A uniform meterstick pivoted at its center has a 100
g mass suspended at the 21.0
cm position. At what position should a 95.0 g mass be suspended to put the system in equilibrium? What mass would have to be suspended at the 80.0
cm position for the system to be in equilibrium?
To find the position where a 95.0 g mass should be suspended to put the system in equilibrium, we need to consider the principle of lever balance. In this case, the meterstick acts as a lever, and the masses are like forces acting at different distances from the pivot point.
Let's assume that the pivot point is the center of the meterstick. The torque (or turning effect) produced by a force is given by the equation:
Torque = (Force) x (Distance from the pivot)
For equilibrium, the sum of the torques on both sides of the pivot should be zero. In other words, the clockwise torques should balance the counterclockwise torques.
Now, let's solve the problem step by step:
1. Given that a 100 g mass is suspended at the 21.0 cm position, we can calculate the torque produced by this mass:
Torque1 = (100 g) x (21.0 cm)
2. Let x cm be the position where the 95.0 g mass should be suspended. The torque produced by this mass can be calculated as:
Torque2 = (95.0 g) x (x cm)
3. Since the system is in equilibrium, the sum of the torques should be zero:
Torque1 + Torque2 = 0
(100 g) x (21.0 cm) + (95.0 g) x (x cm) = 0
4. Now, solve the equation for x to find the position of the 95.0 g mass:
(100 g) x (21.0 cm) = (-95.0 g) x (x cm)
(100 g) x (21.0 cm) = (-95.0 g) x (x cm)
x cm = (100 g) x (21.0 cm) / (-95.0 g)
x cm ≈ 22.1 cm (rounded to one decimal place)
Therefore, the 95.0 g mass should be suspended at approximately the 22.1 cm position to put the system in equilibrium.
To find the mass that would have to be suspended at the 80.0 cm position for the system to be in equilibrium, we can use the same principle of lever balance.
1. Assume that the pivot point is still the center of the meterstick. Let y g be the mass that needs to be suspended at the 80.0 cm position.
2. Write the equation for the sum of torques:
(100 g) x (21.0 cm) + (y g) x (80.0 cm) = 0
3. Solve the equation for y:
(100 g) x (21.0 cm) = (-y g) x (80.0 cm)
y g = (100 g) x (21.0 cm) / (80.0 cm)
y g ≈ 26.25 g (rounded to two decimal places)
Therefore, a mass of approximately 26.25 g would have to be suspended at the 80.0 cm position for the system to be in equilibrium.