Engineering statics equilibrium problem

In summary: The answer was positive, hm, anyway I got it right. So for finding the reaction between bead B and rod AC, it would be just the force that is perpendicular to P right?
  • #1
Alison A.
86
2

Homework Statement


Here is the prompt/picture
FBD.png


The hint given is a FBD of the bead is recommended to being this problem. Find the coordinates of B so that both the magnitude and orientation of the elastic cord force can be properly represented. Also, two mutually orthogonal normal force directions (to bar AC) need to be included to permit a general representation of normal forces acting on the bead. A shortcut to solving for the elastic cord force can be obtained by writing the vector equilibrium equation, then taking the dot product of that equation with a unit vector pointing in the direction of bar AC. Since the normal forces are by definition perpendicular to the bar, their contribution is zero, and a single scalar equations remains for the for P.

Homework Equations


ΣFx=0
ΣFy=0
ΣFz=0
Basic trigonometry

The Attempt at a Solution


Alright so I started off by finding the coordinates of all the points
A (124, 0, 0)mm
B (?, 31, 21)mm
C (0, 62, 42)mm
D (62, 0, 62)mm

To find B I made a right triangle with A and C to find the magnitude. Since B is in the middle of AC as stated, I divided the magnitude I found and got 65.5. So B (65.5, 31, 21)mm.

I found the unit vector of AC like the hint suggested and found that to be (-0.856, 0.428, 0.290).

I am stuck here, I don't know how to go about finding the vector equilibrium.

Thank you for any help!
 
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  • #2
Alison A. said:

The Attempt at a Solution


Alright so I started off by finding the coordinates of all the points
A (124, 0, 0)mm
B (?, 31, 21)mm
C (0, 62, 42)mm
D (62, 0, 62)mm

To find B I made a right triangle with A and C to find the magnitude. Since B is in the middle of AC as stated, I divided the magnitude I found and got 65.5. So B (65.5, 31, 21)mm.

Why did you do this? If the y coordinate and z coordinate of point B are both halfway between the y and z coordinates of points A and C, it stands to reason that the x coordinate of B will also be halfway between the x coordinates for these two points.
I found the unit vector of AC like the hint suggested and found that to be (-0.856, 0.428, 0.290).

I am stuck here, I don't know how to go about finding the vector equilibrium.

Thank you for any help!

Find the correct location for point B.

Once you know this, you can then find the tension force in the cord BD, given the spring constant and its unstretched length.
 
  • #3
SteamKing said:
Why did you do this? If the y coordinate and z coordinate of point B are both halfway between the y and z coordinates of points A and C, it stands to reason that the x coordinate of B will also be halfway between the x coordinates for these two points.Find the correct location for point B.

Once you know this, you can then find the tension force in the cord BD, given the spring constant and its unstretched length.

A (124, 0, 0)mm
B (62, 31, 21)mm
C (0, 62, 42)mm
D (62, 0, 62)mm

TBD=k(l-l0) = 4(20-51.4) = -125.6
 
  • #4
Alison A. said:
A (124, 0, 0)mm
B (62, 31, 21)mm
C (0, 62, 42)mm
D (62, 0, 62)mm

TBD=k(l-l0) = 4(20-51.4) = -125.6
Tensions are usually taken to be positive. The stretch in the spring would be calculated as stretched length - unstretched length = 51.4 - 20 = 31.4 mm
The tension would therefore be 4 N/mm * 31.4 mm = 125.6 N.
 
  • #5
So I add this value with the unit vector in the direction of AC?
 
  • #6
Alison A. said:
So I add this value with the unit vector in the direction of AC?
The tension in the cord is pulling the bead at B against the bar AC. You want to convert the magnitude of this tension, 125.6 N, into a force vector FBD parallel to BD.

Then, according to the directions in the problem statement, you want to take the dot product of FBD and the unit vector uAC to find out what magnitude P must be.
 
  • #7
So the unit vector of BD is (0, -0.6031, 0.7977), then multiplied by the tension is (0, -75.75, 100.2)

The dot product of that and uAC is -3.367 which doesn't seem right...
 
  • #8
Alison A. said:
So the unit vector of BD is (0, -0.6031, 0.7977), then multiplied by the tension is (0, -75.75, 100.2)

This looks OK.

The dot product of that and uAC is -3.367 which doesn't seem right...

I think it's correct. I think the fact that FBDuAC is negative means that P is in the direction of point C.
 
  • #9
The answer was positive, hm, anyway I got it right. So for finding the reaction between bead B and rod AC, it would be just the force that is perpendicular to P right? I know the hint says nothing about it but wouldn't the cross product between them?
 
  • #10
I know there are infinitely number of vectors perpendicular to bar AC, but I don't know how to find it specifically where bead B is. Thank you so much for your help, I need to figure out these problems within the next hour. :cry:
 

1. What is the purpose of solving an engineering statics equilibrium problem?

The purpose of solving an engineering statics equilibrium problem is to determine the forces acting on a system in order to ensure that it is in a state of equilibrium. This is important for designing structures and machines that can support their own weight and any external forces without collapsing or breaking.

2. What are the steps to solving an engineering statics equilibrium problem?

The steps to solving an engineering statics equilibrium problem are:

  • Draw a free body diagram of the system, showing all the forces acting on it.
  • Apply Newton's first and second law to write equations for the sum of forces and moments in both the horizontal and vertical directions.
  • Solve the equations simultaneously to find the unknown forces and moments.
  • Check the solution for consistency and accuracy.

3. What are the key assumptions made in solving an engineering statics equilibrium problem?

The key assumptions made in solving an engineering statics equilibrium problem are:

  • The system is in a state of static equilibrium, meaning that all forces and moments are balanced.
  • The system is rigid and does not deform under load.
  • The forces acting on the system are concurrent (intersecting at a single point) or parallel.
  • The system is in a two-dimensional plane, meaning that all forces and moments are acting in either the horizontal or vertical direction.

4. How do real-world factors such as friction and elasticity affect the solution of an engineering statics equilibrium problem?

Real-world factors such as friction and elasticity can affect the solution of an engineering statics equilibrium problem by introducing additional forces and moments that need to be considered in the equations. For example, friction can cause a force in the opposite direction to the motion of an object, and elasticity can cause a spring force that changes with the displacement of an object. These factors can make the problem more complex and may require additional calculations or assumptions to be made in order to find an accurate solution.

5. What are some common applications of engineering statics equilibrium problems?

Engineering statics equilibrium problems are commonly used in the design and analysis of structures such as bridges, buildings, and dams. They are also used in the design of mechanical systems such as engines, cranes, and vehicles. Additionally, these principles are applied in the fields of civil, mechanical, and aerospace engineering, and are essential in ensuring the safety and stability of various structures and systems.

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