Equilibrium of Vertical Load P on Rod BC

[ttiger2k7]

Homework Statement

A vertical load P is applied at end B of rod BC. The constant of the spring is k and the spring is unstretched when $$\theta$$ = 0. (a) Neglecting the weight of the rod, express the angle $$\theta$$ corresponding to the equilibrium position in terms of P, k, and l. (b) determine the value of $$\theta$$ corresponding to equilibrium if P = 2kl.

http://img229.imageshack.us/img229/3746/77369190ws9.jpg

Homework Equations

$$\sum F_{x}=0$$

$$\sum F_{y}=0$$

Force of Spring = ks

The Attempt at a Solution

There are three forces acting on the spring: P, $$F_{B}$$,$$F_{C}$$. The force at C has an x component pointing to the left, and a y component pointing up. The force at B is equal to a force of a spring, ks.

I guess where I am running into trouble is when I am trying to solve for the components of the Force at B.

 

[rl.bhat]

By using Lami's theorem you can find the relation between P, ks and theta. By using the properties of triangle you can find the relation between s, l and theta. Try.

 

[ttiger2k7]

Do I need to know the original unstretched length of the spring?

and btw: the correct answers are supposed to be

a) $$tan^{-1} \frac{P}{kl}$$
b) 63.4 degs

 

[rl.bhat]

Along AB force ks acts. So according to Lami's theorem P/sin(angleABC) = ks/sin(angleCBP). Now from properties of triangle s/sin(angleACB) = l/sin(angleBAC). Solve these equations. You will get the result.

 

[ttiger2k7]

deleted

 

[ttiger2k7]

So I am assuming you want me to solve these equations for $$\theta$$, right?

Well, this is what I have:

$$\frac{s}{sin( \theta)} = \frac{l}{sin( \angle BAC)}$$

$$s = \frac {l sin (\theta)}{sin( \angle BAC)}$$

then plug s into:

$$\frac{P}{sin( \angle ABC)} = \frac{ks}{sin( \angle CBP)}$$

to get:

$$\frac{P}{sin( \angle ABC)} = \frac{klsin(\theta)}{sin( \angle BAC) sin( \angle CBP)}$$

and so

$$P sin( \angle BAC) = klsin(\theta)$$

$$\frac{P sin( \angle BAC)}{kl} = sin(\theta)$$

and so, I solve for theta:

$$\theta = sin^{-1} \frac{P sin( \angle BAC)}{kl}$$

which isn't the right answer. Am I doing something wrong, or am I not understanding what you wrote clearly?

 

[rl.bhat]

Angle ABC = 90- theta/2, Angle CBP = 90 + theta, Angle ACB = theta and Angle CAB = 90 - theta/2. Remember that in triangleABC AC = BC. Substitute these values in the equations.

 

[rl.bhat]

In triangle ABC angle ABC =Angle BAC, and Angle CBP = 90 + theta. SO sin(angle CBP) = sin( 90 + theta) = cos(theta). Substitute these values in your fourth step in 6th mail, you will get your answer.

 

[ttiger2k7]

are you sure that sin(90+theta) = cos(theta)? It's very close, but I don't think that's a correct statement.

edit...nevermind. I'm assuming you meant pi/2 instead of 90.

THANKS! I cannot tell you how much of a help you've been.