two people carry a heavy electric motor by placing it on a light board 2m long.one person lifts at one end with a force of 700N, and the other lifts the opposite end with a force of 400N.Calculate the weight of the motor and determine where its center of Gravity is Located
To calculate the weight of the motor, we can use the principle of moments. The total moment about any point on the board should be zero since the motor is in equilibrium.
Let's assume that the center of gravity of the motor is located at a distance 'x' from the end where the 700N force is applied and at a distance (2m - x) from the other end.
The moment due to the 700N force is given by:
Moment1 = Force1 * Distance1
Moment1 = 700N * x
Similarly, the moment due to the 400N force is given by:
Moment2 = Force2 * Distance2
Moment2 = 400N * (2m - x)
Since the motor is in equilibrium, the total moment is zero:
Moment1 + Moment2 = 0
700N * x + 400N * (2m - x) = 0
Simplifying the equation:
700N * x + 800N m - 400N * x = 0
300N * x = 800N * m
x = (800N * m) / 300N
x = 2.67m
Now, to calculate the weight of the motor, we know that the total force acting on the motor is the sum of the two forces.
Weight = Force1 + Force2
Weight = 700N + 400N
Weight = 1100N
Therefore, the weight of the motor is 1100N and its center of gravity is located at a distance of 2.67m from the end where the 700N force is applied.
To calculate the weight of the motor, we need to find the net force acting on it. Since the motor is in equilibrium (not accelerating), the sum of the forces acting on it must be zero.
In this case, we have two forces acting on the motor: 700N upward force and 400N upward force. Since the motor is in equilibrium, there must be an equal and opposite force acting downward. The weight of the motor is this downward force.
The net force can be found by subtracting the smaller force from the larger force: 700N - 400N = 300N. Therefore, the weight of the motor is 300N.
Now, to determine the center of gravity (CG) of the motor, we need to find the point at which the weight is evenly distributed. In a symmetrical object like the motor, the center of gravity is usually located at the center of the object.
Since the motor is placed on a 2m long board, we can assume that the center of gravity is located at the midpoint of the board, which is 1m from each end. Therefore, the center of gravity of the motor is located 1m from each end of the board.
The sums of torques respectively the edges of the board are:
W•x-400•2 = 0
W• (2-x) -700•2=0
W•x=800
2W-W•x -1400 = 0
2W-800-1400=0
W=1100 N
x=800/W =0.73 m