[HELP]- Determine the Resultant of the Force System.

Click For Summary

Homework Help Overview

The discussion revolves around determining the resultant of a force system in a mechanics problem, specifically focusing on the summation of forces and moments. Participants are exploring the implications of the term "resultant" and its application in the context of static equilibrium.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of the resultant and whether it refers to the sum of the forces. There are attempts to clarify the process of calculating the resultant and moments, with some participants questioning the specific forces to include and the points about which to take moments.

Discussion Status

The discussion is active, with various participants providing insights and seeking clarification on the steps involved in calculating the resultant and moments. Some guidance has been offered regarding the need to consider reaction forces and the correct application of the summation equations, though there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are navigating potential confusion stemming from textbook examples and their application to the current problem. There are mentions of specific distances and forces that need to be accurately accounted for in the calculations, as well as the importance of identifying actions and reactions in the system.

  • #31
Doc Al said:
No, still not right. Since the 160 lb force is spread out uniformly over 8 ft, measure the distance from the midpoint of that spread to point A.

Oh I was reading from the roller all the way to the midpoint, so you're saying for the 160 read from the midpoint of the line load to the Point A?, so it is 4?(as I said above) or do you use half the distance 4ft in the equation times the other 8 ft?


\SigmaM= -100( 0 ) -700(8) - 160( 4 )(8)=
 
Last edited:
Physics news on Phys.org
  • #32
Dellis said:
Oh I was reading from the roller all the way to the midpoint, so you're saying read from the midpoint of the line load to the Point A?,
Yes.
like this?


\SigmaM= -100( 0 ) -700(8) - 160( 4 )=
No. On your diagram, draw a dot at the midpoint of the 160 lb line load. How far is that midpoint from point A?
 
  • #33
Doc Al said:
Yes.

No. On your diagram, draw a dot at the midpoint of the 160 lb line load. How far is that midpoint from point A?

But you need to know the midpoint, which I assume is half the distance which is 4ft and then you read from that 4ft all the way to Point A?.


\SigmaM= -100( 0 ) -700(8) - 160( 4 )(8)=
 
  • #34
Dellis said:
But you need to know the midpoint, which I assume is half the distance which is 4ft and then you read from that 4ft all the way to Point A?
The line load stretches from 8 ft from point A (on its left side) to 16 ft from pont A (on its right side). The midpoint is right in the middle of the line load. That midpoint is certainly not 4 ft away from point A! :eek:

Please mark the location of the midpoint of that line load on your diagram.
 
  • #35
Doc Al said:
The line load stretches from 8 ft from point A (on its left side) to 16 ft from point A (on its right side). The midpoint is right in the middle of the line load. That midpoint is certainly not 4 ft away from point A! :eek:

Please mark the location of the midpoint of that line load on your diagram.

I didn't say it was 4 feet from the midpoint of the line load to point A, I wondered if the midpoint you said to mark is half the distance of 8 ft, which is 4ft.
 
  • #36
Dellis said:
I didn't say it was 4 feet from the midpoint of the line load to point A, I wondered if the midpoint you said to mark is half the distance of 8 ft, which is 4ft.
Yes, the midpoint is right in the middle of the 8 ft span, thus 4 ft from each end of that span.
 
  • #37
Doc Al said:
Yes, the midpoint is right in the middle of the 8 ft span, thus 4 ft from each end of that span.

Ok I knew I had something right there, so after that would the equation be this?, btw what about the Roller Distance toward POINT A?


\SigmaM= -100( 0 ) -700(8) - 160( 4 )(8) - 160(16) =Can't believe this is 3 pages deep, can't believe this is confusing me so bad.
 
Last edited:
  • #38
Got your PM...

Are you still having problems with this?

CS
 
  • #39
If you take the moment about A, then
ΣM = -100(0) - 700(8) - 160(8 + 4)
R = -(100 + 700 + 160)
X(bar) =ΣM/R.
 
  • #40
Dellis said:
Ok I knew I had something right there, so after that would the equation be this?, btw what about the Roller Distance toward POINT A?


