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Two people are carrying a uniform wooden board that is \(3.00 \mathrm{~m}\) long and weighs \(160 \mathrm{~N}\). If one person applies an upward force equal to \(60 \mathrm{~N}\) at one end, at what point does the other person lift? Begin with a free-body diagram of the board.

Short Answer

Expert verified
The second person lifts the board at a distance of 2.4m away from the end where the first person is applying an upward force.

Step by step solution

01

Identify the forces acting on the board

There are three forces acting on the board: the force applied by the first person \( F_1 = 60 \mathrm{~N} \) at one end of the board, the force applied by the second person \( F_2 \) at the other end, and the weight of the board \( W = 160 \mathrm{~N} \) that acts in the middle of the board (1.5m from either end).
02

Set up the equation for the moments about a point

Since the board is in equilibrium, the sum of moments about any point is zero. The principle of moments states that for a body to be in equilibrium, the sum of the clockwise moments about a point should be equal to the sum of the anticlockwise moments about the same point. Choosing the end where \( F_1 \) is applied as the pivot (so \( F_1 \) contributes no moment because the distance is 0), we have \( F_1 \times 0 + W \times 1.5 = F_2 \times d \), where \( d \) is the distance from the pivot to the point \( F_2 \) is applied.
03

Solve for the unknown

Rearranging the equation to find \( d \) gives us \( d = \frac{W \times 1.5}{F_2} \). But we know \( F_2 \) is the remaining force required to balance the weight after \( F_1 \) has acted, so \( F_2 = W - F_1 = 160 \mathrm{~N} - 60 \mathrm{~N} = 100 \mathrm{~N} \). Substituting the values gives \( d = \frac{160 \mathrm{~N} \times 1.5 \mathrm{~m}}{100 \mathrm{~N}} = 2.4 \mathrm{~m} \).

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Key Concepts

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

Free-Body Diagram
Understanding the forces at play in a physical system is critical for analyzing its motion or equilibrium. This is where a free-body diagram becomes an invaluable tool. It is a simple graphical illustration used to show all the external forces acting upon an object or system. In the case of the uniform wooden board carried by two people, the diagram should clearly depict the board at rest, with the downward gravitational force (\(W = 160 \text{ N}\)) and the upward forces (\(F_1 = 60 \text{ N}\) and an unknown force \(F_2\)) applied at both ends. The free-body diagram helps to identify these forces and to visualize where they act, setting the stage for further analysis using the principles of equilibrium and moments.

To improve the student’s grasp on free-body diagrams, it’s pivotal to remind them that such diagrams represent all the active forces and should indicate the object's orientation and the direction of forces. Also, consider explicitly labeling the forces with their magnitudes and directions to eliminate any ambiguities. By mastering the creation of a free-body diagram, students take a significant step towards understanding complex physical systems.
Principle of Moments
The equilibrium of an object subjected to multiple forces is analytically explored through the principle of moments, a fundamental concept that emerges from Newton's laws of motion. According to this principle, for a system in rotational equilibrium, the sum of clockwise moments about any pivot point must equal the sum of anticlockwise moments about that same point. A moment, in physics, is the product of the force applied and the perpendicular distance from the pivot to the line of action of the force, expressed as \(M = F \times d\).

In the textbook exercise, we use the principle of moments to determine the position \(d\) where the second person must apply force \(F_2\) to maintain equilibrium. By considering the moments about the end where \(F_1\) is applied, we have only two contributing forces to balance: the board's weight and the force from the second person. By setting the sum of these moments to zero, we can solve for the unknown distance \(d\) which mathematically models the physical reality of the board's equilibrium. It's crucial to explain to students that the pivot point choice is arbitrary and that the principle will hold true regardless of the pivot chosen, which reinforces the universality and the power of this concept as an analytical tool.
Uniform Board Equilibrium
Equilibrium scenarios, especially in the context of a uniform board, present unique challenges and learning opportunities. A uniform board, by nature, has its weight evenly distributed along its length, which implies that the gravitational force—its weight—acts directly in the center. When we talk about equilibrium for such an object, we’re referring to a state where the sum of all forces and the sum of all moments (torques) are zero. This static situation means that the object is either at rest or moving with constant velocity, and there is no net rotation.

For the two-person board-carrying exercise, understanding that despite the weight being centralized, the position where the second force \(F_2\) is applied, need not be symmetrical due to the differing magnitudes of applied forces. The equilibrium is achieved not by equal distances but by equal moments. To help students improve their understanding, it’s essential to emphasize the distinction between force magnitude and the moment it creates. A greater force can balance a smaller force by applying it closer to the pivot, resulting in equal moments and hence, equilibrium. It’s this interplay between force and distance that is pivotal in accomplishing a state of uniform board equilibrium.

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Most popular questions from this chapter

A uniform \(300 \mathrm{~N}\) trapdoor in a floor is hinged at one side. Find the net upward force needed to begin to open it and the total force exerted on the door by the hinges (a) if the upward force is applied at the center and (b) if the upward force is applied at the center of the edge opposite the hinges.

A lead sphere has volume \(6.0 \mathrm{~cm}^{3}\) when it is resting on a lab table, where the pressure applied to the sphere is atmospheric pressure. The sphere is then placed in the fluid of a hydraulic press. What increase in the pressure above atmospheric pressure produces a \(0.50 \%\) decrease in the volume of the sphere?

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?

A solid gold bar is pulled up from the hold of the sunken RMS Titanic. (a) What happens to its volume as it goes from the pressure at the ship to the lower pressure at the ocean's surface? (b) The pressure difference is proportional to the depth. How many times greater would the volume change have been had the ship been twice as deep? (c) The bulk modulus of lead is one- fourth that of gold. Find the ratio of the volume change of a solid lead bar to that of a gold bar of equal volume for the same pressure change.

A \(60.0 \mathrm{~cm}\), uniform, \(50.0 \mathrm{~N}\) shelf is supported horizontally by two vertical wires attached to the sloping ceiling (Fig. E11.10). A very small \(25.0 \mathrm{~N}\) tool is placed on the shelf midway between the points where the wires are attached to it. Find the tension in each wire. Begin by making a free-body diagram of the shelf.

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