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Chapter 9: Problem 13

A hiker, who weighs 985 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs \(3610 \mathrm{N}\), and rests on two concrete supports, one at each end. He stops one-fifth of the way along the bridge. What is the magnitude of the force that a concrete support exerts on the bridge (a) at the near end and (b) at the far end?

Short Answer

Expert verified
Near end force is 2593 N; far end force is 2002 N.

Step by step solution

01

Understand the problem

We need to find the reaction forces at both ends of the bridge when a hiker is standing on it. The bridge itself is uniformly distributed in terms of weight and has two supports at its ends. The hiker stops one-fifth along the bridge, and we need to calculate the forces at the near end and the far end.
02

Establish the system

Let's denote the length of the bridge as \(L\). The weight of the bridge \(W_b = 3610 \mathrm{N}\), and the weight of the hiker \(W_h = 985 \mathrm{N}\). The walker stops one-fifth along \(L\), so the distance the walker is from the near end is \(\frac{L}{5}\). Let \(R_1\) and \(R_2\) be the reactions at the near and far end supports, respectively.
03

Set up the equilibrium conditions

For the system to be in equilibrium, the sum of vertical forces should be zero and the sum of moments about any point should be zero.**Vertical Force Balance:** \[R_1 + R_2 = 3610 + 985\]**Moment Balance about the near end:** The moments produced by the hiker and the weight of the bridge should balance with the reaction at the far end.\[R_2 \cdot L = 985 \cdot \frac{L}{5} + 3610 \cdot \frac{L}{2}\]
04

Solve for the reaction forces

We'll first solve the moment balance equation for \(R_2\):\[R_2 \cdot L = 985 \cdot \frac{L}{5} + 3610 \cdot \frac{L}{2}\]Solving, we find\[R_2 = \frac{985}{5} + \frac{3610}{2} = 197 + 1805 = 2002 \mathrm{N}\]Now substitute \(R_2\) back into the vertical force balance to find \(R_1\):\[R_1 + 2002 = 4595\]\[R_1 = 4595 - 2002 = 2593 \mathrm{N}\]
05

Conclusion

The magnitude of the force that the concrete support exerts on the bridge at the near end is \(2593 \mathrm{N}\), and at the far end is \(2002 \mathrm{N}\).

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Key Concepts

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

Reaction Forces
Reaction forces occur at the points where the bridge is supported by concrete on each end. These forces are responsible for keeping the bridge stable as it supports the weight of the hiker. To understand reaction forces, it's essential to remember that:
  • They act in response to applied loads, such as the weight of the bridge and the hiker's weight.
  • Their direction is vertical, counteracting the downward force of gravity.
  • They ensure the entire system remains balanced.
In this problem, we denote these forces as \( R_1 \) and \( R_2 \), with \( R_1 \) being the reaction at the near end and \( R_2 \) at the far end of the bridge. Calculating reaction forces helps us to understand how support structures work in real-world applications.
Torque
Torque is crucial when analyzing the equilibrium of the bridge. It鈥檚 a measure of how much a force causes an object to rotate around an axis. In simple terms:
  • Torque depends on the force applied and the distance from the axis of rotation (moment arm).
  • If the force is applied further from the pivot point, the torque increases.
  • Torque can cause an object to rotate clockwise or counterclockwise.
For the bridge, the torque is computed about the near end to determine the force reaction \( R_2 \). This involves analyzing the torque contributed by the hiker \( 985 \mathrm{N} \times \frac{L}{5} \) and the bridge's weight \( 3610 \mathrm{N} \times \frac{L}{2} \), balancing it with \( R_2 \times L \). This balance ensures the bridge does not rotate, maintaining equilibrium.
Uniform Distribution
Uniform distribution refers to the even spread of the bridge鈥檚 weight along its entire length. Understanding this helps to:
  • Simplify calculations in equilibrium problems because weight distribution affects reaction forces.
  • Ensure that structural integrity is maintained under consistent loading conditions.
  • Calculate the center of mass, which for the bridge is at its midpoint.
For the bridge weighing \( 3610 \mathrm{N} \), uniform distribution means that its weight can be thought of as acting at its midpoint \( \frac{L}{2} \). This helps to balance the system by understanding how weight is distributed and ultimately affects the reaction forces at both ends.
Moment Balance
Moment balance involves ensuring that the sum of moments (torques) around any pivot point is zero. This concept is vital in ensuring the system remains in static equilibrium. Here鈥檚 how it works:
  • By choosing a pivot point, you can isolate forces acting at other points to solve for unknowns.
  • The sum of moments about any point must equal zero for the system to be balanced.
  • It involves considering all force contributors, including weights and distances from the pivot.
In the exercise, moments are balanced around the near end to solve for \( R_2 \). This step is essential to ensure no net rotation and properly counteracted forces, maintaining the force equilibrium across the bridge as the hiker stands on it.

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Most popular questions from this chapter

A 15.0-m length of hose is wound around a reel, which is initially at rest. The moment of inertia of the reel is \(0.44 \mathrm{kg} \cdot \mathrm{m}^{2}\), and its radius is \(0.160 \mathrm{m} .\) When the reel is turning. friction at the axle exerts a torque of magnitude \(3.40 \mathrm{N} \cdot \mathrm{m}\) on the reel. If the hose is pulled so that the tension in it remains a constant \(25.0 \mathrm{N}\), how long does it take to completely unwind the hose from the reel? Neglect the mass and thickness of the hose on the reel, and assume that the hose unwinds without slipping.

Three objects lie in the \(x, y\) plane. Each rotates about the \(z\) axis with an angular speed of 6.00 rad/s. The mass \(m\) of each object and its perpendicular distance \(r\) from the \(z\) axis are as follows: (1) \(m_{1}=6.00 \mathrm{kg}\) and \(r_{1}=2.00 \mathrm{m},(2) m_{2}=4.00 \mathrm{kg}\) and \(r_{2}=1.50 \mathrm{m}\) (3) \(m_{3}=3.00 \mathrm{kg}\) and \(r_{3}=3.00 \mathrm{m} .\) (a) Find the tangential speed of each object. (b) Determine the total kinetic energy of this system using the expression \(\mathrm{KE}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}+\frac{1}{2} m_{3} v_{3}^{2}\) (c) Obtain the moment of inertia of the system. (d) Find the rotational kinetic energy of the system using the relation \(\mathrm{KE}_{4}=\frac{1}{2} \mathrm{I\omega}^{2}\) to verify that the answer is the same as the answer to (b).

A thin rod has a length of \(0.25 \mathrm{m}\) and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of \(0.32 \mathrm{rad} / \mathrm{s}\) and a moment of inertia of \(1.1 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}\), A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (mass \(=4.2 \times 10^{-3} \mathrm{kg}\) ) gets where it's going, what is the angular velocity of the rod?

A clay vase on a potter's wheel expericnces an angular acceleration of \(8.00 \mathrm{rad} / \mathrm{s}^{2}\) due to the application of a \(10.0-\mathrm{N} \cdot \mathrm{m}\) net torque. Find the total moment of inertia of the vase and potter's wheel.

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 300 -mile trip in a typical midsize car produces about \(1.2 \times 10^{9}\) J of energy. How fast would a \(13-k g\) flywhcel with a radius of \(0.30 \mathrm{m}\) have to rotate to store this much energy? Give your answer in rev/min.
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