# Finding equation of equlibrium for rod and two blocks

1. Jul 3, 2009

### KEØM

1. The problem statement, all variables and given/known data
A slender rod AB of weight W is attached to blocks A and B which move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when $$\theta$$ = 0. (a) Neglecting the weight of the blocks, derive an equation in W, k, l, and $$\theta$$ which must be satisfied when the rod is in equilibrium. (b) Determine the value of $$\theta$$ when W = 4.5 lb, l = 30 in., and k = 1.8 lb/ft.

Here is a picture of the problem. It is number 4.60

2. Relevant equations
$$\Sigma F_{x} = 0$$

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

$$\Sigma M_{B} = 0$$

3. The attempt at a solution

$$F_{s} = ks = lsin(\theta)$$

$$\Sigma M_{B} = lsin(\theta)(klsin(\theta) - lcos(\theta)A_{y} + W(l/2cos)(\theta) = 0$$ where $$A_{y}$$ is the reaction force at A.

$$\Sigma F_{y} = A_{y} - W = 0$$

$$\Sigma M_{B} = lsin(\theta)(klsin(\theta) - lcos(\theta)W + W(l/2cos)(\theta) = 0$$

$$\Sigma M_{B} = l^2sin^2(\theta)k - (l/2)cos(\theta)W = 0$$

$$\frac{2lk}{W} = \frac{cos(\theta)}{sin^2(\theta)}$$

After inverting both sides and using the identity $$sin^2(\theta) + cos^2(\theta) = 1$$

$$sec(\theta) - cos(\theta) = \frac{W}{2lk}$$

The book's answer is $$tan(\theta) - sin(\theta) = \frac{W}{2lk}$$

Can someone tell me where I went wrong?

KEØM

Last edited by a moderator: May 4, 2017
2. Jul 3, 2009

### LowlyPion

I approached it a little more simply using the Tension in the arm L.

At equilibrium you have in the vertical direction

mg = T*sinθ

In the horizontal direction

kx = T*cosθ

Noting that x = (1 - cosθ)*(L/2)

then solving. I think that yields the answer in the book.

3. Jul 3, 2009

### KEØM

Thanks for the reply LowlyPion. I have a couple of questions. When I solve your equations I get:
$$tan(\theta) - sin(\theta) = \frac{2W}{kl}$$.

I have probably done something wrong though.

I don't understand how you got your x. Your equation is telling me that it is half the distance of the rod minus the horizontal distance that half of the length of the rod covers. I don't understand how that gives you x.

One more question: Shouldn't your forces in the x direction be $$kx = -Tcos(\theta).$$
Or shouldn't one of your equations have a negative?

Thanks again,

KEØM

4. Jul 3, 2009

### LowlyPion

Isn't W acting at L/2 along the rod? It just seems to me that you would want to take the tension from that point to A.

As to the minus sign, the Tension is oppositely directed at A and L/2, I just took the direction that they seemed to be naturally acting.