Engineering Mechanics - Statics

In summary: It seems that there are a few assumptions that need to be made in order to solve it. Overall, the goal is to find the unknown load F and the reaction forces at the joints in order to keep the system in equilibrium. The equations and unknowns provided seem to be excessive and can be simplified or reduced in some way.
  • #1
cordines
15
0

Homework Statement


Please refer to the attached diagram. I am trying to find the unknown load F (along line of action lF) that will keep the system in equilibrium as well as compute all reaction forces developed at the joints.

Homework Equations



Summation of forces in the horizontal and vertical direction as well as moments about a point.


The Attempt at a Solution



I have split the mechanism into these members:

Member AB, Member BC, Member CDE, Member DF and Member FG as well as the slider. From these members I can derive the following equations. For:

Member AB - 3 equations (summation of forces in horizontal and vertical direction and moments)
Member BC - 3 equations
Member CDE - 3 equations
Member DF - 3 equations
Member FG - 3 equations
Slider E - 2 equations (summation of forces in horizontal and vertical direction)

Total number of equations = 17.

However I have 16 unknown parameters which are:

Joint A (Ax, Ay)
Joint B (Bx, By)
Joint C (Cx, Cy)
Joint D (Dx, Dy)
Joint E (Ex, Ey)
Joint F (Fx, Fy)
Joint G (Gx, Gy)
Unknown force F (along line of action lF)
Slider reaction Sy

Clearly I need the same number of Unknown parameters and equations to solve. What am I doing wrong?
 

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  • #2
"the unknown load F (along line of action lF)" baffles me because on the diagram they are parallel but separated. Is the diagram correct? An additional comment is that, although your attempt is well meant, it is not necessarily the best way to approach the problem because with so many unknowns there is a very high chance of making mistakes. For example, member 3, by inspection carries no moment or shear force; only an axial force. So you could look for simplification before assembling a lot of trivial equations. You can also recognise that joints C and D are hinges. Therefore the moments on one side of those joints must sum to zero... and I believe you could find other simplifications that either reduce the unknowns or at least reveal them gradually through the calculation process, rather than lumpthem into an impractical matrix of equations.
 
  • #3
Could you throw out one equation, and then solve for the 16 unknowns?
 
  • #4
@pongo38. Note that the moment and force in red are both given. What exactly do you mean by "the unknown load F (along line of action lF)" are parallel but separated"?

@nvn. Yes, I threw out one equation and solved 15 equations with 15 unknowns and then used the other equation as a check to see if the LHS matches the RHS. However, that didn't happen.
 
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  • #5
What I mean is: On the wheel labelled 'A' there is a red F with 2 arrows. Presumably the upper arrow is the symbol that F is a vector, and the lower arrow indicates its line of action, whose position is not well defined. Then you have LF at point E. Is the red F supposed to be acting through point E, or point A, or some other point? Sorry to be difficult, but it's not clear to me. Concerning your remark to nvn, if the system is statically determinate, then, in general there will be a spare equation you can use for checking.
 
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  • #6
No, on wheel A there is only one horizontal force (which is taken to the left). This force acts at the midspan of the member AB as denoted by the parameter r/2. The arrow on the force F indicates that it is a vector. The arrow below the force F indicates how it acts and its direction.

The force acting through the line of action lF is taken to act at point E.

I agree that the problem is not clear but that was how it was assigned.
 
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What is Engineering Mechanics - Statics?

Engineering Mechanics - Statics is a branch of engineering that deals with the study of forces and their effects on stationary objects or systems. It is a fundamental subject in engineering that provides the basis for understanding and analyzing the behavior of structures, machines, and other objects under different types of forces.

What are the main principles of Engineering Mechanics - Statics?

The main principles of Engineering Mechanics - Statics include the concept of equilibrium, which states that the sum of all forces acting on an object must be equal to zero for the object to be in a state of rest or constant velocity. Additionally, the principles of Newton's laws of motion, moment of a force, and free body diagrams are also important in statics.

What are some real-world applications of Engineering Mechanics - Statics?

Engineering Mechanics - Statics has a wide range of applications in various fields, including civil engineering, mechanical engineering, aerospace engineering, and more. It is used in the design and analysis of structures such as buildings, bridges, and dams, as well as in the design of machines and mechanical systems.

What are the common types of problems solved in Engineering Mechanics - Statics?

Some common types of problems solved in Engineering Mechanics - Statics include determining the forces acting on an object, finding the center of mass of an object, analyzing the stability of structures, and calculating the stresses and strains in a structure under different loading conditions.

Is it possible to apply the principles of Engineering Mechanics - Statics to real-world situations?

Yes, the principles of Engineering Mechanics - Statics can be applied to real-world situations. By understanding the fundamental concepts and principles of statics, engineers are able to design and analyze structures and machines that can withstand different types of forces and remain in equilibrium. This is essential for ensuring the safety and functionality of various structures and systems in our daily lives.

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