Final answer:
To balance a 5 g beetle sitting at one end of a 30 cm plank, a 10 g grasshopper must be placed 22.5 cm from the left end of the plank to achieve equal torques and balance the system.
Step-by-step explanation:
The scenario you're describing involves the principle of torques in physics, often applied in the context of a seesaw or balance problem. Torque is the product of the force applied and the distance from the pivot point at which the force is applied. To balance two objects on either side of a pivot, the torques must be equal and opposite. This means that the product of the mass and distance from the pivot must be the same for both objects. In your case, a 5 g beetle sits at one end of the 30 cm plank, with the pivot at the center.
In order to balance the plank, the 10 g grasshopper must exert twice the torque of the beetle since it is twice as massive. This means that the grasshopper must sit at half the distance from the pivot point than the beetle. Since the beetle is at the end of the 30 cm plank, and the pivot is in the center, the beetle is 15 cm from the pivot. Therefore, to balance the beetle, the grasshopper should sit 7.5 cm from the pivot. However, since you asked for the distance from the left end, we need to remember that the pivot is 15 cm from each end, so we must add this distance to our answer. Therefore, the grasshopper should be placed at 15 cm + 7.5 cm = 22.5 cm from the left end of the plank.