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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 claw hammer is used to pull a nail out of a board (see Fig. \(\mathrm{P} 11.45\) ). The nail is at an angle of \(60^{\circ}\) to the board, and a force \(\overrightarrow{\boldsymbol{F}}_{1}\) of magnitude \(400 \mathrm{~N}\) applied to the nail is required to pull it from the board. The hammer head contacts the board at point \(A,\) which is \(0.080 \mathrm{~m}\) from where the nail enters the board. A horizontal force \(\vec{F}_{2}\) is applied to the hammer handle at a distance of \(0.300 \mathrm{~m}\) above the board. What magnitude of force \(\overrightarrow{\boldsymbol{F}}_{2}\) is required to apply the required \(400 \mathrm{~N}\) force \(\left(F_{1}\right)\) to the nail? (Ignore the weight of the hammer.)

A \(350 \mathrm{~N}\), uniform, \(1.50 \mathrm{~m}\) bar is suspended horizontally by two vertical cables at each end. Cable \(A\) can support a maximum tension of \(500.0 \mathrm{~N}\) without breaking, and cable \(B\) can support up to \(400.0 \mathrm{~N}\). You want to place a small weight on this bar. (a) What is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?

A uniform ladder \(5.0 \mathrm{~m}\) long rests against a frictionless, vertical wall with its lower end \(3.0 \mathrm{~m}\) from the wall. The ladder weighs \(160 \mathrm{~N}\). The coefficient of static friction between the foot of the ladder and the ground is \(0.40 .\) A man weighing \(740 \mathrm{~N}\) climbs slowly up the ladder. Start by drawing a free-body diagram of the ladder. (a) What is the maximum friction force that the ground can exert on the ladder at its lower end? (b) What is the actual friction force when the man has climbed \(1.0 \mathrm{~m}\) along the ladder? (c) How far along the ladder can the man climb before the ladder starts to slip?

Two people carry a heavy electric motor by placing it on a light board \(2.00 \mathrm{~m}\) long. One person lifts at one end with a force of \(400 \mathrm{~N}\), and the other lifts the opposite end with a force of \(600 \mathrm{~N}\). (a) What is the weight of the motor, and where along the board is its center of gravity located? (b) Suppose the board is not light but weighs \(200 \mathrm{~N},\) with its center of gravity at its center, and the two people exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?

A nylon rope used by mountaineers elongates \(1.10 \mathrm{~m}\) under the weight of a \(65.0 \mathrm{~kg}\) climber. If the rope is \(45.0 \mathrm{~m}\) in length and \(7.0 \mathrm{~mm}\) in diameter, what is Young's modulus for nylon?

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