\SigmaM= -100( 0 ) -700(8) - 160( 4 )(8) - 160(16) =
No, still not correct. The line load force is 160 lb. Where does it act? Effectively at the midpoint of the line load. So to calculate its moment about point A, you need the distance between that midpoint and point A. What is that distance?

The distance between the roller and point A is irrelevant.
 
  • #41
rl.bhat said:
If you take the moment about A, then
ΣM = -100(0) - 700(8) - 160(8 + 4)
R = -(100 + 700 + 160)
X(bar) =ΣM/R.

Wow so is (8 + 4)?, here I was thinking multiplication(8)(4) for some reason

can you explain how you got R=-(100 + 700+ 160)?

because the mentor told me what I had before was good in terms of the summation of

the forces.The X(/bar) part I know how to do that, the issue I was having was the summations of the moments and noticing the reactions, as you can see I am 3 pages deep with that issue.
 
Last edited:
  • #42
stewartcs said:
Got your PM...

Are you still having problems with this?

CS

sadly yes, the Moments part got me going in circles, one things for sure I am learning as I am making all of these mistakes, so it is good experience.
 
  • #43
Doc Al said:
No, still not correct. The line load force is 160 lb. Where does it act? Effectively at the midpoint of the line load. So to calculate its moment about point A, you need the distance between that midpoint and point A. What is that distance?

The distance between the roller and point A is irrelevant.

12...

ΣM = -100(0) - 700(8) - 160(8 + 4)yeah I figured it was irrelevant when you went off line(or looked like you did) then I erased that out of my process.
 
  • #44
Dellis said:
Wow so is (8 + 4)?, here I was thinking multiplication for some reason

can you explain how you got R=-(100 + 700+ 160)?

because the mentor told me what I had before was good in terms of the summation of

the forces.


The X(/bar) part I know how to do that, the issue I was having was the summations of the moments and noticing the reactions as you can see I am 3 pages deep.

The magnitude of the resultant is the summation of the two point loads and the distributed load. Hence, -100 + (-700) + (-20 lb/ft x 8 ft) = -(100 + 700 + 160) lb. The negative sign is due to the convention chosen (i.e. downward is negative).

CS
 
  • #45
Dellis said:
12...

ΣM = -100(0) - 700(8) - 160(8 + 4)


yeah I figured it was irrelevant when you went off line(or looked like you did) then I erased that out of my process.

Looks good.

CS
 
  • #46
stewartcs said:
The magnitude of the resultant is the summation of the two point loads and the distributed load. Hence, -100 + (-700) + (-20 lb/ft x 8 ft) = -(100 + 700 + 160) lb. The negative sign is due to the convention chosen (i.e. downward is negative).

CS
ah ok, well he left that out, so was wondering, thanks for clearing that up.So I should use that to get the resultant?, I thought my summation of the forces was called "good" by the mentor.
Doc Al said:
Good, this will tell you the magnitude of the force.
 
  • #47
Dellis said:
ah ok, well he left that out, so was wondering, thanks for clearing that up.


So I should use that to get the resultant?, I thought my summation of the forces was called "good" by the mentor.

That is the magnitude of the resultant...you will use that to find the point of application of the resultant force (i.e. x_bar).

CS
 
  • #48
Doc Al said:
I know you've already gotten help on this issue, but just for the record:

Good, this will tell you the magnitude of the force.

You are taking moments about point A, which is fine. But you're using the wrong distance for the 160 lb force.

He told you the same thing regarding the resultant...However he told you that your moment calculation was still wrong...which it was at the time.

I'm not sure what you are asking now.

CS
 
  • #49
stewartcs said:
He told you the same thing...I'm not sure what you are asking now.

CS

He told me the summations of forces was good, meaning I can get the Resultant with this

\Sigma F= -100 lb - 700 lb - 160 lb=...


Then I was presented with this a few seconds ago, explained further by you

R = -(100 + 700 + 160)


^That isn't asking me to multiply right?, it isn't asking me to multiply -1 with(100+700+160)
right?

He just put it that way but it still means this \Sigma F= -100 lb - 700 lb - 160 lb=..., what I originally have in my process.
 
