/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A seesaw weighing \(45 \mathrm{N... [FREE SOLUTION] | 91Ó°ÊÓ

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A seesaw weighing \(45 \mathrm{N} / \mathrm{m}\) of length is occupied by two children, each weighing \(400 \mathrm{N}\) (see figure). The center of gravity of each child is \(2.5 \mathrm{m}\) from the fulcrum. The board is \(6 \mathrm{m}\) long, \(200 \mathrm{mm}\) wide, and \(40 \mathrm{mm}\) thick. What is the maximum bending stress in the board?

Short Answer

Expert verified
The maximum bending stress is 375.14 kPa.

Step by step solution

01

Determine the Dimensions of the Seesaw

The seesaw is given to be 6 meters long with a width of 200 mm (0.2 m) and a thickness of 40 mm (0.04 m). These dimensions are crucial because they will be used to calculate the moment of inertia.
02

Calculate the Reactions at the Fulcrum

Given that each child weights 400 N and is positioned 2.5 m from the fulcrum, the total downward force due to the children's weight on each side is 400 N \(\times\) 2 = 800 N. The reaction at the fulcrum would counterbalance the total weight on each side, so the reaction force at the fulcrum is 800 N.
03

Find the Moment of Inertia

The moment of inertia for a rectangular cross-section is given by the formula \(I = \frac{1}{12}bw^3\), where \(b\) is the width and \(w\) is the thickness. For this seesaw, \(b = 0.2\) m and \(w = 0.04\) m. Therefore, \(I = \frac{1}{12}(0.2)(0.04)^3 = 1.0667 \times 10^{-7} \text{ m}^4\).
04

Calculate the Bending Moment

The bending moment at the center (or fulcrum) is critical here. For each child, the force of 400 N spans 2.5 meters. Hence, the bending moment for one child at the center is \(400 \times 2.5 = 1000 \text{ Nm}\). Since there are two children on opposite sides, the total bending moment is 2000 Nm.
05

Compute the Maximum Bending Stress

Bending stress \(\sigma\) is calculated using the bending equation \(\sigma = \frac{M \cdot c}{I}\), where \(M\) is the bending moment, \(c\) is the distance from the neutral axis to the outermost fiber (half of the thickness, which is \(0.02 \text{ m}\) in this case), and \(I\) is the moment of inertia. Substituting these values, \(\sigma = \frac{2000 \cdot 0.02}{1.0667 \times 10^{-7}} = 375140 \, \text{N/m}^2\).
06

Conclusion

The maximum bending stress experienced by the seesaw board is \(375140 \text{ N/m}^2\) or 375.14 kPa.

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Key Concepts

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

Moment of Inertia
The moment of inertia is an important concept in the context of bending stress and seesaw mechanics. It measures an object's resistance to changes in its rotation. Think about it as how difficult it is to start or stop spinning an object.
For flat shapes like the board of a seesaw, moment of inertia is related to its dimensions. It's given by the formula \( I = \frac{1}{12} bw^3 \), where \( b \) is the width and \( w \) is the thickness of the board. These measurements tell us how mass is distributed, which affects how the board reacts under force. In our exercise, the seesaw board has a width of 0.2 meters and a thickness of 0.04 meters. These values result in a calculated moment of inertia of \( 1.0667 \times 10^{-7} \, \text{m}^4 \).
  • This value helps understand how the board will bend.
  • A higher moment of inertia implies the board is more resistant to bending.
  • Seesaw mechanics hinge on this concept, making it crucial for designing safe play equipment.
Bending Moment
The bending moment is another essential concept when understanding seesaw mechanics and the bending stress experienced by the board. It refers to the tendency of a force to rotate an object around an axis, in this case, the seesaw's fulcrum.
Bending moment can be calculated with the formula \( M = F \cdot d \), where \( F \) is the force and \( d \) is the perpendicular distance from the fulcrum to where the force is applied.
In our scenario, each child exerts a force of 400 N situated 2.5 meters from the fulcrum, creating a moment of \( 1000 \, \text{Nm} \) per child. With two children, the total bending moment becomes \( 2000 \, \text{Nm} \). This tells us how much rotational force the seesaw board experiences.
  • Bending moment gives insight into potential rotation points.
  • It allows for calculating the stress the board can handle before bending.
  • In seesaw designs, understanding this helps balance and ensure structural integrity.
Stress Analysis
Stress analysis is vital for assessing the strength and durability of structures like a seesaw board. It relates to how internal forces develop within a material to resist external forces.
The core formula for bending stress is \( \sigma = \frac{M \cdot c}{I} \), where \( \sigma \) is the stress, \( M \) is the bending moment, \( c \) is the distance from the neutral axis to the outermost surface (half the board's thickness in this context), and \( I \) is the moment of inertia.
In the seesaw example, the maximum bending stress reaches \( 375140 \, \text{N/m}^2 \) or 375.14 kPa. This calculation is paramount in determining the safety and performance limits of the board.
  • Helps predict points of failure under load.
  • Provides data to ensure materials withstand expected forces.
  • Key in verifying structural components in engineering and safety standards.
Seesaw Mechanics
Understanding seesaw mechanics is crucial for solving problems involving balance and motion. A seesaw functions by balancing forces at its fulcrum, a critical pivot point.
In the context of our exercise, seesaw mechanics involve calculating how forces from the children's weights and the board's weight distribute. This helps maintain balance and predicts stresses at the fulcrum.
  • The board's design, including dimensions and material, impacts its performance.
  • These mechanics ensure the seesaw balances evenly, allowing for safe play.
  • Professionals use these calculations to design seesaws that neither flip over nor crack under pressure.
When studying seesaw mechanics, always consider the force equilibrium and the moment equilibrium to ensure the seesaw functions effectively and safely.

