# Finding the Distance for Static Equilibrium of a Suspended Rod

• flash2
In summary, a uniform .122kg rod of .90 m length is used to suspend two masses (.20kg and .50kg) at different distances from the support point. To achieve static equilibrium, the sum of torques clockwise must be equal to the sum of torques counterclockwise. The mass of the rod matters because it applies a torque on the system, and it can be assumed that the weight of the rod is concentrated at its center. Using this information, the distance at which the .20kg mass should be placed to achieve equilibrium is found to be 0.503 meters, taking into account the torque from the rod.
flash2
A uniform .122kg rod of .90 m length is used to suspend two masses as shown below. At what distance x should the .20kg mass be place to achieve static equilibrium?

Okay so I got the Torque of the 0.2kg mass

T= (0.2)(9.8)(x) with x being the length I hafta find.

Then I got the Torque of the 0.50kg mass

T= (0.5)(9.8)(0.2)

Then I plugged it into this...

sum of Torques CW= Sum of Torques CCW and the answer was 0.5m and I got that but originally for the 0.50 mass, I put in (0.25) for the length because that's what it is but I didn't get teh answer so I did trial and error with the numbers I had and I figured 0.2m was the right length to get the right answer. So I'm wondering why it's 0.2 m and if the mass of the rod matters.. cause I dind't use it.

Well the answer I'm arriving at is .503 meters, or just .5 if you want to keep it simple. When they give you mass of the rod, that usually means it matters, and they make it simple when they tell you it's uniform. Since the mass-density is uniform you can assume that the rod applies all it's weight as though it were concentrated at the length of the rod divided by two, or .45. Since .45 is to the left of the support point, you can say the rod applies a torque of (.122*9.81*.2)

So the equation becomes (.2*9.81*x) + (.122*9.81*.2) = (.5*9.81*.25)
Solving for x yields .986886/(.2*9.81) = .503

When we plug this value into our original summation of torques equation we arrive at 1.22625 = 1.22625 so the bar must now be in static equilibrium.

The mass of the bar matters because even without the weights the bar will tilt in the direction of the longer side.

Oh man I get it so much more now. So is there a torque from the rod exerted on the right side of the balance too? Or does taht even matter?

I know this is prbly really easy but it's extremely hard when your teacher doesn't teach the lesson.

Thanks!

-f

flash2 said:
Oh man I get it so much more now. So is there a torque from the rod exerted on the right side of the balance too? Or does taht even matter?

I know this is prbly really easy but it's extremely hard when your teacher doesn't teach the lesson.

Thanks!

-f
You only need on torque from the rod. You may take all the mass of the rod as if it were located at the center of the rod (assuming the rod has uniform density).

## 1. What is static equilibrium and why is it important for a suspended rod?

Static equilibrium refers to the state when an object is at rest and all the forces acting on it are balanced. It is important for a suspended rod because it ensures that the rod remains stable and does not move or fall due to imbalanced forces.

## 2. How do you find the distance for static equilibrium of a suspended rod?

The distance for static equilibrium of a suspended rod can be found using the principle of moments. This involves finding the sum of all the clockwise moments and the sum of all the counterclockwise moments acting on the rod. The distance for static equilibrium is the point at which these two sums are equal.

## 3. What factors affect the distance for static equilibrium of a suspended rod?

The distance for static equilibrium of a suspended rod is affected by the weight of the rod, the weight of any attached objects, the length of the rod, and the angle at which the rod is suspended.

## 4. How can the distance for static equilibrium of a suspended rod be applied in real-life scenarios?

The concept of static equilibrium and finding the distance for a suspended rod can be applied in various real-life scenarios, such as building structures and bridges, determining the center of mass for objects, and designing cranes and other lifting equipment.

## 5. What are some common mistakes to avoid when finding the distance for static equilibrium of a suspended rod?

Some common mistakes to avoid when finding the distance for static equilibrium of a suspended rod include not taking into account all the forces acting on the rod, not considering the correct direction of the forces, and not using the correct units for calculations.

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