Help with Physics

[Tavar]

Okay, I'm taking Physics, and I'm having a bit of an issue with the current subject matter. Specifically, we're dealing with equilibrium of forces/torques. For example:
Calculate FA and FB for the uniform cantilever shown in the figure whose mass is 2450 kg.

http://www.solutioninn.com/images/P-M-EE%20(156).PNG

How do I solve this?

[Cespenar]

I could post the solution, but that probably wouldn't help. So, to guide you a little (and pardon me if I sound off, as I didn't learn this in English):

Equilibrium of forces suggests: FA + FB - W (Weight) = 0

Equilibrium of torques suggests:

Total torque at point A = 0
Total torque at point B = 0

To calculate total torque at point A, ignore FA, look at the other two forces*. Same thing for point B (ignore FB this time, of course). I say ignore because a force has no torque on the point it applies to, since F * 0 = 0. Then check if the equilibrium of forces holds up. A negative value in force means the force is to the opposite direction. Keep in mind which point you take as anchor point when calculating torques, and which direction the forces have to be to validate the equilibrium.

I hope this helps.

*: (FB * 20m) = (W * 25m)

P.S. As the question gives you weight, not force, start the question by converting it.

[Miklus]

OK...Going into physics mode....

CG is the mass multiplied by the gravitational acceleration

2450Kg * 9.82m/s2 = somthing something Newton

The forces have to balance out, otherwise the beam would be accelerating in some direction. I assume that the beam is fixed onto the two pillars.

Fa + Fb + CG = 0 (I)

I guess you have to decide if "plus" is up or down. Let's say plus is down. That might make Fa and Fb negative.

The moment around the middle pillar must be zero too, otherwise the beam would start to rotate. Fb has no arm here. We set zero at the middle pillar and plus to the right.

Fa*(-25m) + Fb*0m + CG*5m = 0 (II)

And the same deal for the pillar on the left

Fa*0m + Fb*20m + CG*25m = 0 (III)

So three equations, that should do it. The first one (I) gives:

Fa + Fb = negative something newton (IV).

And the two other: (by dividing (III) by 5 and subtracting it from (II))

Fa*(-25/4) = Fb (V)

Note here that Fa has opposite sign from Fb. The middle pillar push up on the beam, but the left one pulls it down.

Anyway, substitute (V) into (IV) to get Fa in Newton. Get Fb the same way.

EDIT: Damn, ninja'ed!

EDIT EDIT: DO'H, You can get Fa directly from (II) and Fb from (III). Way to make things overly complicated :smallwink:

[PirateMonk]

I could post the solution, but that probably wouldn't help. So, to guide you a little (and pardon me if I sound off, as I didn't learn this in English):

Equilibrium of forces suggests: FA + FB - W (Weight) = 0

Equilibrium of torques suggests:

Total torque at point A = 0
Total torque at point B = 0

To calculate total torque at point A, ignore FA, look at the other two forces*. Same thing for point B (ignore FB this time, of course). I say ignore because a force has no torque on the point it applies to, since F * 0 = 0. Then check if the equilibrium of forces holds up. A negative value in force means the force is to the opposite direction. Keep in mind which point you take as anchor point when calculating torques, and which direction the forces have to be to validate the equilibrium.

I hope this helps.

*: (FB * 20m) = (W * 25m)

P.S. As the question gives you weight, not force, start the question by converting it.

This is mostly correct, but you should only need to calculate torque at one point. I would suggest A. Also, it technically gives you mass, not weight.

[Tavar]

Thank you, the help was exactly what I was looking for. Of course, now I feel silly for not seeing it earlier. Oh well.

[Cespenar]

This is mostly correct, but you should only need to calculate torque at one point. I would suggest A. Also, it technically gives you mass, not weight.

True, I should have said "either, or".

[Keld Denar]

For future reference, probably the best solution for problems like this is a term called "Superposition". Basically, most books have formula for "base" cases calculated, such as simply supported with a point load at a given location.

Here's a simple demonstration I found on google. (http://www.optics.arizona.edu/optomech/references/OPTI_222/OPTI_222_W12.pdf)

[Sir Enigma]

For future reference, probably the best solution for problems like this is a term called "Superposition". Basically, most books have formula for "base" cases calculated, such as simply supported with a point load at a given location.

Here's a simple demonstration I found on google. (http://www.optics.arizona.edu/optomech/references/OPTI_222/OPTI_222_W12.pdf)

The method you linked to is for dealing with small deformations - the problem here is just static mechanics, with all deformations ignored, so the only things that need to be taken into account are balance of forces and balance of torques.

[Keld Denar]

Ah, true. My bad. I'm just so used to the upper level stuff...

Yea, thats basic statics. Simply resolve sum of forces in the X (none in this case), forces in the Y, and the moments about any of the reactionary points. You might have to solve larger systems of equations, depending on how many variables you have.