Question

a person is balancing a massless meter stick on two fingers, one at the 0m mark and the other at the 1m mark. a soda can of mass 0.015 kg is resting on the 0.25 m mark. how much support force is the person exerting at the 0m mark?

Answer 1

The support force that the person must exert at the 0m mark to balance a 0.015 kg soda can at the 0.25m mark on a massless meter stick is 0.11025 Newtons.

Step-by-step explanation:

The student is asking about the support force exerted at the 0m mark of a massless meter stick, with a 0.015 kg soda can placed at the 0.25m mark. To solve this, we will use the principles of torque and equilibrium. Since the meter stick is massless and in equilibrium, the sum of torques around any pivot point must be zero. We can start by setting the pivot point at the 1m mark, which is where the other finger is placed.

First, we calculate the torque due to the soda can:

At equilibrium, the sum of torques = 0, which means the support force at the 0m mark must produce an equal and opposite torque to the one produced by the soda can. So, the support force (F) at the 0m mark times its distance from the pivot (1m) must equal the Torque (soda can).

Thus, the support force the person must exert at the 0m mark is 0.11025 Newtons.