Two identical, massive rods are hinged at one end and connected together by a massless ideal spring.

If the system is in static equilibrium, what is the angle $\theta$ (in degrees)?

**
Details and Assumptions:
**

- Each rod is $1 \text{ m}$ long and has a mass of $1 \text{ kg}.$
- The spring constant is $k = 50 \text{ N/m}$ , and the spring's unstretched length is $1 \text{ m}.$
- The two hinges are $3 \text{ m}$ apart horizontally.
- The strength of gravity is $10 \text{ m/s}^2$ in the downward, vertical direction.

The answer is 26.107.

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Nice approach. Another option is a brute parameter sweep.

Steven Chase
- 3 years, 7 months ago

You can also use the well-known Newton's Method for finding roots of any function $f(x)$ . It requires finding the derivative of the function which is very simple for this particular function. It is very fast and efficient, provided the initial guess is "close enough" to the actual root.

Hosam Hajjir
- 3 years, 7 months ago

Question on the Solution posted, is not the net displacement of the spring caused by forces acting on both rods. But the moment of the force has been taken with respect to only one rod. What has been done is
(weight of the left hand side rod * length of the rod /2 * cos(theta)) = force on spring * sin(theta) * length of the rod. Should it not be calculated as half the displacement due to tension on the left hand side rod
(weight of the left hand side rod * length of the rod /2 * cos(theta)) =

force on spring * sin(theta) * length of the rod/2.

Srinivasa Gopal
- 3 years, 6 months ago

In the equation above take cos(theta) as t , sin(theta) is sqrt(1-t
*
t).
So (g/2) * t = 100 *(1-t)
*
(sqrt(1-t
*
t) or 5t = 100(1-t)
*
sqrt(1-t
*
t). or t = 20
*
(1-t)
*
sqrt(1-t
*
t)
squaring both sides
t
*
t = 400 * (1- 2
*
t + t
*
t) *(1-t
*
t) ; t
*
t = 400
*
(1-t
*
t -2
*
t + 2
*
t
*
t
*
t +t
*
t -t
*
t
*
t
*
t);
400
*
t

Srinivasa Gopal
- 3 years, 6 months ago

sorry, I got confused, why is it g/2 on the left?

Licheng Zhang
- 3 years, 6 months ago

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The vertical line that goes through the center of the mass of the rod is at cos(t)/2 meters distance from the hinge

Laszlo Kocsis
- 3 years, 6 months ago

I have come to the same result with an energetic approach: spring's potential elastic energy - rods' potential gravitational energy. Imposing that the derivative equals zero (minimum energy) I've come to your formula

Filippo Cona
- 3 years, 6 months ago

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I think that would be interesting to see as a formal solution. Would you please post it?

Steven Chase
- 3 years, 6 months ago

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Hp1: gravitational potential energy Ug = 0 when rods are horizontal Hp2: elastic potential energy Ue = 0 when spring is at rest (1m long) In the following L = 1m

(1) Ug = -2
*
m
*
g
*
L
*
sin(theta)/2 --> formula is m
*
g
*
h, - sign because rods' CoMs move downwards, 2* because the rods are two, L
*
sin(theta) is the vertical coordinate of the lowest tips of the rods, /2 because rods' CoMs are at half the rods's lenghts)
(2) Ue = 1/2
*
k
*
(3
*
L-2
*
L
*
cos(theta)-L)^2 --> formula is 1/2
*
k
*
DELTAx^2, DELTAx is the spring lenght minus its length at rest (-L), the spring length is the hinges distance (3
*
L) minus two times the horizontal projection of the rods (2
*
L*cos(theta))

Now the system will go at rest in the point of minimum total energy, which gives (a little algebra applied with respect to equations (1) and (2))
(3) Utot = Ug+Ue = 2
*
k
*
L^2
*
(1-cos(theta))^2-m
*
g
*
L
*
sin(theta)

