Static determinancy (lifting mechanism)

[louza8]

Homework Statement

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Homework Equations

Sum Horizontal forces = 0
Sum vertical forces = 0
Sum moments = 0
Sum Horizontal Forces=0

The Attempt at a Solution

Dx-Ax+Bx+Cx=0
Dx-Ax+Bx+981
Sum Vertical Forces
Dy+Ay-By=0
Sum Moments=0
Ma+Md=0
Ma=-Md
Ma=Dx(0.5)+By(0.8) +981(0.1)

D and A are fixed, compute reaction forces at B. I don't know if I have the Cx bit right about how it distributes forces within the structure. I seem to have too many variables and not enough equations. Apologies for being a newbie at this stuff. Thanks in advance.

 

[louza8]

If somebody could help to set up the equations properly or explain where I am lacking understanding I will happily do the rest.

 

[I like Serena]

Well, usually we start with looking "only" at the external forces.
Those have to be in equilibrium, meaning all your equations apply.
Note that the internal forces can be ignored (for now), because each internal force will be countered by an equal and opposite reaction force.

Then we break the thing up in the individual rigid bodies.
For each individual rigid body all equilibrium equations apply as well.

So let's start with the external forces.
These are:
Ax, Ay, Dx, Dy, Wy.

With Wy I mean the downward gravity force on the 100 kg mass, which is 981 N.

If we look at the moment sum relative to A this is (depending on your choices of directions):

Edit back again (*smacks forehead*):

∑M_A = 0; Dx x 0.5 - Wy x 1.7 = 0

Can you see how this yields Dx?

 

[louza8]

Thanks I like Serena! I can see how the equation yields Dx but I am still lacking some understanding if you would be so kind?

Why are the forces Ax, Ay, Dx, Dy, Wy considered external forces? By this I mean how do I identify what constitutes an internal force and what constitutes an external force in this instance?

So with the Moment sum relative to A, I can see that this moment sum is a result of taking into account only the external forces listed. What would the moment equation about D be? ∑MD = 0; Ax*0.5+Wy*1.7=0 (?)

So knowing the provided moment equation:

∑MA = 0; Dx * 0.5 - Wy * 1.7 = 0 where Wy=100kg*9.81=981N
Dx=3335.4N
∑Fx = 0; Dx + Ax
Ax = -3335.4N
∑Fy = 0; Wy + Ay + Dy = 0
981N+Ay+Dy = 0

How do I find either Ay or Dy? Moment about C or something? Is C a part of the 'internal forces'

Thanks so much for your help!

 

[I like Serena]

Thanks I like Serena! I can see how the equation yields Dx but I am still lacking some understanding if you would be so kind?

Why are the forces Ax, Ay, Dx, Dy, Wy considered external forces? By this I mean how do I identify what constitutes an internal force and what constitutes an external force in this instance?

Your system is basically built up from 5 components: section AC, section BD, the pulley at C, the 100 kg mass, and the rope.

The wall is considered outside of your system, and the Earth (exerting gravity) as well.
The external forces (in this case) are the forces exerted by the wall (force at A and force at D) and by gravity (force on the mass).

An internal force is the force that one component exerts on another component of the system.
For instance, section AC exerts a force on section BD at point B.
And as the 3rd law of Newton states: action = - reaction.
So section BD exerts exactly the same but opposite force on section AC at point B.

So with the Moment sum relative to A, I can see that this moment sum is a result of taking into account only the external forces listed. What would the moment equation about D be? ∑MD = 0; Ax*0.5+Wy*1.7=0 (?)

So knowing the provided moment equation:

∑MA = 0; Dx * 0.5 - Wy * 1.7 = 0 where Wy=100kg*9.81=981N
Dx=3335.4N
∑Fx = 0; Dx + Ax
Ax = -3335.4N
∑Fy = 0; Wy + Ay + Dy = 0
981N+Ay+Dy = 0

How do I find either Ay or Dy? Moment about C or something? Is C a part of the 'internal forces'

Thanks so much for your help!

As yet, you can not find Ay or Dy from the external forces diagram.

However, if you take e.g. only section BD and check your equations, you'll find you can find Dy straight away.

Btw, it will take awhile before I can give another response, since I have to run.
But perhaps some one else will pick up where I left off.

 

[louza8]

Ahh I'm starting to grasp it a little better now.

So for section BD, I realize the x and y forces exerted at D must be equal and opposite to the forces at B. So therefore I separate section BD.

