xe minus xa is: xe is the x coordinate of the equilibirum force and xa the x coordinate of the reduction point.
I thought that ecuation wasn't useful because I had seen this formule:
Mo= [+/-] Ey*x [+/-] Ex*y
Well... The best way to answer me is doing this exercise. Can anyone resolve it ans show me step by step the way to do it? Thanks in advance :smile:
Exercise:
Determine the equilibrium force to add to the system
http://img458.imageshack.us/img458/9932/system8vq.jpg (system )
Yes. I want to find the point of application of the equilibrium force. In that point, the equilibirum force will have the opposite torque or couple of the resultant force. That will give total equilibirum (Fy, Fx, and Torque).
I want to know how to calcule that point of application.
In a system of forces which forces "cross" or "start" (dunno how to explain properly in english, sorry) in a point is easy to find where to place the equilibrium force because that place is the point the forces cross or start.
Now, in a system of forces that do not cross or start in a point...
When solving a system of forces, if the system has no equilibrium, one has to find the force that gives the equilibrium to that system, which is opposite to the resultant of the system. What I want to know is where to put the equilibrium force, that is its point of application. Setting the...
Sorry for the diagram, but I used some words and letters in spanish, perhaps this can clear the problem:
The bar has a shape like this: |_
........_|
.......|
I used "M" to represent the torques, and I specified the directions of them.
Can you now help me? Thanks :smile:
On the steel bar of the figure acts a system of forces. Check the balance.
http://img140.imageshack.us/img140/3714/dibujo2ch.jpg
F1(2,-1.5) = (1000N,120º)
M1 = 200Nm
F2(2,1.5) = (1000N,240º)
M2 = 400Nm
F3(0,2.5) = (1000N,0º)
M3 = 1900Nm
¿How do I solve it? ¿Can anyone teach me to do...