Question

Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts at one end with a force of 400 N, and the other lifts the opposite end with a force of 600 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 N, with its center of gravity at its center, and the two people each 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

(a) Motor's weight: 1000 N, center at 1.2 m; (b) Motor's weight: 800 N, center at 1.25 m.

Step-by-step solution

Analyze the Problem (a)

In part (a), we have two forces: 400 N and 600 N, lifting a motor placed on a light board. We are given that the board is 2.00 m long and not considered in weight. We need to find the weight of the motor and the position of its center of gravity along the board.

Apply Equilibrium Conditions (a)

Since the motor and board are in static equilibrium, the sum of the upward forces must equal the weight of the motor. Therefore, the weight of the motor is the sum of the forces:\[ W_{ ext{motor}} = 400 ext{ N} + 600 ext{ N} = 1000 ext{ N} \]

Calculate the Center of Gravity (a)

To find the center of gravity, use the moments about one end (let's choose the end where the 400 N force acts). For equilibrium, the moment due to the motor's weight should balance the moment due to the 600 N force:\[ 400 ext{ N} imes 0 + 600 ext{ N} imes 2 = 1000 ext{ N} imes x \]Solving for \(x\), the position of the center of gravity:\[ x = \frac{600 imes 2}{1000} = 1.2 ext{ m} \] from the 400 N end.

Analyze the Problem (b)

In part (b), the board is not light and has a weight of 200 N with its center of gravity at the center of the board (1.00 m from either end). The same lifting forces are applied as in part (a). We need to find the weight of the motor in this case and reassess the center of gravity.

Apply Equilibrium Conditions (b)

As in part (a), calculate the total upward force:\[ 400 ext{ N} + 600 ext{ N} = 1000 ext{ N} \]This must support the combined weight of the board and motor:\[ W_{ ext{motor}} + 200 ext{ N} = 1000 ext{ N} \]Solving for the motor's weight:\[ W_{ ext{motor}} = 800 ext{ N} \]

Calculate the Center of Gravity (b)

Taking moments about the 400 N end again, considering the weight of the board at its center (1.00 m from the 400 N end):\[ 600 ext{ N} imes 2 = 800 ext{ N} imes x + 200 ext{ N} imes 1 \]Solving for \(x\):\[ 1200 = 800x + 200 \]\[ 800x = 1000 \]\[ x = 1.25 ext{ m} \] from the 400 N end.

Key concepts

Center of Gravity

The center of gravity is the point at which the total weight of a body or system is considered to be concentrated. In scenarios involving multiple forces, such as two people carrying a motor on a board, determining the center of gravity helps in understanding how the forces affect motion and balance.
For instance, in the exercise, finding the motor's center of gravity involves using moments, or the rotational equivalent of force. Simply put, if you take moments about one end of the board, the center of gravity is the point where these moments balance out.
In the given problem, for part (a), the board is lightweight, and thus, only the motor's weight needs consideration. The 600 N force creates a moment about the 400 N end, which helps locate the center of gravity 1.2 m from the 400 N end. In part (b), with the board weighing 200 N, its own center of gravity influences the overall position. This results in the center of gravity shifting slightly to 1.25 m from the 400 N end. This adjustment is because the board's weight now contributes to the total moment.

Moment of Force

The moment of force, often referred to as torque, is the tendency of a force to rotate an object about an axis or pivot. It is calculated as the product of the force and the distance from the pivot or reference point to the line of action of the force.
When aiming to maintain static equilibrium, it is critical to balance these moments. In the exercise, you apply this principle by calculating moments to determine the center of gravity.
For part (a) of the problem, at the 400 N end, the moment caused by the 600 N force must be equal and opposite to the moment due to the motor's weight about the same point. By setting up the equation \( 600 \text{ N} \times 2 \text{ m} = 1000 \text{ N} \times x \), you solve for \( x \) to find the motor's center of gravity.
In part (b), with the board having its own weight, a new moment is introduced. Therefore, when calculating, you also add the board's contribution, leading to the equation: \( 600 \text{ N} \times 2 \text{ m} = 800x + 200 \text{ N} \times 1 \text{ m} \). Solving this gives a slightly different center of gravity value.

Weight Calculation

Calculating the total weight is essential in identifying equilibrium conditions. In the context of this problem, you need to calculate the weight of the motor when supported by forces at both ends of the board.
For part (a), because the board itself is light and negligible in weight, the total upward force exerted by both individuals on the motor is equal to the motor's weight. Therefore, summing the forces, \( 400 \text{ N} + 600 \text{ N} \), results in the motor's weight being 1000 N.
In part (b), however, the scenario involves a heavier board. The combined weight of the board and motor must be equal to the total 1000 N of the upward force provided by the people. Subtracting the board's weight of 200 N from the total suggests the motor itself weighs 800 N. This kind of calculation is crucial when dealing with static systems to ensure all forces balance accurately.