Statics/ equilibrium - angle question

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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 [tex]\alpha[/tex] is the given angle of 60° and [tex]\theta[/tex] 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.

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

[tex] L_{1}Fsin \alpha = \frac{1}{2}(L_{1} + L_{2})W_{B}cos \theta + (L_{1} + L_{2})Wcos \theta[/tex]

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

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Answers and Replies

  • #2
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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.
 
  • #3
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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:

[tex]\Sigma F_{y} = 0 = Fsin \alpha - W_{B}cos \theta - Wcos \theta[/tex]

[tex]cos \theta = \frac{Fsin \alpha}{W_{B} + W}[/tex]

but F cancels out when substituting in the torque equation. Taking the sum of the forces in the x-direction introduces yet another unknown
 
  • #4
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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.
 
  • #5
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[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.
 
Last edited:
  • #6
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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.
 
  • #7
PhanthomJay
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You have to assume that theta is given. There is not enough information to solve for it.
 
  • #8
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Hey jay,

Here's my calculations on this question:

Taking moment about pivot p,

[tex] T l_1 sin60 = W_B \frac{1}{2} (l_1 + l_2)sin\theta + W_l (l_1 +l_2)sin\theta [/tex]

Where theta is the angle between the vertical and the beam.
Simplying,
[tex] \sqrt3 T = 1500sin\theta [/tex]

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:

[tex] T cos(120-\theta) = W_B + W_l [/tex]
[tex] T = \frac{200}{cos(120-\theta)} [/tex]

With 2 equations and 2 unknown:

[tex] (2)(\frac{200\sqrt3}{1500}) = 2 cos(120-\theta)sin\theta [/tex]

Using trigo identities, product to sum:

[tex] \frac{4}{15}\sqrt3 = sin120 = sin(2\theta - 120) [/tex]

[tex] sin (2\theta - 120) = -\frac{7}{30}\sqrt3 [/tex]

[tex] \theta = 48.08 deg, T = 644N [/tex]

I somehow am able to compute an answer out. Is there any mistake i made?
 
  • #9
PhanthomJay
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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.
 
  • #11
PhanthomJay
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Although I haven't checked all your work, I believe the angle theta that you have found is the one and only angle where the vertical reaction force of the wall on the beam is 0. This was based on your assumption, keenly noted, that there is no friction between the wall and beam, which you stated earlier. For every value of theta other than the one you have found, there is a vertical reaction of the wall on the beam, and the problem cannot be solved without additional information.
If the intent of the problem was to assume that the wall is frictionless, the problem should have stated that assumption. Or maybe the angle theta was inadvertently left out of the question? I don't know.
 
  • #12
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Yeah. I assumed that there is no friction so as to allow me to carry on computing the value. Thanks for your input. Guess the question is no very clear.
 
  • #13
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hi guys (gals?) thanks for all the help. the question as is is not solvable. just got an email from the prof saying that the angle between the beam and the horizontal should be 30°, which wasn't stated in the original question.
 

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