Calculate the magnitude, direction and sense of the resultant force of the noncurrent

  • Thread starter bcoolbelal
  • Start date
  • #1
15
0

Homework Statement


Calculate the magnitude, direction and sense of the resultant force of the noncurrent force system shown, determine where the resultant intersects the bottom of the shape with respect to Point A



Homework Equations



http://i826.photobucket.com/albums/zz190/bcoolbelal/A.jpg [Broken]

The Attempt at a Solution

 
Last edited by a moderator:

Answers and Replies

  • #2
741
27


Please make some attempt, or identify your difficulty.
 
  • #3
15
0


the problem is i dont get the question, the first attempt is. the 4KN side.

M=F X D

F Y = F Sin Ø = 4 Sin 45 = 2.82
F x = F Sin Ø = 4 Cos 45 = 2.82

ACM = F X D = 2.82 X 1 = 2.82NA
CM = F X D = 2.82 X 0 = 0

then i dont know where to go from there
 
  • #4
741
27


Can you resolve the 8 kN force into its x- and y- components, as you did for the 4 kN force? (By the way you had a typo in #3, but it didn't matter). Also what does "NA" stand for?
 
  • #5
15
0


well i had a go at it, and this iss how it went:

F Y = F Sin Ø = 8 Sin -30 = -4
F x = F Sin Ø = 8 Cos -30 = 6.93

ACM = F X D = -4 X 0.5 = -2
CM = F X D = 6.93X 0 = 0

(In FY and FX i put -30 angle cause the angle is acting anti clock wise) so what to next
 
  • #6
741
27


It is self-evidently wrong because 8 has an upwards component (positive value of Fy) and a leftwards component (negative direction of x). Of course you can change the -30 to +30, but that isn't really the issue. You can either do this by inspection, and say that Fy is upwards. Or, you can take the angle anticlockwise positive from Ox, that is 150 degrees. That gives you the right answer to that part of the analysis. To get the resultant of all the Fx and Fy components in the diagram, what must you now do? If you can get that bit right, we are then ready to discuss where the resultant line of action is. In any case, you would be wise to check the result by drawing (on graph paper, say, or even approximately). Forces are vector quantities and can be added either algebraically or geometrically.
 
  • #7
15
0


OK THANKX FOR THAT BIT, i re don it and came out with the following

F Y = F Sin Ø = 8 Sin 30 = 4
F x = F Sin Ø = 8 Cos 150 = -6.93

ACM = F X D = 4 X 0.5 = 2
CM = F X D = -6.93X 0 = 0

sory i did not get ehat u ment after this bit
 
  • #8
741
27


Your 8 sin 30 should be 8 sin 150, even though that gives the same result. This indicates you need a better understanding of the frame of reference. There are the usual x and y axes in the plane of the paper, and a z axis perpendicular to that plane. To define sin and cosine etc, you take a radius vector of length 1 unit with its tail at the origin, O, and its head lying along the positive x axis. To get an angle, you rotate this radius vector about the axis Oz in an anticlockwise manner. The sin and cos are the projections of this radius vector onto the y- and x-axes respectively. The reason it is taken as anticlockwise positive is that you are looking towards the origin. The frame of reference is a right-handed system in which clockwise is positive to someone located at the origin, looking along one of the axes in a positive direction. That's why it is positive anticlockwise when looking towards the origin, as most of the world does in this type of problem. If you want a rigorous mathematical approach, then you need to understand and apply correctly the sign conventions described above.

OK. You now have 4 forces, and you know their Fx and Fy components. How can you get the value and direction of the resultant force? Clearly, as they are vectors, you can draw them to scale in any order, but nose to tail, to get the resultant force. Or you can do it algebraically. Can you do either or both of those approaches? If not, please explain exactly where your difficulty is.

When we do come to the moments, about which axis are you taking moments?
 
