Static Equilibrium - board problem

[Seraph404]

Homework Statement

A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.0 g stacked over the 12.0 cm mark, the stick is found to balance at the 45.5 cm mark. What is the mass of the meter stick?

Homework Equations

netF = 0
netTorque = 0
The answer is 74 g

The Attempt at a Solution

I know there is an upward force exerted by the knife, a downward force exerted by the weight of the coins, and another downward force exerted by gravity acting on the meter stick itself, which I originally thought was at the .5 mark.

Does there need to be a second upward force at the .445 m mark, or is that where I put my pivot point?

This shouldn't be a hard problem, but I don't think I really understand how the force diagram should look.

 

[tiny-tim]

A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.0 g stacked over the 12.0 cm mark, the stick is found to balance at the 45.5 cm mark. What is the mass of the meter stick?

…

I know there is an upward force exerted by the knife, a downward force exerted by the weight of the coins, and another downward force exerted by gravity acting on the meter stick itself, which I originally thought was at the .5 mark.

Does there need to be a second upward force at the .445 m mark, or is that where I put my pivot point

Hi Seraph!

I think the question is slightly misleading.

It intends you to assume that there is only one knife-edge, and it is moved to the .445 m mark.

So the only force you need worry about at the .5 m mark is the weight of the stick.

 

[Moetasim]

To solve this problem, we first need to understand the concept of static equilibrium. In static equilibrium, an object is at rest and has no net forces acting on it. This also means that the net torque, or rotational force, must also be zero.

In this problem, the meter stick is balanced on a knife-edge, which acts as a pivot point. The forces acting on the meter stick are the upward force from the knife-edge, the downward force from the coins, and the downward force from gravity acting on the meter stick itself.

To find the mass of the meter stick, we can use the equation netF = 0, which means that the sum of all the forces must equal zero. Since the meter stick is balanced, this means that the upward force from the knife-edge must be equal to the downward force from the coins and the meter stick.

We also need to consider the net torque, which is calculated using the equation netTorque = 0. In this case, we need to choose a pivot point to calculate the torque. The most convenient pivot point to use is the knife-edge itself, as it is the point where the meter stick is balanced.

Using the equation netTorque = 0, we can set up the following equation:

(meter stick mass)(distance from pivot to center of mass of meter stick) = (coin mass)(distance from pivot to center of mass of coins)

Solving for the mass of the meter stick, we get:

(meter stick mass) = (coin mass)(distance from pivot to center of mass of coins) / (distance from pivot to center of mass of meter stick)

Substituting in the given values, we get:

(meter stick mass) = (10 g)(12.0 cm) / (45.5 cm - 50.0 cm)

Simplifying, we get a mass of approximately 74 g for the meter stick.

In conclusion, to solve this problem, we need to understand the concept of static equilibrium and use the equations netF = 0 and netTorque = 0. By carefully considering the forces and choosing a suitable pivot point, we can solve for the mass of the meter stick.