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Two people are carrying a uniform wooden board that is 3.00 m long and weighs 160 N. If one person applies an upward force equal to 60 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 point 2.4 m from the end where the 60 N force is applied.

Step by step solution

01

Understand the Problem

We need to determine the point where the second person applies a force on a wooden board, given the weight of the board and the force applied by the first person. We will use the concept of torque equilibrium to find this position.
02

Visualize with a Free-Body Diagram

Draw a free-body diagram of the board. The board is 3.00 m long. Place the board horizontally, and show the gravitational force (160 N) acting at its center (1.5 m from either end). Mark one person applying a 60 N force at one end. Let's call the other person's position 'x' meters from the same end.
03

Apply the Principle of Torque

For the board to be in rotational equilibrium, the sum of torques around any point must be zero. Choose the point at the location where the 60 N force is applied as the pivot. The torque due to the 160 N weight is \( 160 imes 1.5 \) clockwise and the torque due to the unknown force \( F \) applied by the other person at distance \( x \) is \( F \times x \) counterclockwise.
04

Set Up the Torque Equation

The net torque around the point where the first person applies the force should sum to zero: \[ 60 \times 0 + F \times x = 160 \times 1.5 \]Since the first torque is zero (pivot point), simplify the equation to only the non-zero terms.
05

Calculate the Second Force

Since we know the total upward force must equal the weight of the board for vertical equilibrium, we have: \[ F + 60 = 160 \]Solve for \( F \):\[ F = 100 \text{ N} \]
06

Solve for Position 'x'

Plug \( F = 100 \) N back into the torque equation:\[ 100 \times x = 160 \times 1.5 \]Solve for \( x \):\[ x = \frac{240}{100} = 2.4 \text{ m} \] from the same end where the 60 N force is applied.

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Key Concepts

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

Torque Equilibrium
To understand how objects remain stable when forces are applied, we delve into the concept of torque equilibrium. Torque is the force that causes an object to rotate around an axis. For an object to remain in rotational equilibrium, the sum of all torques acting on it must be zero. In our exercise, the wooden board is balanced by two people lifting at different points. The first person applies a force at one end and creates a torque based on the distance from the pivot point (rotation point).
For rotational equilibrium, the torques created by the gravitational force of the board and the force by the second person must cancel each other. This means the counterclockwise torques must equal the clockwise torques. By ensuring this balance, the board remains level and doesn't rotate.
In practical application, it is essential to calculate the distances and forces accurately to maintain torque equilibrium, especially in structures and balancing problems.
Free-Body Diagram
A simple yet powerful tool in physics problem-solving is the free-body diagram. It's a visual representation that helps us understand all the forces acting upon an object. For the wooden board scenario, drawing a free-body diagram allows us to identify where each force acts and their magnitudes.
In this example, the board itself is sketched, showing the center where gravity acts, marked with a 160 N force. Additionally, the 60 N upward force by one person is indicated at one end. The position where the second person lifts is marked as 'x'.
Free-body diagrams simplify complex problems and are crucial for identifying the forces and calculating the resulting torques. By making the forces visible, solving for unknown values like the position 'x' in this problem becomes more approachable.
Rotational Equilibrium
When we talk about rotational equilibrium, we're referring to an object's ability to stay steady without rotating. For this, the sum of all torques around a given point should equal zero. This principle is crucial in situations where multiple forces are acting at different points.
In the case of the wooden board, to maintain rotational equilibrium, the torques produced by each applied force (both gravitational and by the people lifting) must offset one another. The gravitational force pulls down at the center of mass, creating a torque. The second person must apply their lifting force at the right point 'x' so that this force's torque counteracts the gravitational torque.
Understanding rotational equilibrium helps engineers design stable structures and helps us in daily balancing tasks, ensuring objects remain securely in place without tipping over.
Force Analysis
Force analysis is the breakdown of forces acting on a system to understand how they interact and affect motion or stability. In the wooden board exercise, force analysis involves determining both the specific points where forces are applied and their magnitudes.
Firstly, the total force applied upwards by both individuals must balance the downward gravitational force of the board. This is vertical equilibrium, ensuring the board doesn't move up or down. Here, we calculated that the second person must exert a force of 100 N.
Secondly, this force analysis helps in setting and solving the torque equation. By analyzing the forces correctly, one can determine the unknown variables, like the correct lifting position 'x' for the second individual, ensuring the board stays balanced horizontally.
Force analysis is integral in physics for predicting and explaining how forces will affect the motion or stability of objects in practical situations.

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Most popular questions from this chapter

A relaxed biceps muscle requires a force of 25.0 N for an elongation of 3.0 cm; the same muscle under maximum tension requires a force of 500 N for the same elongation. Find Young's modulus for the muscle tissue under each of these conditions if the muscle is assumed to be a uniform cylinder with length 0.200 m and cross-sectional area 50.0 cm\(^2\).

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