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

[Doc Al]

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.

 

[Dellis]

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?


$$\Sigma$$M= -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.

 

[stewartcs]

Got your PM...

Are you still having problems with this?

CS

 

[rl.bhat]

If you take the moment about A, then
ΣM = -100(0) - 700(8) - 160(8 + 4)
R = -(100 + 700 + 160)
X(bar) =ΣM/R.

 

[Doc Al]

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?


$$\Sigma$$M= -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.

 

[Dellis]

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.

 

[Dellis]

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.

 

[Dellis]

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.

 

[stewartcs]

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

 

[stewartcs]

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

 

[Dellis]

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.

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

 

[stewartcs]

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

 

[stewartcs]

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

 

[Dellis]

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.

 

[stewartcs]

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

 

[Dellis]

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

 

[stewartcs]

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

 

[Dellis]

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

 

[stewartcs]

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

 

[Dellis]

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

 

[stewartcs]

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

 

[Dellis]

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 :).