The minimum energy can be found by imposing the first derivative of Utot with respect to theta is equal to 0:
(4) partial(Utot)/partial(theta) = -4
*
k
*
L^2
*
(1-cos(theta))
*
sin(theta)-m
*
g
*
L
*
cos(theta) = 0
Thus,
(5) m
*
g
*
cos(theta) = -4
*
k
*
L
*
(1-cos(theta))*sin(theta)

Substituting the numerical value it can be turned into the recursive formula (6) theta = acos(1-(1/20*tan(theta))) even if I preferred to solve directly (5) numerically

Filippo Cona
- 3 years, 6 months ago

Could you show the analysis on how you solved the problem?

Guiseppi Butel
- 3 years, 6 months ago

With an angle of 26 degrees I compute the tension on the spring to be in the neighbourhood of 5.1 Nt. Is this sufficient to hold the bar of weight 9.8 Nt in this position??

Guiseppi Butel
- 3 years, 6 months ago

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The hinge can supply any forces needed to keep the rods in stasis. But the hinge cannot supply moments to the rods. That is why the gravity and spring moments have to sum to zero.

Steven Chase
- 3 years, 6 months ago

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What would the solution to this problem be if the spring constant k is changed to 10 N/m?

Guiseppi Butel
- 3 years, 6 months ago

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@Guiseppi Butel – With k = 10, the angle is 42.974 degrees

Steven Chase
- 3 years, 6 months ago

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@Steven Chase – I may be dense but I would like to see the actual figures that justify your answer of 26 degrees.

Guiseppi Butel
- 3 years, 6 months ago

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@Guiseppi Butel – I put up a solution. Hopefully that will help

Steven Chase
- 3 years, 6 months ago

Is not the net displacement of the spring caused by forces acting on both the rods. But in the analysis above the moment of the weight of the rod and the force in the spring (KX) have been taken with respect to one of the hinges. Ie., (weight of the left hand side rod * length of the rod /2 * cos(theta)) = force on spring * sin(theta) * length of the rod. If the extension in the spring is considered to be caused due to a component of the weight of the both the rods then should the force balancing equation not be (weight of the left hand side rod * length of the rod /2 * cos(theta)) = force on spring * sin(theta) * length of the rod/2.

Srinivasa Gopal
- 3 years, 6 months ago

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There is no force balancing equation in this problem, because the hinge can supply any necessary forces for stasis. We need only consider torque balance.

Steven Chase
- 3 years, 6 months ago

why is the g on the left is divided by 2?

Daniel Dan
- 3 years, 6 months ago

i don't think this is correct. the extension on two ends of the spring are equally on two different directions, so hookes' law can't be applied blindly on the entire extension, but on the extension per side. so the restoring force on the spring is k.X/2, where X is the total extension. so the governing equation is 1/10 = [1 - cos(theta)]
*
tan(theta) instead of 1/20 = [1 - cos(theta)]
*
tan(theta). therefore, theta = 32.5 degrees.

Bedabrata Pain
- 3 years, 6 months ago

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Springs will
*
always
*
apply equal and opposite forces to their two ends (Newton's third law). Hooke's Law gives the magnitude of this force (same for both ends) given the total extension of the spring.

Samuel Li
- 3 years, 5 months ago

11 Helpful
0 Interesting
0 Brilliant
0 Confused

Yes, I ended up with the same expression

Steven Chase
- 3 years, 7 months ago

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Okay. Then Am Correct

Md Zuhair
- 3 years, 7 months ago

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Yes, I got that equation too. Using the "Table" of the graphics calculator, it seems the answer is just under 26.1 though.

Bob.

Robert Nelder
- 3 years, 7 months ago

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@Robert Nelder – Hmm.. thats quite a tough equation to solve

Md Zuhair
- 3 years, 7 months ago

How do you isolate theta?

