Final answer:
To determine the weight of the meter stick, we used torque balance, setting the counterclockwise torque (the weight of the stick acting 15 cm from the pivot) equal to the clockwise torque (0.85N acting 29.5 cm from the pivot). Solving for the weight, we found it to be 1.765N.
Step-by-step explanation:
To compute the weight of the meter stick, we use the concept of torque balance. Torque is the product of the force and the distance from the pivot. For a system in equilibrium, the clockwise torques equal the counterclockwise torques. The weight of the meter stick acts at its center of mass, which is at the 50 cm mark if it's uniform.
Let w be the weight of the meter stick. To balance the 0.85N weight at the 5.5cm mark, we have:
Counterclockwise torque (due to the weight of the stick) = w x (50cm - 35cm)
Clockwise torque (due to the 0.85N weight) = 0.85N x (35cm - 5.5cm)
Setting the torques equal for balance:
w x 15cm = 0.85N x 29.5cm
Now we solve for w:
w = (0.85N x 29.5cm) / 15cm
w = 1.765N
The weight of the meter stick is 1.765N.