What Are the Internal and External Forces Acting on a Steel Rod?

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Homework Help Overview

The discussion revolves around understanding the internal and external forces acting on a steel rod, particularly focusing on section BC. Participants are trying to clarify how to account for various forces when analyzing the rod's equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the concept of making imaginary cuts in the rod to analyze forces, questioning how to account for forces on either side of the cut. There is discussion about whether to include certain forces in the free-body diagram and how to interpret the forces acting on section BC.

Discussion Status

The discussion is ongoing, with participants offering different perspectives on how to approach the problem. Some guidance has been provided regarding the treatment of forces on either side of the cut, but there is still some confusion about the definitions of internal and external forces.

Contextual Notes

Participants are grappling with the implications of the free-body diagram and the specific forces acting on the rod, including external forces and how they relate to internal forces. There is mention of specific values for forces, but the context of these values remains under discussion.

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Homework Statement


I couldn't understand why the forces act on section Bc of steel rod is 715-325 = 390kN ?

Homework Equations

The Attempt at a Solution


Since there's also 215kN to the right in the rod, so forces act on section BC should be 715-325 +215 =605kN ?
 

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You only sum the forces acting on the section. These are shown to you completely. Thus, other forces are already taken into account.
 
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You could make an imaginary cut through the middle of section BC and create a free-body diagram. You then have 715 kN pulling to the left, and 325 kN pulling to the right. How much internal force does section BC need to supply to keep it in equilibrium?
 
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David Lewis said:
You could make an imaginary cut through the middle of section BC and create a free-body diagram. You then have 715 kN pulling to the left, and 325 kN pulling to the right. How much internal force does section BC need to supply to keep it in equilibrium?
There's also 215kN force acting to the right of section BC, right?
 
Simon Bridge said:
You only sum the forces acting on the section. These are shown to you completely. Thus, other forces are already taken into account.
How about the 215kN force? It's also acting on section BC, right?
 
Quite so, but remember when you make your imaginary cut, you ignore everything that's going on to the right of the cut.
 
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David Lewis said:
Quite so, but remember when you make your imaginary cut, you ignore everything that's going on to the right of the cut.
If ignore the force on the right, we should also ignore the forces on the right?
 
Yes. You don't pay any attention to it. You can cover the right hand side of the structure with a piece of paper so it doesn't distract you.
 
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David Lewis said:
Yes. You don't pay any attention to it. You can cover the right hand side of the structure with a piece of paper so it doesn't distract you.
if i make an imaginary cut, the forces on the left and right of the rod are ignored, then i only have 325kN , how could that be?
 
  • #10
All forces to the left of the cut must be included in your diagram and accounted for, and all forces acting on the face of the cut* must be accounted for. Anything other than that you don't worry about.

* The face of the cut that is on the right end of the body.
 
Last edited:
  • #11
David Lewis said:
All forces to the left of the cut must be included in your diagram and accounted for, and all forces acting on the face of the cut* must be accounted for. Anything other than that you don't worry about.

* The face of the cut that is on the right end of the member.
do you mean we only need to cut at C?i thought we must cut at B and C to get section BC ?
 
  • #12
Actually the cut I hypothesized is approximately halfway between B and C.
You would then have two parts (everything to the left of the cut, and everything to the right of the cut).
 
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  • #13
David Lewis said:
All forces to the left of the cut must be included in your diagram and accounted for, and all forces acting on the face of the cut* must be accounted for. Anything other than that you don't worry about.
* The face of the cut that is on the right end of the member.
deleted
 
  • #14
David Lewis said:
Actually the cut I hypothesized is approximately halfway between B and C.
You would then have two parts (everything to the left of the cut, and everything to the right of the cut).
If so, not all the forces are included for section

BC, then we only have the forces on the left, the forces on the right are ignored?
Why is it so? In the diagram, we could see that the author make imaginary cuts at B and C...not middle of section BC...
 
  • #15
You are correct but the author has omitted some steps. You are allowed to make an imaginary cut wherever is convenient for you. In the example I've put forward, you make only one cut. You end up with all of part A, and half of part B. This is the free-body we are analyzing. The other body (half of part B, all of part C, and all forces appertaining thereto) we ignore for the time being.
 
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  • #16
chetzread said:
If so, not all the forces are included for section

BC, then we only have the forces on the left, the forces on the right are ignored?
Why is it so? In the diagram, we could see that the author make imaginary cuts at B and C...not middle of section BC...
What David is explaining is that you must start out by specifying your free body, and then do a force balance on your free body, including only forces acting on the specified free body. In this case, David has specified as the free body the portion of the rod to the left of BC. So you cannot include the forces to the right of BC, since they are not acting on your free body.
 
  • #17
Chestermiller said:
What David is explaining is that you must start out by specifying your free body, and then do a force balance on your free body, including only forces acting on the specified free body. In this case, David has specified as the free body the portion of the rod to the left of BC. So you cannot include the forces to the right of BC, since they are not acting on your free body.
ok , if i want to cut off section BC from the others , then the force act on the left end of the rod is 715-325= 390kN to the left ( point away from the rod) ?
For the right section of rod BC , it's 215 + 175 = 390kN to the right ( point away from the rod) ? Am i right
 
  • #18
chetzread said:
ok , if i want to cut off section BC from the others , then the force act on the left end of the rod is 715-325= 390kN to the left ( point away from the rod) ?
For the right section of rod BC , it's 215 + 175 = 390kN to the right ( point away from the rod) ? Am i right
The internal force acting on the free body to the left is pointing to the right. The internal force acting on the free body to the right is pointing to the left.
 
  • #19
Chestermiller said:
The internal force acting on the free body to the left is pointing to the right. The internal force acting on the free body to the right is pointing to the left.
so , in my working above , it's external force , am i right ?
 
  • #20
chetzread said:
so , in my working above , it's external force , am i right ?
Huh?
 
  • #21
Chestermiller said:
Huh?
the force act on the left end of the rod is 715-325= 390kN to the left ( point away from the rod) ?
For the right section of rod BC , it's 215 + 175 = 390kN to the right ( point away from the rod) ?
These forces are the external forces ? so , internal forces should be opposite to them ?
 
  • #22
chetzread said:
the force act on the left end of the rod is 715-325= 390kN to the left ( point away from the rod) ?
For the right section of rod BC , it's 215 + 175 = 390kN to the right ( point away from the rod) ?
These forces are the external forces ? so , internal forces should be opposite to them ?
Yes.
 
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