How Do You Calculate the Forces and Tension in a Supported Beam System?

[jrrodri7]

A floodlight with a mass of (19,0 Kg) is supported at the end of a horizontal beam that's hinged to a vertical pole, as shown. A cable that makes an angle of (27 degrees) with the beam is attached to the pole to help support the floodlight. Assume the mass of the beam is negligible when compared with the mass of the floodlight. Find Tension provided by the cable. (Answer in N). Find the horizontal force exerted on the beam by the pole? Find the Vertical Force exerted on the beam by the pole.

http://img186.imageshack.us/img186/1382/staticproblemquestionmd3.png

Relevant equations
F = ma = 0
T = I (alpha) = 0


Here's what I tried doing, I set the x,y components of F to 0 to find a T from the normal force, gravity and the beam, set them all to 0.

F = -Tsin(63){x-direction} + Tcos(63){y-direction} = 0, and I tried solving for T? I then realized that like this is if there was a pin in the beam, and the beam had mass, so that can't be correct.

I'm just asking for some direction with some values...? HELP?

 

[Shooting Star]

There is tension in the string, the vertical component of which supports the weight, and the horizontal component exerts a force on the beam. The pole in turn exerts an equal and opposite reaction on the beam.

The weight is given, and you should be able to find the tension.

 

[tiny-tim]

… one direction at a time …

F = -Tsin(63){x-direction} + Tcos(63){y-direction} = 0, and I tried solving for T? I then realized that like this is if there was a pin in the beam, and the beam had mass, so that can't be correct.

Hi jrrodri7!

No no no …

F = ma is a vector equation.

That means it's a different equation for each direction.

So your F = 0 must be done for one direction at a time.

(You've done both directions in the same equation, and also ignored the other forces.)

You can choose the x-direction, the y-direction, the direction of the cable, or the direction perpendicular to the cable.

Which directions make the equations easier to solve?

 

[jrrodri7]

Alright, here's what I attempted.
I set,

m = 19 kg
M = mass of beam (unknown)
T = tension

F_y = mg - Tsin(63)

F_x = Mg - Tcos(63)

I set them both to zero, and just did substitution of systems of equations.

I ended up with, but I'm missing something I think...I'm not sure what though...

T = 209.4 N
M = 9.7 kg

 

[tiny-tim]

F_y = mg - Tsin(63)

F_x = Mg - Tcos(63)

I set them both to zero, and just did substitution of systems of equations.

Hi jrrodri7!

Your F_y equation for the floodlight looks superficially correct (except you've got confused between 63º and 27º - whyever did you change them over?), but you haven't included the force on the floodlight from the beam (let's call that force B).

The main problem is, we don't know the direction of B, nor the direction of the force H on the beam from the hinge with the pole.

(We only know the direction of T because a string can only have force along its length - that's called tension- but a beam can have sideways forces also.)

So the first thing is not to do a ordinary force equation for the floodlight, but a torque equation for the beam.

(btw, the question tells you the mass of the beam is negligible, which means M = 0.)

Take torques (moments) about the hinge, for forces on the beam.

What do you get?

Then do an ordinary force equation for the beam (which direction?) to get H.

 

[jrrodri7]

Alright, The moment's obviously \frac{1}{3}ML^{2} , there's no R or \alpha Along with the original Force equations, they would stay the same, but there'd be another factor to put into play, most likely a substitution.

I do know that \alpha = \frac{a}{R} or L in this case. Do i use g as the acceleration in this case and apply it as a \frac{g}{R} instead of \alpha?

 

[tiny-tim]

Alright, The moment's obviously \frac{1}{3}ML^{2} , there's no R or \alpha Along with the original Force equations, they would stay the same, but there'd be another factor to put into play, most likely a substitution.

I do know that \alpha = \frac{a}{R} or L in this case. Do i use g as the acceleration in this case and apply it as a \frac{g}{R} instead of \alpha?

Hi jrrodri7!

I don't understand any of this.

M = 0, from the question.

There's no acceleration: this diagram is static.

And I don't know what alpha a R or L are.

Look, this is very simple - the beam has only three forces acting on it: W (the weight of the floodlight), T (the tension) and H (the reaction from the hinge).

So draw a diagram (roughly) showing these three forces, where they apply to the beam, and their directions.

Then take the torque of those three forces about a sensible point, and also their x-components and their y-components (separately).

What do you have?

 

[jrrodri7]

Hi jrrodri7!

I don't understand any of this.

M = 0, from the question.

There's no acceleration: this diagram is static.

And I don't know what alpha a R or L are.

Look, this is very simple - the beam has only three forces acting on it: W (the weight of the floodlight), T (the tension) and H (the reaction from the hinge).

So draw a diagram (roughly) showing these three forces, where they apply to the beam, and their directions.

Then take the torque of those three forces about a sensible point, and also their x-components and their y-components (separately).

What do you have?

I decided to pick the point where the light is hanging from the beam, from there, I plotted those 3 forces (W, T, and H). W facing South, T facing (180 - 27) from horizontal (the right) and the force H, I presume is perpendicular to the weight of the floodlight.

From there, sensing that R what the length is supposed to be, and alpha being 0, so I can assume the torque is 0, or the sums are at least, because it is static correct, along with the mass of the beam being negligible. SO, Leaving me with, the sum of the torques = 0, and i'll be left using the magnitudes of torque = rFsin(27) where F is my resultant force vector, which will probably be either a cross product OR it will just the tension of the wire.

 

[jrrodri7]

If that doesn't work I COULD do something from point of the beam that's against the wall, but I don't have really any relative evidence given about the beam other that it doesn't weigh anything, no length, nada, so I thought the prior was the better of the two choices, even though the answer would still be the same if worked out correctly.

 

[tiny-tim]

I'm going to sleep now, so this has to be short. :zzz:

the force H, I presume is perpendicular to the weight of the floodlight.

No it isn't - H can be any direction.

It's at a hinge!

torque = rFsin(27) where F is my resultant force vector, which will probably be either a cross product OR it will just the tension of the wire.

I don't understand. Whatever is F?

 

[jrrodri7]

F is the magnitude of the Force causing the torque, but I did resolve to attain something form the H idea.

Fy = Tsin - mg
Fx = H - Tcos

?

 

[tongpu]

just do sum of the moments or torques at the bottom of the pole to get one equation one unknown consisting of Tension...

 

[jrrodri7]

OOOOOH! you set two separate x and y coords, i thought it was combining the single system, but you substituted that for what you have there. I did get the 410.4 numeral though through the mg/sin for T_2. Thanks.

 

[tron_2.0]

no problemo, if the thanks was directed towards me. I am mildly intoxicated atm.

 

[jrrodri7]

ya you and tinytim