  • #50
Dellis said:
He told me the summations of forces was good, meaning I can get the Resultant with this

\Sigma F= -100 lb - 700 lb - 160 lb=...


Then I was presented with this a few seconds ago, explained further by you

R = -(100 + 700 + 160)


^That isn't asking me to multiply right?, it isn't asking me to multiply -1 with(100+700+160)
right?

He just put it that way but it still means this \Sigma F= -100 lb - 700 lb - 160 lb=..., what I originally have in my process.

Those two are equivalent. One just factors out the negative sign...

CS
 
  • #51
stewartcs said:
Those two are equivalent. One just factors out the negative sign...

CS

Sorry I just got a bit confused, he presented it in a way not shown to me in the book but yeah it is the same...

post after this one has my calculations done
 
Last edited:
  • #52
Dellis said:
Sorry I just got a bit confused, he presented it in a way not shown to me in the book but yeah it is the same...

\Sigma F= -100 lb - 700 lb - 160 lb= -960



R = -(100 + 700 + 160)=-960

No problem...that's why we are here...

CS
 
  • #53
stewartcs said:
No problem...that's why we are here...

CS

I appreciate it, I learned more reading these forums then using this 150 dollar buck, thanks again.

Ok so here the finish line right?...

ΣF= -100 lb - 700 lb - 160 lb= -960 lb

ΣM = -100(0) - 700(8) - 160(8 + 4)= - 7520 lbX(bar) =ΣM/R.

-7520 lb/-960lb= 7.83 lb
 
  • #54
Dellis said:
Ok so here the finish line...

\Sigma F= -100 lb - 700 lb - 160 lb= -960 lb

ΣM = -100(0) - 700(8) - 160(8 + 4)= - 7520 lb


X(bar) =ΣM/R.

-7520 lb/-960lb= 7.83 lb

Check your units...

CS
 
  • #55
stewartcs said:
Check your units...

CS

Woah I almost forgot about that lil detail, so this is it right.ΣF= -100 lb - 700 lb - 160 lb= -960 lb

ΣM = -100 lb(0) - 700 lb(8) - 160lb/ft(12)= -7520 lb/ftX(bar) =ΣM/R

-7520 lb/ft (divided by) -960lb= 7.83 ft
 
Last edited:
  • #56
Dellis said:
Woah I had forgot about that lil detail, so it is like this

ΣF= -100 lb - 700 lb - 160 lb= -960 lb

ΣM = -100 lb(0) - 700 lb(8) - 160lb/ft(12)= -7520 lb/ft


X(bar) =ΣM/R.

-7520 lb/ft (divided by) -960lb= 7.83 ft


Correct.

CS
 
  • #57
stewartcs said:
Correct.

CS

Yes!:), thank you all for the guidance in regards to this problem, see now I can do these type of exercises since I understand the process, that is my problem I need to get familiar with the processes.Like I said before I do have two other problems I tried doing by myself and I need them checked/corrected, I will make a new thread since this one is too much already, hopefully you can provide further guidance for those two :).
 
Last edited:

Similar threads

Where on a rigid body is a resultant force applied?
  • · Replies 11 ·
Replies
11
Views
2K
Determine Force on 6ft Door Due to Water & Air Pressure
  • · Replies 7 ·
Replies
7
Views
2K
Find moment arm of resultant force of two forces applied at two points
  • · Replies 3 ·
Replies
3
Views
1K
How can I find the length of this pole?
  • · Replies 1 ·
Replies
1
Views
2K
Conducting rod in equilibrium due to magnetic force
  • · Replies 13 ·
Replies
13
Views
2K
How High Can Water Rise Before Tipping a Dam?
  • · Replies 4 ·
Replies
4
Views
2K
Help With Understanding Resultant Forces By Connecting Vectors
  • · Replies 5 ·
Replies
5
Views
3K
Problems about magnetic field, electromotive force, induced current
  • · Replies 1 ·
Replies
1
Views
1K
To find the resultant of forces on a lamina
  • · Replies 5 ·
Replies
5
Views
3K
Solving a Rod-Particle System: Troubles with Forces
  • · Replies 7 ·
Replies
7
Views
2K