I find using trigonometry (+Newton's law) and the equations of static equilibrium:

Dx=D*cos(32deg)
3335.4/cos(32deg)=D
D=3933N
Therefore,
Dsin(32deg) = Dy
Dy=-2084.18N
Therefore,
∑Fy=0; Dy+By = 0
-2084.18+By = 0
By= 2084.18N
and
∑Fx=0; Dx+Bx = 0
Bx = -Dx
Bx=-3335.4N

Knowing Dy I can go back to my original equation:
∑Fy = 0; Wy + Ay + Dy = 0
981N+Ay+Dy = 0
Ay=+2084.18N-981N
Ay=1103.18N

How does this sound? Way off base or not too far off?

PS thanks for your help so far it has been invaluable

 

[I like Serena]

Not too far off :)

There's one thing left to factor in: the rope.
It pulls section BD to the right at the point where it is attached with the same force as Wy, since the power is transmitted through the rope.
I'm afraid you'll have to correct your equations to add this (known) force.

Cheers!

 

[louza8]

Thanks I Like Serena! I think I've got it now!

 

[louza8]

Edit:

Still having some trouble with the direction of By and Dy :(

I have it in my head that the forces must be compressive or tensile, but they don't have to be do they? Could you explain this point a little more or check the following to see if I've got it right?

∑Fx=0; Dx+Bx+981
Bx=-3335.4 - 981N
Bx= -4316.4N
Bx=B*cos(32deg)
B=Bx/cos(32deg)
B=-5089.81/cos(32deg)
By=sin(32deg)*B
By=-26997.18N ------------------> is this the right direction? like from top to bottom
Therefore
∑Fy = 0; By+Dy
Dy=2697.18N ----------->is this the right direction? like from bottom to top?

Or have I muddled up my directions here?

Hopefully these will be my last questions ;)

 

[I like Serena]

Edit:

Still having some trouble with the direction of By and Dy :(

I have it in my head that the forces must be compressive or tensile, but they don't have to be do they? Could you explain this point a little more or check the following to see if I've got it right?

∑Fx=0; Dx+Bx+981
Bx=-3335.4 - 981N
Bx= -4316.4N
Bx=B*cos(32deg)
B=Bx/cos(32deg)
B=-5089.81/cos(32deg)
By=sin(32deg)*B
By=-26997.18N ------------------> is this the right direction? like from top to bottom
Therefore
∑Fy = 0; By+Dy
Dy=2697.18N ----------->is this the right direction? like from bottom to top?

Or have I muddled up my directions here?

Hopefully these will be my last questions ;)

You seem to be a bit uncareful with your signs, but it's hard to tell without a picture.

I think your first equation should be:
∑Fx=0; -Dx+Bx+981=0
based on how you chose Dx earlier.

Furthermore, you seem to assume that the force at B is aligned with the section, which in this case isn't. The moment sum will show that.

And take care that the force at B is actually 2 forces, which are equal and opposite.
BD pulls AC up, and simultaneously AC pulls BD down, in such a way the the joint remains in the same place.

It should work out that the By on section BD is pointed down.
And Bx on section BD is pointed left.
Dy (on section BD as well as on the entire system) would indeed be pointing up.

I'd suggest you draw a diagram of the entire structure with the external forces, showing explicitly what directions you have chosen for those forces.

And I'd suggest you draw a diagram of BD with the forces acting on it.

And finally a diagram of AC.
Note that in the diagram of AC the (chosen) forces at B are the same as in the diagram of BD but exactly opposite.

If you have calculated everything correctly, you'll see that when you work out section AC works, that all forces are already known, making it possible to verify that you made no mistakes.
If section AC turns out not to be in equilibrium, it's very likely you made a mistake with a minus-sign.

Cheers! :)

 

[pongo38]

If I may chip into this, adding to I Like Serena's excellent ideas, it's not necessarily external or internal to worry about. You have to identify the object about which you are making equilibrium statements. For example member BD, with a hinge at each end and a horizontal force 0.1 m up from B. In this case the rope force is external, but looking at the whole frame to the right of AD, that rope force is internal. You have to 'cut' the structure away that you want to consider, and replace the 'cuts' with appropriate forces and moments. This frame is a variation on the 3-hinged arch, for which there is a standard process: Moments about D for whole frame. Moments about A for whole frame. Moments about B for BD only. That gives you enough equations. The key to getting an extra equation is to say, in effect, the bending moment in BD at B is zero (even though the bending moment at B in ABC is not zero.