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Most popular questions from this chapter

A wood beam \(A B C\) with simple supports at \(A\) and \(B\) and an overhang \(B C\) has height \(h=300 \mathrm{mm}\) (see figure). The length of the main span of the beam is \(L=3.6 \mathrm{m}\) and the length of the overhang is \(L / 3=1.2 \mathrm{m}\). The beam supports a concentrated load \(3 P=18 \mathrm{kN}\) at the midpoint of the main span and a moment \(P L / 2=10.8 \mathrm{kN} \cdot \mathrm{m}\) at the free end of the overhang. The wood has weight density \(\gamma=5.5 \mathrm{kN} / \mathrm{m}^{3}\) (a) Determine the required width \(b\) of the beam based upon an allowable bending stress of \(8.2 \mathrm{MPa}\) (b) Determine the required width based upon an allowable shear stress of \(0.7 \mathrm{MPa}\)

A two-axle carriage that is part of an overhead traveling crane in a testing laboratory moves slowly across a simple beam \(A B\) (see figure). The load transmitted to the beam from the front axle is \(9 \mathrm{kN}\) and from the rear axle is \(18 \mathrm{kN} .\) The weight of the beam itself may be disregarded. (a) Determine the minimum required section modulus \(S\) for the beam if the allowable bending stress is \(110 \mathrm{MPa}\) the length of the beam is \(5 \mathrm{m},\) and the wheelbase of the carriage is \(1.5 \mathrm{m}\) (b) Select the most economical I-beam (IPN shape) from Table E-2, Appendix E.

The wood joists supporting a plank floor (see figure) \(\operatorname{arc} 38 \mathrm{mm} \times 220 \mathrm{mm}\) in cross scction (actual dimensions) and have a span length of \(L=4.0 \mathrm{m}\). The floor load is \(5.0 \mathrm{kPa},\) which includes the weight of the joists and the floor. (a) Calculate the maximum permissible spacing \(s\) of the joists if the allowable bending stress is 14 MPa. (Assume that each joist may be represented as a simple beam carrying a uniform load.) (b) If spacing \(s=406 \mathrm{mm},\) what is the required depth \(h\) of the joist? Assume all other variables remain unchanged.

A cantilever beam \(A B\) with a rectangular cross section has a longitudinal hole drilled throughout its length (see figure). The beam supports a load \(P=600 \mathrm{N}\). The cross section is \(25 \mathrm{mm}\) wide and \(50 \mathrm{mm}\) high, and the hole has a diameter of \(10 \mathrm{mm}\) Find the bending stresses at the top of the beam, at the top of the hole, and at the bottom of the beam.

A square wood platform, \(2.4 \mathrm{m} \times 2.4 \mathrm{m}\) in area rests on masonry walls (see figure). The deck of the platform is constructed of \(50 \mathrm{mm}\) nominal thickness tongue-andgroove planks (actual thickness \(47 \mathrm{mm}\); see Appendix \(\mathrm{F}\) ) supported on two \(2.4-\mathrm{m}\) long beams. The beams have \(100 \mathrm{mm} \times 150 \mathrm{mm}\) nominal dimensions (actual dimensions \(97 \mathrm{mm} \times 147 \mathrm{mm}\) ). The planks are designed to support a uniformly distributed load \(w\left(\mathrm{kN} / \mathrm{m}^{2}\right)\) acting over the entire top surface of the platform. The allowable bending stress for the planks is \(17 \mathrm{MPa}\) and the allowable shear stress is 0.7 MPa. When analyzing the planks, disregard their weights and assume that their reactions are uniformly distributed over the top surfaces of the supporting beams. \(\begin{array}{lllll}\text { (a) } & \text { Determine } & \text { the } & \text { allowable } & \text { platform } & \text { load }\end{array}\) \(w_{1}\left(\mathrm{kN} / \mathrm{m}^{2}\right)\) based upon the bending stress in the planks. (b) Determine the allowable platform load \(w_{2}\left(\mathrm{kN} / \mathrm{m}^{2}\right)\) based upon the shear stress in the planks. (c) Which of the preceding values becomes the allowable load \(w_{\text {allow }}\) on the platform? (Hints: Use care in constructing the loading diagram for the planks, noting especially that the reactions are distributed loads instead of concentrated loads. Also, note that the maximum shear forces occur at the inside faces of the supporting beams.)

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