- #1
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
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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