Find the mass in static equilibrium problem involving a spring

In summary, the object hangs from the beam and has a spring attached to it. The spring has a horizontal component and the beam is at rest. If I define x as the extension of the spring, the spring is not perpendicular to the beam and the spring length is not in the same straight line as the original length of the beam.
  • #1
sleepymia
5
1

Homework Statement


Consider a rigid 2.60-m-long beam (see figure) that is supported by a fixed 1.30-m-high post through its center and pivots on a frictionless bearing at its center atop the vertical 1.30-m-high post. One end of the beam is connected to the floor by a spring that has a force constant k = 1100 N/m. When the beam is horizontal, the spring is vertical and unstressed. If an object is hung from the opposite end of the beam, the beam settles into an equilibrium position where it makes an angle of 22.6° with the horizontal. What is the mass of the object?
12-51.gif

Homework Equations


∑Fx=0
∑Fy=0
∑torque = 0

The Attempt at a Solution


I don't have any forces working horizontally, and I don't think I need to set up an equation for the vertical forces because I don't have the mass of any of the beams. So that leaves me to set up a torque equation:
mgcos(22.6)(L/2) = kx(L/2)
m = (kx)/(gcos(22.6))
I need to solve for x, and when I originally attempted this problem I solved for x like this:
x = (L/2)tan(22.6)
But I didn't get the correct answer for m with this x, and I think it is because I solved for x as if it was perpendicular to the dashed line in the diagram. I am currently stuck trying to find x(change in length of spring from rest)
 

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  • #2
sleepymia said:
don't have any forces working horizontally,
The spring force has a horizontal component, but I agree it is not interesting because there is an unknown horizontal force from the support. Likewise, that is why vertical force balance is uninteresting.
sleepymia said:
kx(L/2)
I assume you are defining x as the extension. The spring is neither vertical nor perpendicular to the beam.
 
  • #3
sleepymia said:
When the beam is horizontal, the spring is vertical and unstressed.
Welcome to the PF. :smile:

It doesn't look to me that the diagram matches this statement. Is the diagram correct (and if so, what is the offset of the spring anchor point), or is the statement correct?

EDIT -- Oops, sorry, it looks like you both are on top of it.
 
  • #4
haruspex said:
The spring force has a horizontal component, but I agree it is not interesting because there is an unknown horizontal force from the support. Likewise, that is why vertical force balance is uninteresting.

I assume you are defining x as the extension. The spring is neither vertical nor perpendicular to the beam.
Yes I am defining x as extension of the spring, and the spring is at a non-right angle to the beam and not vertical since it is not at rest.
 
  • #5
sleepymia said:
Yes I am defining x as extension of the spring, and the spring is at a non-right angle to the beam and not vertical since it is not at rest.
Well, it is at rest, but it is not in the original position.
Since the spring is not perpendicular to the beam, this is not the right torque:
sleepymia said:
kx(L/2)
And this is not the right extension:
sleepymia said:
x = (L/2)tan(22.6)
How long is the spring in the diagram position?
 
  • #6
haruspex said:
Well, it is at fest, but it is not in the original position.
Since the spring is not perpendicular to the beam, this is not the right torque:

And this is not the right extension:

How long is the spring in the diagram position?
You're right it is at rest, my bad just not horizontal. In diagram the spring length I think would be (L/2)+x
Corrected torque of spring kx(L/2)sin(Φ), where Φ is the angle labeled in the drawing below
upload_2017-11-21_22-33-25.png
 

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  • #7
sleepymia said:
In diagram the spring length I think would be (L/2)+x
No. The x is not in the same straight line as the original L/2.
Consider the point where the spring attaches to the beam in relation to its base. What is the horizontal distance? What is the vertical distance?
sleepymia said:
where Φ is the angle
Ok, but you need to express sin(Φ) in terms of L and θ.
 
  • #8
haruspex said:
No. The x is not in the same straight line as the original L/2.
Consider the point where the spring attaches to the beam in relation to its base. What is the horizontal distance? What is the vertical distance?

Ok, but you need to express sin(Φ) in terms of L and θ.
Horizontal distance = (L/2)cos(22.6)
Vertical distance = (L/2)sin(22.6) + (L/2)
Edit: so torque from spring would be the perpendicular force of the spring in relation to the beam multiplied by the horizontal distance I found?
 
  • #9
sleepymia said:
Horizontal distance = (L/2)cos(22.6)
No, that would be from the base of the support to the end of the beam. What is it from the base of the spring?
Having got that, what is the total length of the spring?
sleepymia said:
so torque from spring would be the perpendicular force of the spring in relation to the beam multiplied by the horizontal distance I found?
Yes, or equivalently, the force times the perpendicular distance from the spring to the axis.
 
  • #10
haruspex said:
No, that would be from the base of the support to the end of the beam. What is it from the base of the spring?
Having got that, what is the total length of the spring?

Yes, or equivalently, the force times the perpendicular distance from the spring to the axis.
Ah I understand now, thank you!
 

Related to Find the mass in static equilibrium problem involving a spring

What is the definition of static equilibrium?

Static equilibrium is a state in which all forces acting on an object are balanced, resulting in no net force and no movement.

How does a spring contribute to a static equilibrium problem?

A spring is a common component in static equilibrium problems as its elasticity allows it to exert a force that can balance out other forces acting on an object.

What factors affect the mass in a static equilibrium problem involving a spring?

The mass in a static equilibrium problem involving a spring is affected by the force applied to the spring, the spring's stiffness or spring constant, and the displacement of the spring.

How can the mass be determined in a static equilibrium problem involving a spring?

The mass can be determined by setting up an equation of forces and solving for the unknown mass. This equation will include the force applied to the spring, the spring constant, and the displacement of the spring.

What is the significance of finding the mass in a static equilibrium problem involving a spring?

Finding the mass in a static equilibrium problem involving a spring allows for a better understanding of the forces at play and can help determine the stability and balance of the system. It also allows for the prediction of how the system will behave under different conditions.

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