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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

An amusement park ride consists of airplane-shaped cars attached to steel rods (Fig. \(\mathbf{P} \mathbf{1 1 . 8 4}\) ). Each rod has a length of \(15.0 \mathrm{~m}\) and a cross-sectional area of \(8.00 \mathrm{~cm}^{2}\). The rods are attached to a frictionless hinge at the top, so that the cars can swing outward when the ride rotates. (a) How much is each rod stretched when it is vertical and the ride is at rest? (Assume that each car plus two people seated in it has a total weight of \(1900 \mathrm{~N}\).) (b) When operating, the ride has a maximum angular speed of 12.0 rev \(/\) min. How much is the rod stretched then?

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The left-hand end of a slender uniform rod of mass \(m\) is placed against a vertical wall. The rod is held in a horizontal position by friction at the wall and by a light wire that runs from the right-hand end of the rod to a point on the wall above the rod. The wire makes an angle \(\theta\) with the rod. (a) What must the magnitude of the friction force be in order for the rod to remain at rest? (b) If the coefficient of static friction between the rod and the wall is \(\mu_{\mathrm{s}},\) what is the maximum angle between the wire and the rod at which the rod doesn't slip at the wall?

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