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 400.0 N, and the other lifts at the opposite end with a force of 600.0 N. What is the weight of the motor? and where along the board is the center of gravity?
To find the weight of the motor and the location of its center of gravity, we first need to understand the concept of torque.
Torque is the rotational equivalent of force. It is defined as the product of force and the perpendicular distance from the point of rotation to the line of action of the force. Mathematically, torque (τ) can be calculated using the formula:
τ = F × r
Where:
τ = torque
F = force applied
r = perpendicular distance from the point of rotation to the line of action of the force
In this case, the torque applied by the person lifting with a force of 400.0 N can be calculated as follows:
τ1 = F1 × r1
Given:
F1 = 400.0 N
r1 = 1.00 m (half the length of the board)
τ1 = 400.0 N × 1.00 m
τ1 = 400.0 N⋅m
Similarly, the torque applied by the person lifting with a force of 600.0 N can be calculated as:
τ2 = F2 × r2
Given:
F2 = 600.0 N
r2 = 1.00 m (half the length of the board)
τ2 = 600.0 N × 1.00 m
τ2 = 600.0 N⋅m
Now, we know that the total torque acting on the motor is the sum of the torques applied by each person:
τtotal = τ1 + τ2
τtotal = 400.0 N⋅m + 600.0 N⋅m
τtotal = 1000 N⋅m
Since torque is equal to the product of force and perpendicular distance, we can determine the total force acting on the motor by dividing the total torque by the perpendicular distance from the point of rotation to the line of action of the total force:
τtotal = Ftotal × rtotal
Given:
τtotal = 1000 N⋅m
rtotal = 2.00 m (the length of the board)
1000 N⋅m = Ftotal × 2.00 m
Ftotal = 1000 N⋅m / 2.00 m
Ftotal = 500 N
Therefore, the weight of the motor is 500.0 N.
To find the center of gravity of the motor along the board, we assume that the motor has uniform density. In this case, the center of gravity will be exactly at the middle point of the board, which is 1.00 m from both ends.
Hence, the center of gravity of the motor is located at the midpoint of the board.