# Find the mass in static equilibrium problem involving a spring

#### sleepymia

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
∑Fx=0
∑Fy=0
∑torque = 0

3. 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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#### haruspex

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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.
I assume you are defining x as the extension. The spring is neither vertical nor perpendicular to the beam.

#### berkeman

Mentor
When the beam is horizontal, the spring is vertical and unstressed.
Welcome to the PF.

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.

#### sleepymia

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.

#### haruspex

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Gold Member
2018 Award
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:
And this is not the right extension:
x = (L/2)tan(22.6)
How long is the spring in the diagram position?

#### sleepymia

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

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#### haruspex

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2018 Award
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?
where Φ is the angle
Ok, but you need to express sin(Φ) in terms of L and θ.

#### sleepymia

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?

#### haruspex

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2018 Award
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?
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.

#### sleepymia

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!!

"Find the mass in static equilibrium problem involving a spring"

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