The center of gravity of a meter stick that weighs 1.950 N is at the 50.00 cm mark. A 3.54 N weight is suspended from the 20.0 cm mark and an 8.00 N weight is suspended from the 70.0 cm mark on the meter stick. Where would you place the fulcrum to balance this system?
To find the position of the fulcrum to balance this system, you need to consider the moments (torques) acting on the meter stick.
The moment (torque) of a force about a point is given by the equation:
Moment = Force × Distance
In this case, the center of gravity of the meter stick is at the 50.00 cm mark.
Let's first calculate the moment caused by the weight of the meter stick itself. The weight of the meter stick is 1.950 N (newtons), and its distance from the fulcrum (center of gravity) is 50.00 cm.
Moment of the meter stick = 1.950 N × 50.00 cm
Next, consider the moments caused by the two weights suspended from the meter stick. The 3.54 N weight is suspended at the 20.0 cm mark, and the 8.00 N weight is suspended at the 70.0 cm mark.
Moment of the 3.54 N weight = 3.54 N × (50.00 cm - 20.0 cm)
Moment of the 8.00 N weight = 8.00 N × (70.0 cm - 50.00 cm)
Now, to balance the system, the total clockwise moments should be equal to the total counterclockwise moments.
Sum of clockwise moments = Moment of the meter stick
Sum of counterclockwise moments = Moment of the 3.54 N weight + Moment of the 8.00 N weight
Setting these two sums equal, we can solve for the position of the fulcrum:
1.950 N × 50.00 cm = 3.54 N × (50.00 cm - 20.0 cm) + 8.00 N × (70.0 cm - 50.00 cm)
Simplifying the equation:
97.50 N·cm = 3.54 N × 30.00 cm + 8.00 N × 20.0 cm
Now, we can rearrange the equation to solve for the position of the fulcrum:
97.50 N·cm - 3.54 N × 30.00 cm = 8.00 N × 20.0 cm
74.50 N·cm = 8.00 N × 20.0 cm
Dividing both sides by 8.00 N:
9.3135 cm = 20.0 cm - fulcrum position
Subtracting 20.0 cm from both sides, we find:
-10.6865 cm = -fulcrum position
Therefore, the fulcrum should be placed at approximately 10.69 cm to balance the system.