a uniform meter stick is balanced at its midpoint with a single support. a 50 N weight is suspended at the 30 cm mark. at what point must a 20 N weight be hung to balance the system
To find the point where the 20 N weight must be hung to balance the system, we need to consider the torques acting on the meter stick.
The torque (or moment) of an object is the product of its force and the perpendicular distance from the pivot point (or fulcrum) to the line of action of the force. In this case, the pivot point is the midpoint of the meter stick.
Given information:
- Weight of the meter stick = 0 (since it is balanced)
- Weight of the 50 N weight = 50 N
- Distance of the 50 N weight from the pivot point = 30 cm = 0.3 m
Let's assume the distance of the 20 N weight from the pivot point is 'x' cm or 'x' meters.
To balance the system, we need the total torque on both sides of the meter stick to be equal.
Torque on the left side (from the 20 N weight) = (20 N) * (x m)
Torque on the right side (from the 50 N weight) = (50 N) * (0.3 m)
Since the meter stick is uniformly balanced, the torques should be equal:
(20 N) * (x m) = (50 N) * (0.3 m)
Now we can solve for 'x':
(20 N) * (x m) = (50 N) * (0.3 m)
20x = 15
x = 15 / 20
x = 0.75 meters
Therefore, to balance the system, the 20 N weight must be hung at the 0.75-meter mark on the meter stick.