Help on a free body diagram and equation

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  • #1
kiwimia
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1. Relevant information
This is a two-block-system on top of each other to the right side of a wall, both attached with springs that are also attached to the wall. A force is applied on the bottom block to the right direction, frictions occurred on both bottoms of the two blocks. Here is the diagram illustration:

http://i62.tinypic.com/291ytxt.jpg

(BTW, don't worry about Weight = Mg, this can be ignored)

2. Question
Draw a free body diagram and determine the equations of motion for this system

The Attempt at a Solution


This is the free body diagram I drew:

http://i60.tinypic.com/2zdqhpt.jpg

According to my free body diagram, I have the equation of motion for this system.

M2X2’’ + K2X2 + B2(X1’ – X2’) = 0

M1X1’’ + K1X1 + B1X1’ = f + B2(X1’ – X2’)


Does this seem right? Please let me know. Thanks!
 
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Answers and Replies

  • #2
PhanthomJay
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1. Relevant information
This is a two-block-system on top of each other to the right side of a wall, both attached with springs that are also attached to the wall. A force is applied on the bottom block to the right direction, frictions occurred on both bottoms of the two blocks. Here is the diagram illustration:

http://i62.tinypic.com/291ytxt.jpg

(BTW, don't worry about Weight = Mg, this can be ignored)

2. Question
Draw a free body diagram and determine the equations of motion for this system

The Attempt at a Solution


This is the free body diagram I drew:

http://i60.tinypic.com/2zdqhpt.jpg

According to my free body diagram, I have the equation of motion for this system.

M2X2’’ + K2X2 + B2(X1’ – X2’) = 0

M1X1’’ + K1X1 + B1X1’ = f + B2(X1’ – X2’)


Does this seem right? Please let me know. Thanks!
No weight means no normal force which means no friction, which is a function of the normal force, not some sort of linear drag equation as you seem to show. Also, you are apparently using pseudo forces in your diagrams, which you should label as 'free body diagrams with pseudo forces shown' if you must, but I don't see why. Are you assuming motion?
 
  • #3
kiwimia
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The vertical forces on the mass (the weight Mg and the upward forces exerted by the frictionless bearings) have been omitted because these force are perpendicular to the direction of motion. Only the horizontal forces are considered.

The original question description is stated as the following:
The system is at rest with no energy stored for t < 0. At t = 0, force f is applied to mass M1. Draw a free body diagram and determine the equations of motion for this system.

I got my reference from the Close book titled Modeling Analysis of Dynamic Systems. Here is a similar example,

2hd15ao.jpg


I am trying to use his way to solve my problem, but I am not sure if my free body diagram was drawn correctly.
 
  • #4
PhanthomJay
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The vertical forces on the mass (the weight Mg and the upward forces exerted by the frictionless bearings) have been omitted because these force are perpendicular to the direction of motion. Only the horizontal forces are considered.

The original question description is stated as the following:
The system is at rest with no energy stored for t < 0. At t = 0, force f is applied to mass M1. Draw a free body diagram and determine the equations of motion for this system.

I got my reference from the Close book titled Modeling Analysis of Dynamic Systems. Here is a similar example,



I am trying to use his way to solve my problem, but I am not sure if my free body diagram was drawn correctly.
In your original post, you noted that there was friction between all surfaces, which typically is a function of the normal force between the surfaces. Your diagrams and example indicate some sort of a resistance force from the bearings which is a function of the relative speeds between surfaces (B dx/dt), which I am not familiar with.

Free body diagrams should show all real forces, including the weight and normal forces, whether or not you need to use them. And if you must show the pseudo forces ( M dx^2/dt^2) in your diagrams, label the diagrams appropriately.

You seem to have the wrong directions shown for your B2(x1dot-x2dot) forces.

I am not sure if you understand all the descriptions of the forces and pseudo forces shown. I don't understand the bearing forces.
 
  • #5
kiwimia
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In your original post, you noted that there was friction between all surfaces, which typically is a function of the normal force between the surfaces. Your diagrams and example indicate some sort of a resistance force from the bearings which is a function of the relative speeds between surfaces (B dx/dt), which I am not familiar with.

Free body diagrams should show all real forces, including the weight and normal forces, whether or not you need to use them. And if you must show the pseudo forces ( M dx^2/dt^2) in your diagrams, label the diagrams appropriately.

You seem to have the wrong directions shown for your B2(x1dot-x2dot) forces.

I am not sure if you understand all the descriptions of the forces and pseudo forces shown. I don't understand the bearing forces.

I can't agree with you more on all real forces should be shown on free body diagrams. I think the book is showing a different way and it's because the D' Alembert's Law is applied, which is the law of reaction forces, and the law for displacement variables. That was why I added the pseudo force. However, this confuses me even more because I used to understand the way that you mentioned.

Also, thank you very much for pointing out B2(x1dot-x2dot) forces are in opposite directions. I totally did not realized that.

10xheyp.jpg


From chapter 2 of the book: https://www.scribd.com/doc/241929209/Modeling-and-Analysis-of-Dynamic-Systems-Third-Edition [Broken]
 
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  • #6
PhanthomJay
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Sweet!
 

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