# Statics/ equilibrium - angle question

## Homework Statement

Please see attachment. A uniform massive beam attached to a wall at one end and supported by a cable (also attached to wall). On the free end of the beam is a weight. We are supposed to find the tension in the cable.

## Homework Equations

sum of the torques about "pivot" (point where beam meets wall):

## The Attempt at a Solution

where $$\alpha$$ is the given angle of 60° and $$\theta$$ is the angle of the beam to the horizontal. Also W is the weight of the load, WB anf F is the force in the supporting cable.

$$\Gamma = 0 = L_{1}Fsin \alpha - \frac{1}{2}(L_{1} + L_{2})W_{B}cos \theta - (L_{1} + L_{2})Wcos \theta$$

$$L_{1}Fsin \alpha = \frac{1}{2}(L_{1} + L_{2})W_{B}cos \theta + (L_{1} + L_{2})Wcos \theta$$

From there it should be a simpple step to solve for F. The problem is I can't work out $$\theta$$. Is there a way using the information provided?
thanks

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## Answers and Replies

And you have 2 unknown one equation. Algebra tells us that you can't solve the the unknowns.

Think about what kinds of equilibrium there are.

thanks. I guess you mean that it is static equilibrium. still don't see how this helps in this situation.

Taking the x-direction to be parallel to the beam and the y-direction to be perpendicular to this:

$$\Sigma F_{y} = 0 = Fsin \alpha - W_{B}cos \theta - Wcos \theta$$

$$cos \theta = \frac{Fsin \alpha}{W_{B} + W}$$

but F cancels out when substituting in the torque equation. Taking the sum of the forces in the x-direction introduces yet another unknown

Hey,

I thought for a bit about this question. I don't think the question gave us enough information. The most important information is the reaction force from the wall directed at the beam. Does it include any frictional force? If there is, it would change the calculations radically. If not, there is only one horizontal force and life becomes much easier.

[STRIKE]Oh, and is F sin 60degree the vertical component of the force of F?[/STRIKE]

My mistake. Didn't read which direction you took.

But, i would suggest that you do not take directions in relation to the beam. If not you might end up confusing yourself. And you also have to take into account the reaction force from the wall into your calculations.

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Also, you mentioned that F cancels out. Which is good. Now you can solve for theta, then sub is back into the equation to get your tension.

PhanthomJay
Homework Helper
Gold Member
You have to assume that theta is given. There is not enough information to solve for it.

Hey jay,

Here's my calculations on this question:

Taking moment about pivot p,

$$T l_1 sin60 = W_B \frac{1}{2} (l_1 + l_2)sin\theta + W_l (l_1 +l_2)sin\theta$$

Where theta is the angle between the vertical and the beam.
Simplying,
$$\sqrt3 T = 1500sin\theta$$

Then, assuming there is no friction (no upward force), reaction force from wall is perpendicular to wall.

Summation of forces in the y-direction is zero. Therefore:

$$T cos(120-\theta) = W_B + W_l$$
$$T = \frac{200}{cos(120-\theta)}$$

With 2 equations and 2 unknown:

$$(2)(\frac{200\sqrt3}{1500}) = 2 cos(120-\theta)sin\theta$$

Using trigo identities, product to sum:

$$\frac{4}{15}\sqrt3 = sin120 = sin(2\theta - 120)$$

$$sin (2\theta - 120) = -\frac{7}{30}\sqrt3$$

$$\theta = 48.08 deg, T = 644N$$

I somehow am able to compute an answer out. Is there any mistake i made?

PhanthomJay
Homework Helper
Gold Member
Sorry, my computer needs an updated browser, I can't read LaTeX, it shows up black on my screen, so I can't readily spot your error. But I don't see how you can solve for theta....surely there exists a solution to this problem for any angle theta, not just the one you found.

Here's my print screen (:

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PhanthomJay