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

[Doc Al]

I think the "resultant" means finding the single force and the point of application of that force, that would have the same effect as the given forces.

Exactly. It dawned on me (after I had logged off for the night) that that's what the problem is asking for. (D'oh! :uhh:)

 

[Doc Al]

I know you've already gotten help on this issue, but just for the record:

Ok so far with what you said we can assume this right, I hope this is the right path, if you see something wrong please point it out.


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

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

$$\Sigma$$M= -100( 0 ) - 700 ( 8 )- 160 (16 )=....

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

 

[Dellis]

Hey thanks tiny-tim/Doc Al for clearing that up,

See I knew the mirror image exercise from the book would let me understand but also get me confused, like you said I was going on the wrong distance, now it make sense why the diagram literally just has POINT A in it.

 

[Dellis]

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.

That is the distance that is confusing me, because as I go reading the distances about point A from left to right the distance is 8 ft or is it like the book exercise where in the Uniformly line load they use HALF the distance. For example- 30kN/m x 6 m=180Kn, then they use half the distance in the summation of moments part, like -180( 3 )

this

$$\Sigma$$M= -100( 0 ) -700(8) - 160( 8 )=

or this


$$\Sigma$$M= -100( 0 ) -700(8) - 160( 4 )=

 

[Doc Al]

That is the distance that is confusing me, because as I go reading the distances about point A from left to right, so it is 8 ft. The book, which I did have it before but edit out in all the confusing regarding reactions and what not, so this is the way then?


$$\Sigma$$M= -100( 0 ) -700(8) - 160( 8 )=

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.

 

[Dellis]

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?


$$\Sigma$$M= -100( 0 ) -700(8) - 160( 4 )(8)=

 

[Doc Al]

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?


$$\Sigma$$M= -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?

 

[Dellis]

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?.


$$\Sigma$$M= -100( 0 ) -700(8) - 160( 4 )(8)=

 

[Doc Al]

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!

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

 

[Dellis]

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!

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.

 

[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