With the center of a meter stick place on a knife edge balance and a 2N weight hanging on its 100cm mark, where should a 5N weight be placed to keep the meter stick in static equilibrium.
summing moments about the knife edge:
5(50-x)=2(100-50)
where x is the position
50-x=2(10)
x=30cm mark
To determine where the 5N weight should be placed to keep the meter stick in static equilibrium, we need to consider the principle of torque balance. Torque is the rotational equivalent of force, and in static equilibrium, the total torque acting on an object must be zero.
In this case, the torque generated by the 2N weight hanging on the 100cm mark is determined by multiplying the weight by its distance from the fulcrum (knife edge balance). This can be calculated as follows:
Torque1 = 2N * 100cm
To keep the meter stick in static equilibrium, the torque generated by the 5N weight must counterbalance the torque generated by the 2N weight. Let's assume the position of the 5N weight is at distance x from the fulcrum.
Torque2 = 5N * x
Since the meter stick is in static equilibrium, Torque1 and Torque2 must be equal in magnitude but opposite in direction. Therefore, we can set up the equation:
Torque1 = Torque2
2N * 100cm = 5N * x
To find the value of x, we rearrange the equation:
x = (2N * 100cm) / 5N
x = 200cm / 5
x = 40cm
Therefore, the 5N weight should be placed at the 40cm mark on the meter stick to keep it in static equilibrium.