The meter stick has a fulcrum at the 60.0 cm mark. There is 4.0 N at the 20.0 cm mark. The meter stick is uniform and has a weight of 2 N. How much force must be hanging from the 70.0 cm mark to balance the system? You may refer to Figure 7-34 in the textbook for a similar diagram.
To solve this problem, we can use the principle of torque. The torque is the product of force and the perpendicular distance from the fulcrum to the line of action of the force.
In this case, the torque exerted by the weight of the meter stick itself is given by the product of its weight (2 N) and the distance between the fulcrum and the center of the meter stick (60.0 cm).
The torque exerted by the 4.0 N force at the 20.0 cm mark is given by the product of the force (4.0 N) and the distance between the fulcrum and the line of action of the force (40.0 cm).
To balance the system, the torque exerted by the force hanging from the 70.0 cm mark should be equal to the torque exerted by the weight of the meter stick and the 4.0 N force.
Mathematically, we can express this as:
Torque(weight of the meter stick) + Torque(4.0 N force) = Torque(force hanging from the 70.0 cm mark)
(2 N) * (60.0 cm) + (4.0 N) * (40.0 cm) = (force hanging from the 70.0 cm mark) * (distance from fulcrum)
Now, we can solve for the force hanging from the 70.0 cm mark:
(2 N * 60.0 cm + 4.0 N * 40.0 cm) / (distance from fulcrum) = (force hanging from the 70.0 cm mark)
((2 N * 60.0 cm) + (4.0 N * 40.0 cm)) / (70.0 cm) = (force hanging from the 70.0 cm mark)
(120 N cm + 160 N cm) / (70.0 cm) = (force hanging from the 70.0 cm mark)
280 N cm / 70.0 cm = (force hanging from the 70.0 cm mark)
4.0 N = (force hanging from the 70.0 cm mark)
Therefore, to balance the system, a force of 4.0 N must be hanging from the 70.0 cm mark.