Two people carry a heavy electric motor by placing it on a light board 2 meters in length. One person lifts at one end with a force of 700 N, and the other lifts the opposite end with a force of 500 N. What is the weight of the motor, where along the board is its center of gravity located?
200 Newton and 1 meter along the board
To find the weight of the motor, we can use the concept of torque.
Torque is defined as the force applied multiplied by the distance from the point of application. In this case, each person's force creates a torque on the motor.
Let's assume that the center of gravity of the motor is located at a distance x from the person applying the 700 N force.
Since torque is equal to force multiplied by the distance, the torque created by the person applying the 700 N force is given by:
Torque1 = 700 N * x
Similarly, the torque created by the person applying the 500 N force is:
Torque2 = 500 N * (2 - x)
Now, since the motor is in equilibrium (not rotating), the sum of the torques must be equal to zero:
Torque1 + Torque2 = 0
Substituting the values, we get:
700 N * x + 500 N * (2 - x) = 0
Simplifying the equation gives:
700x + 1000 - 500x = 0
200x + 1000 = 0
200x = -1000
x = -1000 / 200
x = -5
Since the distance cannot be negative, the equation x = -5 implies that the center of gravity is located 5 meters from one end of the board. However, since the board is only 2 meters long, this result indicates an error in our assumptions or measurements.
Therefore, we need to reassess the problem and adjust our calculations to find a valid solution.