What Calculations Are Needed to Achieve Equilibrium in These Statics Problems?

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The discussion focuses on solving statics problems involving equilibrium calculations for two mechanical systems. The first problem involves determining the couple Ma needed to maintain equilibrium in a system of rods, where initial calculations were incorrect due to neglecting perpendicular distances to forces. The second problem concerns a double toggle mechanism, where the user seeks a simpler method to analyze the system instead of generating numerous equations. The user ultimately resolves the first problem but continues to seek guidance on the second, particularly regarding balancing forces and torques. The thread highlights the importance of considering all forces and moments in statics problems to achieve accurate results.
Aaron
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Hi, I have a couple of questions from a statics assignment that I'm in need of help on. Here's what I have so far:

http://216.170.11.226/pub/6131.jpg

Two rods are connected by a slider block as shown. Neglecting the effect of friction, determine the couple Ma required to hold the system in equilibrium.

And this is what I believe the forces are:

http://216.170.11.226/pub/6131-2.jpg

It would seem to me that the external forces are irrelevant to this problem, but I'm guessing not. What I came up with for this problem is:

Sum of moments about B=0 => -25+Nd*271.89=0 => Nd=0.0919
Sum of moments about A=0 => -.0919*150+Ma=0 => Ma=13.79 N*m

I know the answer is Ma=15.22 N*m. I'm guessing that I also need to consider external forces or something.


The next problem is

http://216.170.11.226/pub/6129.jpg

The double toggle mechanism shown is used in a punching machine. Knowing that the links AB and BC are each of length 150 mm, determine the couple M required to hold the system in equilibrium when Phi=20 (degrees).

Once again, my analysis of the problem is:

http://216.170.11.226/pub/6129-2.jpg

Obvious things are:
Dy+Ey+Ay+800=0
Dx+Ex+Ax=0

sum of moments about B of AB=0 => M+By*150*cos(30)+Bx*150*sin(30)=0

Since AB is a two-force body, Bx=B*cos(30), By=B*sin(30) and A=-B.

B'x=-Bx
B'y=-By
Cx=C*cos(30)
Cy=C*sin(30)
sum of moments about C of BC=0 => -B'x*150*sin(30)+By*150*cos(30)=0
sum of moments about B of BC=0 => -Cy*150*cos(30)+Cx*150*sin(30)=0

And so on and so forth. If I continue in this manner I figure I'll end up with something around 24 equations, there has got to be a simplier method. Any idea what that might be?

Any hints or help would greatly be appreciated.

Thanks
-Aaron
 
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Ok, the first problem is easy, I just forgot to get a perpendicular distance to the force, so it works fine now.

As for the second problem, I was thinking about just balancing the Fx, Fy, Ma, Md, and Me, but that doesn't seem to work either. Any suggestions would be appreciated.
 
Is N_D supposed to be an applied force in that drawing?

Can you list the terms you have for the net forces, and net torques, and which point they are about?
 
Sorry, N_D was an internal force from the collar. Thanks anyway, but I managed to figure this one out.
 
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