Joseph Chang
- 3 years, 6 months ago

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???? i didnt get that :P

Md Zuhair
- 3 years, 6 months ago

Got the same problem

Jonathan Gun
- 3 years, 6 months ago

I got the correct formula but there is no way to solve the problem analitically but only numerically thus it is not good problem for Briliant!

Marian Kupczynski
- 3 years, 6 months ago

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Why impose that restriction?

Steven Chase
- 3 years, 6 months ago

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Maybe there shouldn't be such a restriction, but it would be a good thing to mention.

Joni-Pekka Luomala
- 3 years, 6 months ago

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@Joni-Pekka Luomala – I agree that it would be nice if the problem poster says when a problem will need some numerical processing, because otherwise one may end up spending a lot of futile effort trying to solve the thing analytically. To avoid that kind of issue I resolved myself into solving numerically the problems as soon as the analytical solution isn't so obvious, which doesn't actually meet my personal taste...

Filippo Cona
- 3 years, 6 months ago

I got cot (theta)=10 (1-cos((theta)), so where am I wrong....

Xinzhe Ren
- 3 years, 6 months ago

The hinge can supply reaction forces to satisfy the force equilibrium requirements. However, the hinge cannot supply any moments. Thus, the gravity and spring moments must cancel each other:

**
Spring Length
**

$l = 3 - 2 cos\theta$

**
Spring Stretch:
**

$s = l - 1 = 2 - 2 cos\theta$

**
Spring Force:
**

$F_s = k s = 100(1 - cos\theta)$

**
Spring Moment:
**

$M_s = F_s * sin\theta = 100sin\theta (1 - cos\theta)$

**
Gravity Force:
**

$F_g = 1 \times g = 10$

**
Gravity Moment:
**

$M_g = F_g \frac{1}{2} cos\theta = 5 cos\theta$

**
Equating the two:
**

$100sin\theta (1 - cos\theta) = 5 cos\theta \\ cot\theta = 20(1 - cos\theta)$

Solving this equation (by whatever means) for the angle yields an answer of approximately 26.107 degrees

4 Helpful
0 Interesting
0 Brilliant
0 Confused

it'd be, i think, wrong to use the total elongation of the spring to compute the spring force, since the stretch happens in two opposite directions. the restoring force on each side is proportional to the elongation on that side (and of course, in equilibrium, these two forces are equal and opposite). therefore the spring force must be 50
*
(1 - cos(theta)) and 100
*
(1 - cos(theta)). this means the equilibrium angle would be 32.5 degrees and not 26.1.

Bedabrata Pain
- 3 years, 6 months ago

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sorry - i meant NOT 100*(1 - cos(theta))

Bedabrata Pain
- 3 years, 6 months ago

The spring force on each side is a function of the total elongation

Steven Chase
- 3 years, 6 months ago

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Simple trigonometry gives the length of the spring as $3-2\cos{\theta}$ , and so the extension is $2-2\cos{\theta}$ , and the tension in the spring is $100(1-\cos{\theta})$ .

Now consider (say) the left hand rod. Since it is in equilibrium the total turning moment about the hinge must be zero. The only relevant forces are the weight of the rod, acting vertically through the centre of gravity, and the tension force in the spring which acts horizontally. Simple trigonometry gives the length of the lever arm for each of these forces, and we find

$\frac{g}{2}\cos{\theta}=100(1-\cos{\theta})\sin{\theta}$

Substituting in $g=10$ and rearranging gives

$\theta=\cos^{-1}\left(1-\frac{1}{20\tan\theta}\right)$

I solved this quickly by starting at $\theta=50$ and iterating this formula. it quickly converges to

$\boxed{26.107}$

note.

I tried three other rearrangements of the formula first, which did not converge. However this numerical approach is so easy that even with these false starts it is (at least for me!) much faster than finding an exact solution using trig and algebra!

With a pocket calculator just press "50=" then enter $\cos^{-1}\left(1-\frac{1}{20\tan{ANS}}\right)$ and keep tapping "=" until the numbers converge!