Last edited:
  • #9
15
0


and i also had a go at the 3 kN one. abd it turnd out as the following:

M= F X D

F Y = F Sin Ø = 3 Sin 0 = 0
F x = F Sin Ø = 3Cos 0 = 3

ACM = F X D = 0 X 0.5 = 0
CM = F X D = 3 X 0 =
∑ FX = 3
∑ FY = 0
∑ M = 0
------------------------------------------------------------------------------------
however i got stuck on the 5kN one, cause the angle was not stated,
 
  • #10
15
0


thankx for the explination however that was only a typing problem, anyway im currently stuck on the 5kN fx and yx as it does not state an angle
 
  • #11
741
27


In post #7 you said for the 8 kN force "CM = F X D = -6.93X 0 = 0"
I repeat: About which point (or axis, rather) were you taking moments?
In post #9, another typo: F x = F Sin Ø = 3Cos 0 = 3
It really is worth making sure everything is correct.
 
  • #12
15
0


and i belive the way of finsing the resultant force is:

R^2 = (∑ FX )^2 + (∑ Fy)^2
R = √((∑ FX) ^2 + ((∑ Fy)^2

is that equation right
 
  • #13
15
0


thankx
 
Last edited:
  • #14
15
0


oh, the X axis is where the 6.93 comes from
 
  • #15
15
0


typing errors sorted, so where to now, can u please guide me on the 5kn
 
  • #16
15
0


i think i got it now, caan u please comfirm that the following are corect and that all i need help with:
The calculations for the 4kN
M=F X D

F Y = F Sin Ø = 4 Sin 45 = 2.82
F x = F Cos Ø = 4 Cos 45 = 2.82

ACM = F X D = 2.82 X 1 = 2.82
CM = F X D = 2.82 X 0 = 0

∑ FX = 2.82
∑ FY = 2.82
∑ M = 0
--------------------------------------------------------------------------------------------------------------------------------------
The calculations for the 8kN



M= F X D

F Y = F Sin Ø = 8 Sin 150 = 4
F x = F Cos Ø = 8 Cos 30 = 6.93

ACM = F X D = 4 X 0.5 = 2
CM = F X D = -6.93X 0 = 0
∑ FX = -6.93
∑ FY = 4
∑ M = 0

--------------------------------------------------------------------------------------------------------------------------------------
The calculations for the3kN


M= F X D

F Y = F Sin Ø = 3 Sin 0 = 0
F x = F Cos Ø = 3 Cos 0 = 3

ACM = F X D = 0 X 0.5 = 0
CM = F X D = 3 X 0 = 0
∑ FX = 3
∑ FY = 0
∑ M = 0
--------------------------------------------------------------------------------------------------------------------------------------
The calculations for the 5kN

M= F X D

F Y = F Sin Ø = 5 Sin 0 = 0
F x = F Cos Ø = 5 Cos 0 = 5

ACM = F X D = 0 X 1 = 0
CM = F X D = 3 X 0 = 0
∑ FX = 0
∑ FY = 5
∑ M = 0

thankx
 
  • #17
741
27


your equ for R is correct. You also need to know the direction of R. Do you have a technique for that?
Your calcs for moment are inconsistent. You still hven't identified a single point about which you are taking moments. If the moments for the 3 kN and the 5 kN force are both zero, as you claim, then it must be their point of intersection at the bottom left of the diagram that is the point I am trying to get you to identify. However, if that is the case, the moments for the 4 kN and the 8 k forces cannot also be zero, as you say they are. It doesn't matter which point you use as long as it is the same for all the forces.
 
  • #18
15
0


No I don't have the equ for direction of R. So u mean I need a single point where i takey moments from for all 4. Oh it's starting to make more sence now thankx. I'll have a go at it and reply with the result thankx
 
  • #19
741
27


To get R, you used sigma Fx and sigma Fy. Can you sketch those and conclude the direction of R? I emphasise drawing in parallel to algebra/trigonometry, partly to aid your understanding, and partly as a useful check that a silly sign error hasn't been made.
 

Related Threads on Calculate the magnitude, direction and sense of the resultant force of the noncurrent

Top