Equilibrium: What is the Minimum Spring Constant for a Secure Ladder Angle?

[Hernaner28]

Homework Statement

I have that situation in which I'm told that the blocks each support a maximum force of 20N. Neglect the natural state of the spring. And I'm asked to find the MINIMUM value of the spring constant K for the man to be secure with an angle $$\displaystyle \theta =30{}^\text{o}$$.

This is known data:
$$\displaystyle \begin{align}<br /> & m=70,0kg \\ <br /> & L=3,00m \\ <br /> \end{align}$$

Homework Equations

Newton and Torque.

The Attempt at a Solution

What I did first was to work with half of the problem (only one side of the ladder) and worked with the half of the man weight as well.
So I wrote:

NEWTON:
$$\displaystyle \sum{{{F}_{y}}=\frac{-mg}{2}+{{N}_{1}}\Rightarrow {{N}_{1}}=\frac{mg}{2}}$$

Where N1 is the vertical normal.

The equation in axis x is irrelevant.

TORQUE:

$$\displaystyle \sum{\tau }={{N}_{2}}\cos 15{}^\text{o}H+F\cos 15{}^\text{o}\frac{H}{2}-{{N}_{1}}\sin 15H=0$$

Where N2 is the horizontal normal caused by the block, H is the distance of the ladder:
$$\displaystyle H=\frac{L}{\cos 15{}^\text{o}}$$

and F is the force done by the spring. So:
$$\displaystyle F=kx$$
where x is:

$$\displaystyle x=\frac{H}{2}\sin 15{}^\text{o}$$

Solving for K I get the minimum value should be 360 N/m but that's not the correct option. The one marked as correct is 179N/m. What was my mistake?

Thanks!

 

[ehild]

$$\displaystyle x=\frac{H}{2}\sin 15{}^\text{o}$$

The force depends on the length of the whole spring, twice the value you calculated with.


ehild

 

[Hernaner28]

I didin't understand what you've said. It says to neglect the equilibrium position of the spring so I suppose that the equilibrium position is at the middle of the ladder (on the vertical that divides the ladder in two halfs). Therefore, the distance the spring streched of one side is H/2*sin(120º)Thanks!

 

[ehild]

I didin't understand what you've said. It says to neglect the equilibrium position of the spring so I suppose that the equilibrium position is at the middle of the ladder (on the vertical that divides the ladder in two halfs). Therefore, the distance the spring streched of one side is H/2*sin(120º)


Thanks!

The force a spring exerts (at both ends) is kΔx. You need to find the spring constant of the whole spring. If you consider the spring as two halves, you get the spring constant of a half spring.

ehild

 

[Hernaner28]

But I thought that that was taken into account when I considered half of the man mass. Why it has nothing to do?

 

[ehild]

From the half of the weight of the man you got the force the spring exerts on one leg of the ladder. The same force is exerted on the other leg.

The spring constant is defined as the force exerted on the spring divided by the change of length. k=F/x. H/2 sin(15) is not x. It is only half length of the spring.

 

[Hernaner28]

Ahhh I get it now! Thank you!

 

[ehild]

You are welcome. And the picture was amazing!

ehild