A meter stick is positioned on a horizontal table top such that the 100cm end extends over the edge of the table. A 200 g mass is produced on the 85 cm mark of the meter stick. When the 70 cm mark of the meter stick is on the edge of the table, the meter stick and 200 g mass system is perfectly balanced.
a. What is the mass of the meter stick?
b. What is the magnitude of the normal force (from the table) acting on the meter stick?
How do I solve these two parts?
the center of mass of the stick is at 50 cm , 20 cm from the balance point
a. 15 cm * 200 g = 20 cm * m
b. the normal force equals the weight of the stick and the 200 g mass
... n = g (m + 200 g)
To solve these two parts, we can use the principle of torque equilibrium. The torque is a measure of how effectively a force can rotate an object around a given axis. For a system to be in equilibrium, the sum of the torques acting on it must be zero.
a. To find the mass of the meter stick, we need to consider the torque about the edge of the table when the system is balanced. The torque due to the mass (200 g) can be calculated using the formula:
Torque = Force x Distance
The distance is given by the difference between the position of the mass (85 cm) and the center of mass of the meter stick (50 cm). To balance the system, the torque due to the mass must be equal to the torque due to the meter stick. The meter stick is balanced at the 70 cm mark, so the distance from the edge of the table to the center of mass is 70 cm.
Therefore, we have:
(200 g) x (85 cm - 50 cm) = (mass of meter stick) x (70 cm - 0 cm)
Simplifying the equation:
200 g x 35 cm = (mass of meter stick) x 70 cm
mass of meter stick = (200 g x 35 cm) / 70 cm
mass of meter stick = 100 g
Therefore, the mass of the meter stick is 100 g.
b. To find the magnitude of the normal force acting on the meter stick, we need to consider the forces acting on the meter stick when it is in equilibrium. In this case, the normal force from the table is the force that balances the weight of the meter stick and the mass.
When the system is balanced, the sum of the vertical forces (weight and normal force) is zero. The weight of the meter stick is given by:
Weight = mass x acceleration due to gravity
Weight = (100 g) x (9.8 m/s^2) = 0.98 N
Since the sum of the vertical forces is zero, the magnitude of the normal force is also 0.98 N.
Therefore, the magnitude of the normal force acting on the meter stick is 0.98 N.