(R\) are constants. (a) Show that if the gas is compressed while the temperature \(T\) is held constant, the bulk modulus is equal to the pressure. (b) When an ideal gas is compressed without the transfer of any heat into or out of it, the pressure and volume are related by \(p V^{\gamma}=\) constant, where \(\gamma\) is a constant having different values for different gases. Show that, in this case, the bulk modulus is given by \(B=\gamma p\)

Flying Buttress. (a) A symmetric building has a roof sloping upward at \(35.0^{\circ}\) above the horizontal on each side. If each side of the uniform roof weighs \(10,000 \mathrm{N}\) , find the horizontal force that this roof exerts at the top of the wall, which tends to push out the walls. Which type of building would be more in danger of collapsing: one with tall walls or one with short walls? Explain. (b) As you saw in part (a), tall walls are in danger of collapsing from the weight of the roof. This problem plagued the ancient builders of large structures. A solution used in the great Gothic cathedrals during the 1200 s was the flying buttress, a stone support running between the walls and the ground that helped to hold in the walls. A Gothic church has a uniform roof weighing a total of \(20,000 \mathrm{N}\) and rising at \(40^{\circ}\) above the horizontal at each wall. The walls are A Gothic church has a uniform roof weighing a total of \(20,000 \mathrm{N}\) and rising at \(40^{\circ}\) above the horizontal at each wall. The walls are 40 \(\mathrm{m}\) tall, and a flying buttress meets each wall 10 \(\mathrm{m}\) below the base of the roof. What horizontal force must this flying buttress apply to the wall?

A metal wire 3.50 \(\mathrm{m}\) long and 0.70 \(\mathrm{mm}\) in diameter was given the following test. A load weighing 20 \(\mathrm{N}\) was originally hung from the wire to keep it taut. The position of the lower end of the wire was read on a scale as load was added (a) Graph these values, plotting the increase in length horizontally and the added load vertically. (b) Calculate the value of Young's modulus. (c) The proportional limit occurred at a scale reading of 3.34 \(\mathrm{cm} .\) What was the stress at this point?

In a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of 90.8 \(\mathrm{N}\) is applied perpendicular to each end. If the diameter of the wire is \(1.84 \mathrm{mm},\) what is the breaking stress of the alloy?
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