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.