Pivoting beam supported by a spring - equilibrium

[Nihuepana]

Homework Statement

I have the following setup

http://yfrog.com/f1opgave4j

The beam turns without friction around the pivot support and is supported by a vertical, massless spring. When in equilibrium the spring is compressed by the the length b.

My target variables are the spring constant, k and the forces from the support on the beam during equilibrium.

Homework Equations

I state that the beam is at an angle,$$\theta$$ with horizontal.
I suppose that the system is in a static equilibrium and I'm using the second equilibrium condition.

The Attempt at a Solution

I find my angle
$$\theta=\arcsin \left( {\frac {b}{L}} \right)$$

First off I have my second equilibrium condition

$$F_{{{\it elastiv},{\it vertical}}}\cos \left( \theta \right) L+1/2\,W \cos \left( \theta \right) L+F_{{{\it elastic},{\it horizontal}}}b=0$$

I can see that the spring necessarily must follow the end of the beam in the x direction, which is why I conclude that it attacks the end of the beam with a force in the x direction that will have it's opposite counterpart at the pivot.

I'm also pretty sure that if there had been no deflection in the direction for the spring, then the force would have been -k*b and I could easily have found k. As it is I have no idea how the spring will act when it bends or how to get further on with the problem in general, so I really hope you guys can help me.Simon

P.S
The image won't show unless it's opened in a new tab - must be doing something wrong :(

 

[mark726]

Hello Simon,

Thank you for sharing your setup and equations. It seems like you have a good understanding of the problem and have made some progress in finding the angle and setting up the equilibrium conditions.

To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the amount it is stretched or compressed. In this case, we can assume that the spring is only experiencing a vertical displacement, so we can use the vertical equilibrium equation you have set up to solve for the spring constant:

F_{{{\it elastic},{\it vertical}}}= -k\Delta y

Where Δy is the change in length of the spring, which in this case is equal to the compression length b. So we have:

F_{{{\it elastic},{\it vertical}}}=-kb

We can then substitute this into the vertical equilibrium equation and solve for k:

-kb\cos \left( \theta \right) L+1/2\,W
\cos \left( \theta \right) L+F_{{{\it elastic},{\it horizontal}}}b=0

Solving for k:

k= \frac {1/2\,W\cos \left( \theta \right) L+F_{{{\it elastic},{\it horizontal}}}b}{b\cos \left( \theta \right) L}

To find the forces from the support on the beam during equilibrium, we can use the horizontal equilibrium equation you have set up. This will give us the force at the pivot and the force at the end of the beam:

F_{{{\it elastic},{\it horizontal}}}=-F_{{{\it support},{\it pivot}}}

F_{{{\it elastic},{\it horizontal}}}= -F_{{{\it support},{\it end}}}

We can then substitute these values into the equations and solve for the forces at the support.

I hope this helps you with your problem. Let me know if you have any further questions or need clarification on anything. Keep up the good work!