A uniform meterstick pivoted at its center has a 80.0 mass suspended at the 27.0 position.
At what position should a 85.0 mass be suspended to put the system in equilibrium?
What mass would have to be suspended at the 90.0 position for the system to be in equilibrium?
6.5
To find the position at which a 85.0 mass should be suspended to put the system in equilibrium, we need to apply the principle of torque balance. The torque is the product of the force applied and the distance from the pivot point.
In this case, the torques on the meterstick due to the masses should balance each other out for the system to be in equilibrium.
Let's denote the distance of the 85.0 mass from the pivot point as x. To maintain the equilibrium, the product of the 85.0 mass and its distance from the pivot point (x) must be equal to the product of the 80.0 mass and its distance from the pivot point (27.0 cm).
Mathematically, we can express this as:
85.0 kg * x = 80.0 kg * 27.0 cm
To solve for x, we can rearrange the equation:
x = (80.0 kg * 27.0 cm) / 85.0 kg
Calculating this expression, we find that x is approximately 25.52 cm.
Therefore, to put the system in equilibrium, a 85.0 mass should be suspended at approximately the 25.52 position.
To find the mass that would have to be suspended at the 90.0 position for the system to be in equilibrium, we can use the same principle of torque balance.
Let's denote the mass that needs to be suspended at the 90.0 position as M. Now we have two torques to balance: one from the 80.0 kg mass at 27.0 cm and the other from the unknown mass (M) at 90.0 cm.
Mathematically, we can set up the equation as:
80.0 kg * 27.0 cm = M * 90.0 cm
To solve for M, we can rearrange the equation:
M = (80.0 kg * 27.0 cm) / 90.0 cm
Simplifying this expression, we find that M is approximately 24.0 kg.
Therefore, to put the system in equilibrium, a mass of approximately 24.0 kg should be suspended at the 90.0 position.