Two people carry a heavy electric motor by placing it
on a light board 2 meter in length. One person lifts at
one end with a force of 700N, and the other lifts the
opposite end with a force of 500N. What is the weight
of the motor, where along the board is its centre of
gravity located?
weight = 700N + 500N
measuring distance (d) from the 500N end ... d(700 + 500) = 2 * 700
the weight is equal and opposite to the total force up.
Hey, enough already, read the text.
To find the weight of the motor and determine the location of its center of gravity, we can use the principle of moments.
The principle of moments states that the sum of the clockwise moments about any point is equal to the sum of the anti-clockwise moments about the same point, for an object in rotational equilibrium.
Let's assume that the center of gravity (CG) of the motor is located at a distance x from the person exerting the 700N force. The person exerting the 500N force is at the opposite end of the board (2m away from the person exerting the 700N force).
Using the principle of moments, we can set up the equation:
(clockwise moment about CG) = (anti-clockwise moment about CG)
(700N) * x = (500N) * (2m - x)
Now, let's solve the equation for x to find the location of the center of gravity.
700x = 1000 - 500x
700x + 500x = 1000
1200x = 1000
x = 1000 / 1200
x = 0.8333 m
Therefore, the center of gravity of the motor is located approximately 0.8333 meters from the person exerting the 700N force.
To find the weight of the motor, we need to calculate the resultant force:
Resultant force = force exerted by person 1 + force exerted by person 2
Resultant force = 700N + 500N
Resultant force = 1200N
Hence, the weight of the motor is 1200 Newtons.
To find the weight of the motor and the location of its center of gravity along the board, we need to use the principle of moments or torques.
The principle of moments states that when an object is in equilibrium (not rotating), the sum of the clockwise moments is equal to the sum of the anticlockwise moments.
In this case, we have two forces acting on the light board - one force of 700N at one end and another force of 500N at the opposite end. The weight of the motor will act downward from its center of gravity, and we need to find that weight and its location.
Let's assume that the center of gravity of the motor is located at a distance 'x' from the person applying the force of 700N and at a distance '2 - x' from the person applying the force of 500N.
Now, using the principle of moments:
Clockwise moments = 700N * x
Anti-clockwise moments = 500N * (2 - x)
To balance the moments and achieve equilibrium, we set the clockwise moments equal to the anti-clockwise moments:
700N * x = 500N * (2 - x)
Now, we can solve this equation to find the value of 'x', which will give us the location of the center of gravity.
700x = 1000 - 500x
1200x = 1000
x = 1000 / 1200
x = 0.8333 meters
So, the center of gravity of the motor is located approximately at 0.8333 meters from the person applying the force of 700N.
To find the weight of the motor, we can now calculate the total torque acting on the board:
Total torque = 700N * 0.8333m - 500N * (2 - 0.8333m)
Total torque = 583.31N·m - 750N + 416.65N·m
Total torque = 583.31N·m + 416.65N·m - 750N
Total torque = 999.96N·m - 750N
Total torque = 249.96N·m
Since the motor is in equilibrium, the total torque acting on the board must be zero. Thus:
249.96N·m = 0
Therefore, the weight of the motor is 249.96N, and its center of gravity is located approximately at 0.8333 meters from the person applying the force of 700N.