# Replace 3 forces by force-couple system and x-intercept

Northbysouth

## Homework Statement

Replace the three forces which act on the bent bar by a force-couple system at the support point A. Then determine the x-intercept of the line of action of the stand-alone resultant force R. The couple is positive if counterclockwise, negative if clockwise.

## Homework Equations

Ma = Sum of the individual moments

## The Attempt at a Solution

For calculating R:

R = -2140 lb j - 1280 lb j + 730sin(25) i + 730cos(25) j
= 308.511 i - 2758.4 j

I know my R is correct, but my MAis incorrect:

MA= (-2140)(4) - (1280)(4 + 3.6cos(25) + 3.4) + (730)(3.6cos(25) +4)
= -16906.5 lb ft

For the x intercept I then:

(xi + yj) X (308.511 i - 2758.4 j)
= 2758.4c + 308.511y = -16906.5

Solving for x I got:

x = 0.111844y + 6.1291
hence the x intercept should be 6.1291 but it says its wrong.

Any suggestions?

#### Attachments

• forcecouple and x intercept.png
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cmmcnamara
I also got the same value for R as you. However take a look at your formula for the moment. The 730lb force at C acts is located at only 1.8cos(25) not 3.6cos(25). Additionally for the 730lb force you have not included the small contribution to the moment from the y-component. I think that may help you there.

As for the x-intercept, that leaves me a bit confused. It's been a bit since I took statics but correct me if I'm wrong but assuming A is the origin, the resultant's point of application is at A and therefore makes the x-intercept 0. But if I go by your method (I honestly haven't seen this before, teach me something new?), I had calculated the moment to be -22626 lbft which would give an x-intercept of -8.2.

Northbysouth
To be honest I'm not quite sure how finding the x intercept works but what I've done with previous questions, and this seems to work, is I take my value of the resultant R = 308.511i - 2758.4j

I then make this equal to the moment and solve for x.

With the 730lb I'll admit that I'm not sure how to handle it because I wasn't sure how to interpret the angle. Why is it 25 degrees? For some reason I thought that if I could find the horizontal distance from B to D that this would give me my value of d for M = Fd, but I can see now that that doesn't make sense because the 730 lb force acts half way along the slanted beam.

All the same I'm confused about what the correct angle is for the 730lb force. Could someone explain this to me?

Northbysouth
I've just looked over it again I think I understand now. When I originally put 3.6cos(25) for finding the x distance where the 730lb force is acting, the 3.6cos(25) would give me the whole x distance from B to D, which doesn't make sense because the 730lb force only acts at half that. So 1.8cos(25) gives me the x distance from B to C and then 1.8sin(25) would give me the y distance (vertical distance).

But would the 1.8cos(25) count as a value? Because it looks to me like 1.8cos(25)*730lb passes through the center of rotation A, which would makes this value 0. Am I reading this wrong?

cmmcnamara
I also am not sure about the whole x intercept because intuitively, what you are doing is solving for a moment arm vector using your known force and moment information. This basically gives you an equation describing all possible moment arms that have the correct distance, and plugging in y=0 basically just gives you the moment arm whose point of application lies on the xaxis, however as I said before, since you are creating the force couple system, moving the force to A means the point of application of the resultant is A which also means the line of action for the resultant passes through A which in a problem like this I would interpret as the origin.

Finding the x distance along the pipe isn't too difficult just think of the geometry. You know the distance on the non bent sections are 4.0' and 3.4' respectively. Then all your left with is the distance traversed by the bent segment. Since you know the angle of declination is 25 degrees, that means your x distance is the length along the pipe traverses multiplied by cos(25). The same goes for y except you use sine. So in the case for the force at C the, C is at 1.8' along the pipe so the coordinates for C using A as the origin would be C=(4.0+1.8cos(25), -1.8sin(25)).

As for handling the force and using the proper angle, you can look at it via trig and find it fairly quickly but I like to use a quick little trick that I learned with inclined surfaces. If you look at the force a C you notice that it is normal to the bent segment. If the bent piece were along the xaxis, the force would make a 90 degree angle with the xaxis (because it is normal). However since the bent piece has been "rotated" -25 degrees, the angle made with the xaxis after rotation is 90-25=65. Therefore the 730 lb force acts at an angle of 65 degrees

cmmcnamara
Yes your second look is correct. However yes the value counts. Imagine the 730lb force decomposed into its xy components and located at C. Since M=Fd, where d is the perpendicular distance you still need to find the distance perpendicular to each component. Since A is your point of interest, you take the moment about it which means for the x component, that is: M_x=F_xd=(730cos(65))(1.8sin(65)). Similarly for the y component: M_y=F_yd=(730sin(65))(4+1.8cos(65)).

If you were to trace out the line of action, it would not pass through A and would therefore create a moment about A.

Northbysouth
So I tried this for MA:

(-2140lb)(4.0 ft) - 1280lb(4+3.4+3.6cos(25)) + 730lb(4+1.8cos(25) +1.8sin(25))

MA = -140377

But it says my answer is wrong. What am I doing wrong?

cmmcnamara
Take a look again at your 730lb term. You are multiplying the whole force by some odd combination of distances. What you should do is break it into components forces:

Fx=730cos65
Fy=730sin65

Now from here find your perpendicular distances from A to each components line of action.

Dx=4+1.8cos25
Dy=1.8sin25

Note these are the x and y distances from A to C. You need to pair them properly to each force in accordance to perpendicular distance rule.

Putting it all together:

M=(730cos65)(1.8sin25)+(730sin65)(4+1.8cos25)

This should be the proper amount of moment created by the 730lb force. I personally do everything in vector notation because I feel it keeps me organized. I usually only do it by component breakup when the problem is very trivial.