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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 \(400 \mathrm{~N}\), 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?

Short Answer

Expert verified
(a) The motor's weight is 1000 N, and its center of gravity is 0.67 m away from the person applying a force of 600 N. (b) The motor's weight is 800 N and its center of gravity is 0.67 m away from the person applying a force of 600 N.

Step by step solution

01

Identify the forces

There are two forces acting on the board. One is at one end with a magnitude of 400 N, and the other is at the opposite end with a magnitude of 600 N.
02

Calculate the weight of the motor in scenario (a)

In equilibrium, the sum of all forces is zero. Hence, the sum of the two forces given must be equal to the weight of the motor. Using the formula for the sum of forces, which is \(\sum F = 0\), we get \(400N + 600N - W = 0 \Rightarrow W = 400N + 600N = 1000 N\)
03

Determine the location of the center of gravity in scenario (a)

In equilibrium, the sum of all torques is also zero. Hence, the torques due to the two forces must cancel out the torque due to the weight of the engine. The torque is calculated as the product of force and distance from the axis of rotation, which in our case is at the center of the board, so the distance is \(d = 1.00 m\). Setting the clockwise torque equal to the counter-clockwise torque gives: \(400N * 1.00m = 600N * d \Rightarrow d = 400N * 1.00m / 600N = 0.67 m\). In other words, the center of gravity is located 0.67 m from the person exerting a force of 600 N.
04

Calculate the weight of the motor in scenario (b)

In this scenario, we also have the weight of the board acting at its center, which is 200 N. Thus, \(400N + 600N - W - 200N = 0 \Rightarrow W = 400N + 600N - 200N = 800 N\)
05

Determine the location of the center of gravity in scenario (b)

As before, the sum of all torques is zero. However, now the torques due to the three forces need to cancel out. Setting the clockwise torque equal to the counter-clockwise torque gives: \(400N * 1.00m + 200N * 0.00m = 600N * d \Rightarrow d = 400N * 1.00m / 600N = 0.67 m\). Thus, the center of gravity is still located 0.67 m from the person exerting a force of 600 N.

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Key Concepts

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

Center of Gravity
Understanding the center of gravity (CoG) is crucial when analyzing static equilibrium scenarios. It is the point where the weight of an object is considered to be concentrated and around which the force of gravity appears to act. When an object like the heavy electric motor mentioned in the exercise is balanced and at rest, its center of gravity plays a key role.

In the scenario given, the board and motor system is in static equilibrium, which means that the CoG must be directly under the point of support, or in this case, along the line of action of the forces exerted by the two people. Calculating the center of gravity allows us to understand how the object’s weight is distributed and influences the system's stability.

For a uniform object, the CoG is typically at the geometric center. However, when additional weights are applied, as with the motor in our exercise, the CoG shifts. This shift is calculated by considering the magnitudes and points of application of all forces. In the example, the location of the motor's CoG was calculated by assessing the balance of torques provided by the forces the two people apply.
Torque and Rotational Equilibrium
The concept of torque is pivotal when discussing rotational equilibrium. Torque is a measure of the force that can cause an object to rotate around an axis. It is the product of the force applied and the distance from the point of application to the axis of rotation. In the exercise, torque was calculated to assess the rotational equilibrium of the board-and-motor system.

Mathematically, it is given by the equation \(\tau = F \times d\), where \(\tau\) is the torque, \(F\) is the force, and \(d\) is the distance from the axis. For an object to be in rotational equilibrium, the sum of all torques acting on it must be zero. In simple words, the clockwise torques should balance out the counter-clockwise torques, leading to no net rotational motion.

The exercise required the computation of torques to find the motor's center of gravity by setting the board in rotational equilibrium, thereby illustrating the close relationship between torque and the stability of objects in static situations.
Forces in Equilibrium
The state of forces in equilibrium is fundamental for objects in static equilibrium where there is no movement. In our exercise, the heavy electric motor remains stationary on a board, indicating that the forces acting upon it are in balance.

When discussing forces in equilibrium, we're specifically referring to the situation where the sum of all forces acting on an object is zero, as described by the equation \(\sum F = 0\). Applying this principle, we determined the weight of the motor by considering the forces applied by the two individuals carrying it.

In scenario (a), the sum of the forces exerted by the two people was equal to the weight of the motor. When the scenario changed to include the board's weight, we adapted our calculations accordingly. The principle of forces in equilibrium not only helps determine an object's weight but also ensures its stability and provides insight into force distribution within the system. This is critical in practical applications such as engineering, where the balance of forces ensures the integrity and safety of structures.

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