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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 \(400.0 \mathrm{~N}\), and the other lifts at the opposite end with a force of \(600.0 \mathrm{~N}\). (a) Start by making a free-body diagram of the motor. (b) What is the weight of the motor? (c) Where along the board is its center of gravity located?

Short Answer

Expert verified
The weight of the motor is 1000.0 N, and its center of gravity is 1.20 m from the end where the 400.0 N force is applied.

Step by step solution

01

Understand the Problem

We have a board that is 2.00 m long with two forces applied at the ends. One person applies a 400.0 N force, and the other applies a 600.0 N force. The motor exerts a weight force downward on the board, and the board is balanced by the forces on each end.
02

Identify Forces and Create Free-Body Diagram

The free-body diagram will show three forces: \( F_1 = 400.0 \text{ N} \) up on one end, \( F_2 = 600.0 \text{ N} \) up on the other end, and the weight \( W \) of the motor acting downward at a distance \( x \) from the end where \( F_1 \) is applied. The board is in static equilibrium.
03

Calculate the Weight of the Motor

Since the board is in equilibrium, the sum of the forces must be zero. This means \( F_1 + F_2 - W = 0 \). Substituting the given forces: \[ 400.0 \text{ N} + 600.0 \text{ N} = W \]. Therefore, \( W = 1000.0 \text{ N} \).
04

Determine the Position of the Center of Gravity

To find the center of gravity, we use the concept of torque equilibrium about any point. Let's calculate using the left end where \( F_1 \) is applied to avoid introducing unknowns. The torques must balance: \( F_1 \cdot 0 \text{ m} + W \cdot x - F_2 \cdot 2.00 \text{ m} = 0 \). This simplifies to \( 1000.0 \cdot x = 600.0 \cdot 2.00 \). Solving for \( x \), we find \( x = \frac{600.0 \times 2.00}{1000.0} = 1.20 \text{ m} \).

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Key Concepts

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

Free-body Diagram
When faced with a physics problem involving forces, a free-body diagram is a valuable tool. It visually represents all the forces acting on an object, making it easier to understand the physical situation. Let's consider the scenario with the electric motor on a board. In our diagram, we have three forces:
  • The force of 400 N exerted upward at one end by the first person.
  • The force of 600 N exerted upward at the opposite end by the second person.
  • The gravitational force due to the motor's weight, denoted as W, acting downwards.
By representing these forces, the diagram helps us visualize how they balance each other out, holding the board and motor in place without movement. It's essential for analyzing equilibrium problems, as it lays the groundwork for further calculation of force magnitudes and positions.
Torque Balance
As the board holding the motor is in a state of static equilibrium, the torques acting on it must balance. Torque, which is the rotational equivalent of force, depends on both the force applied and the distance from the pivot point. This is crucial for determining how forces affect rotational equilibrium.In the motor problem, we use torque balance to find where the center of gravity lies. By choosing to calculate torques around one end of the board, we simplify the problem, eliminating unknowns. At equilibrium, the clockwise torques equal the counterclockwise torques:- The torque due to the 400 N force is zero, as the pivot point is at the force location.- The torque due to the 600 N force is \( T_{600} = 600 \text{ N} \times 2.00 \text{ m} \).- The torque from the motor's weight is \( W \times x \), where \( x \) is its distance from the pivot.By setting the sum of these torques to zero, we can solve for the center of gravity's location using the formula: \[ 1000 \times x = 600 \times 2.00 \].
Center of Gravity
The center of gravity (CoG) is a critical point on an object where its entire weight seems to act. Understanding where the CoG is located helps in managing an object's balance and stability. In this exercise, the board and motor system are in equilibrium. By using the torque balance method, we find the CoG position by considering the forces acting on the motor. We found that by applying the equilibrium condition of torques, the distance from the pivot point on one end is 1.20 m. This position shows us that the motor's weight is evenly distributed concerning the forces applied. The CoG doesn't necessarily have to be at the geometric center; rather, it reflects the balancing point considering the forces applied.
Force Analysis
Force analysis in physics involves determining the different forces affecting a body, both in magnitude and direction. It is a foundational step in solving equilibrium problems. In our scenario, we analyzed the forces by:- Identifying forces acting on the board: two upward forces (400 N and 600 N) and the downward weight (\( W \)) of the motor.- Using the principle of equilibrium: if an object is at rest, the sum of all vertical forces must be zero.Therefore, the sum of the upward forces must equal the downward force due to the motor. We derived that\( W = 400 + 600 = 1000 \) N.This process gives us a clear understanding of how static equilibrium conditions regulate and comprehend motionlessness in real-world systems. By balancing forces, we ensure that our solutions reflect practical and predictable outcomes.

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Most popular questions from this chapter

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