Static Equilibrium - board problem

In summary, the question asks for the mass of a meter stick that balances horizontally on a knife-edge at the 50.0 cm mark, with two 5.0 g coins stacked over the 12.0 cm mark and the knife-edge moved to the 45.5 cm mark. The solution involves considering the forces of the knife-edge, the weight of the coins, and the weight of the stick itself, with the assumption that there is only one knife-edge at the 45.5 cm mark.
  • #1
Seraph404
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0

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.
 
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  • #2
Seraph404 said:
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! :smile:

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. :smile:
 
  • #3


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.
 

1. What is static equilibrium in a board problem?

Static equilibrium in a board problem refers to the state where all the forces acting on a board are balanced, resulting in no net force or acceleration. This means that the board is not moving and is in a state of rest.

2. How do you determine if a board is in static equilibrium?

To determine if a board is in static equilibrium, you must analyze the forces acting on the board and their directions. If the sum of all the forces acting on the board is equal to zero, then the board is in static equilibrium. This is known as the first condition of equilibrium.

3. What is the difference between static and dynamic equilibrium?

Static equilibrium refers to a state of balance where there is no net force or acceleration, meaning the object is at rest. On the other hand, dynamic equilibrium refers to a state of balance where the object is moving at a constant velocity. In both cases, the sum of all forces acting on the object is equal to zero.

4. How do you calculate the center of mass for a board in static equilibrium?

To calculate the center of mass for a board in static equilibrium, you must first determine the individual masses of each part of the board. Then, you can use the formula: center of mass = (m1x1 + m2x2 + ... + mnxn) / (m1 + m2 + ... + mn), where m represents the mass of each part and x represents the distance from each part to a chosen reference point.

5. What are some common examples of static equilibrium in everyday life?

Some common examples of static equilibrium in everyday life include a book sitting on a table, a pencil balanced on its tip, and a ladder leaning against a wall. In each of these cases, the forces acting on the object are balanced, resulting in a state of rest.

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