A meter stick is supported at its center. If a load of 150 dynes is placed at the 20 cm mark, where should a 100 dynes load be placed in order that the stick remains in the horizontal position?

Torque= force • distance

150 • (0.5-0.2) = 100 • x
x= 150•0.3/100=0.45 m (from the center or at the point 0.95 m mark)

To answer this question, we need to understand the principle of torque. Torque is the rotational force that causes an object to rotate about an axis. In this case, the meter stick is being supported at its center, so the axis of rotation is at the center of the stick.

To keep the meter stick in the horizontal position, the torques on each side of the axis must balance each other out. Torque is calculated by multiplying the force applied by the distance from the axis of rotation.

Let's consider the torque created by the 150 dynes load at the 20 cm mark. Since the load is on the right side of the axis, the torque it creates is positive. The torque (τ) is calculated by multiplying the force (F) by the distance (d) from the axis:

τ = F × d

τ = 150 dynes × 20 cm

τ = 3000 dynes.cm

Now, let's find the position where the 100 dynes load should be placed in order to balance the torques. Since the 100 dynes load is on the left side of the axis, the torque it creates will be negative:

τ = -F × d'

Where d' is the unknown distance from the axis.

We want the torques to balance each other, so we can set up the equation:

3000 dynes.cm = -100 dynes × d'

Now we can solve for d':

d' = (3000 dynes.cm) / (-100 dynes)

d' = -30 cm

Therefore, the 100 dynes load should be placed 30 cm to the left of the center to keep the stick in the horizontal position.

Note that the negative sign indicates that the distance is measured from the other side of the axis.