a uniform meter stick of mass 100g is pivoted about the 30 cm mark. a mass of 200 g is placed at the 10 cm mark. where should a mass of 50 g be placed so that the meter stick balance horizontally?
200•20 +30•15 = 70•35 +50•x
x = 4000+450 -2450/50 = 40 cm ( from the pivot point)
To determine the position where a 50 g mass should be placed so that the meter stick balances horizontally, we need to consider the principle of torque equilibrium.
Torque is the rotational force responsible for causing an object to rotate, and it is given by the formula:
Torque = Force × Distance
In this case, since the meter stick is balanced horizontally, the torques on both sides of the fulcrum point must be equal:
Torque on the left side = Torque on the right side
Now let's calculate the torques:
1. Torque on the left side (due to the 100 g mass at the 30 cm mark):
Torque_left = (100 g) × (distance from fulcrum) = (100 g) × (30 cm)
2. Torque on the right side (due to the 200 g mass at the 10 cm mark):
Torque_right = (200 g) × (distance from fulcrum) = (200 g) × (10 cm)
Since these two torques must be equal for the meter stick to balance horizontally, we can set up the equation:
Torque_left = Torque_right
(100 g) × (30 cm) = (200 g) × (10 cm)
Now, let's solve this equation to find the distance from the fulcrum where the 50 g mass should be placed:
(100 g) × (30 cm) = (200 g) × (10 cm)
Dividing both sides of the equation by 50 g:
(30 cm) / 2 = (10 cm)
15 cm = 10 cm
So, to balance the meter stick horizontally, the 50 g mass should be placed at the 15 cm mark from the fulcrum.